RS Aggarwal Class 10 Mathematics Solutions Chapter 9 Mean Median Mode of Grouped Data Cumulative Frequency Graph and Ogive

 

Question 1. Find the value of x if the mean of the five observations x, x + 2, x + 4, x + 6, and x + 8 is 11.
Answer: The mean formula states that we add all observations and divide by the total count. Setting up the equation with the five values:
\( 11 = \frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} \)
Simplifying: \( 11 = \frac{5x + 20}{5} \)
\( \Rightarrow 55 = 5x + 20 \)
\( \Rightarrow 5x = 35 \)
\( \Rightarrow x = 7 \)
The value of x is 7.

Exam Tip: Always expand and collect like terms before applying the mean formula - this avoids calculation errors.

 

Question 2. The mean of 25 observations is 27. If 7 is subtracted from each observation, find the new mean.
Answer: First, calculate the sum of the 25 observations. Since the mean is 27, the total sum equals:
\( \text{Sum} = 27 \times 25 = 675 \)
When we subtract 7 from each of the 25 observations, the total reduction in the sum is:
\( 25 \times 7 = 175 \)
The new sum becomes: \( 675 - 175 = 500 \)
The new mean is: \( \frac{500}{25} = 20 \)
Therefore, the new mean will be 20.

Exam Tip: When the same value is added or subtracted from all observations, the mean changes by that exact value.

 

Question 3. Calculate the mean from the following grouped data.

Answer: Apply the direct method for grouped data. The mean formula is:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{426}{80} = 5.325 \)
Therefore, the mean of the data is 5.325.

Exam Tip: Always verify that all frequencies sum to the total and that class marks are calculated correctly as the midpoint of each interval.

 

Question 4. Find the mean from the frequency distribution table.

Answer: Using the direct method formula:
\( \bar{x} = \frac{\sum(f_i \times x_i)}{\sum f_i} = \frac{1150}{40} = 28.75 \)
The mean is 28.75.

Exam Tip: Double-check the sum of all frequencies and the total of fi × xi before dividing to get the final mean value.

 

Question 5. Find the mean of the following distribution.

Answer: Apply the mean formula for grouped data:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1980}{40} = 49.5 \)
The mean of the distribution is 49.5.

Exam Tip: For distributions with wider class intervals, ensure that class marks are the true midpoints of each interval to maintain accuracy.

 

Question 6. Calculate the mean for the given frequency distribution.

Answer: Calculate the mean using the formula:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{13200}{50} = 264 \)
The mean is 264.

Exam Tip: When dealing with large class intervals, calculate all products fi × xi carefully to avoid arithmetic mistakes that can compound in the final answer.

 

Question 7. Find the mean of the following grouped data.

Answer: Using the direct method:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{8244}{80} = 103.05 \)
The mean of the data is 103.05.

Exam Tip: For problems with a large total frequency, use systematic methods and check your arithmetic at key points to ensure the final answer is accurate.

 

Question 8. The mean of the following data is 24. Find the missing frequency p.

Answer: Set up the mean equation with the given mean of 24:
\( 24 = \frac{270 + 25p}{12 + p} \)
Cross multiply: \( 24(12 + p) = 270 + 25p \)
\( 288 + 24p = 270 + 25p \)
\( 288 - 270 = 25p - 24p \)
\( 18 = p \)
The value of p is 18.

Exam Tip: When finding missing frequencies, cross-multiply carefully and solve the resulting linear equation step by step to find the unknown value.

 

Question 9. The mean daily pocket allowance is 18. Find the missing frequency f from the table.

Answer: Write the mean formula with the given mean of 18:
\( 18 = \frac{752 + 20f}{44 + f} \)
Cross multiply: \( 18(44 + f) = 752 + 20f \)
\( 792 + 18f = 752 + 20f \)
\( 792 - 752 = 20f - 18f \)
\( 40 = 2f \)
\( f = 20 \)
The value of f is 20.

Exam Tip: Always group all terms containing the variable on one side and constants on the other before solving for the unknown frequency.

 

Question 10. The mean of the following distribution is 54. Find the missing frequency p.

Answer: Set up the equation with the mean of 54:
\( 54 = \frac{2370 + 30p}{39 + p} \)
Cross multiply: \( 54(39 + p) = 2370 + 30p \)
\( 2106 + 54p = 2370 + 30p \)
\( 54p - 30p = 2370 - 2106 \)
\( 24p = 264 \)
\( p = 11 \)
The value of p is 11.

Exam Tip: Collect all variable terms on one side and all constants on the other, then divide to isolate the unknown frequency.

 

Question 11. The sum of the frequencies is 100 and the mean is 42. Find the values of x and y from the given distribution.

Answer: First, use the frequency condition. The total of all frequencies must equal 100:
\( 63 + x + y = 100 \)
\( x + y = 37 \)
\( y = 37 - x \quad \text{...(1)} \)
Next, use the mean condition with mean = 42:
\( 42 = \frac{2775 + 25x + 45y}{100} \)
\( 4200 = 2775 + 25x + 45y \)
\( 1425 = 25x + 45y \)
Substitute equation (1):
\( 1425 = 25x + 45(37 - x) \)
\( 1425 = 25x + 1665 - 45x \)
\( -240 = -20x \)
\( x = 12 \)
If x = 12, then y = 37 - 12 = 25
The values are x = 12 and y = 25.

Exam Tip: When finding two unknowns, establish two separate equations from the given conditions and solve the system methodically using substitution.

 

Question 12. The sum of all frequencies is 100 and the mean is 188. Find the values of f₁ and f₂.

Answer: From the total frequency condition:
\( 35 + f_1 + f_2 = 100 \)
\( f_1 + f_2 = 65 \)
\( f_2 = 65 - f_1 \quad \text{...(1)} \)
From the mean condition with mean = 188:
\( 188 = \frac{6150 + 190f_1 + 210f_2}{100} \)
\( 18800 = 6150 + 190f_1 + 210f_2 \)
\( 12650 = 190f_1 + 210f_2 \)
Substitute equation (1):
\( 12650 = 190f_1 + 210(65 - f_1) \)
\( 12650 = 190f_1 + 13650 - 210f_1 \)
\( -1000 = -20f_1 \)
\( f_1 = 50 \)
If f₁ = 50, then f₂ = 65 - 50 = 15
The values are f₁ = 50 and f₂ = 15.

Exam Tip: When solving a system of two equations in two unknowns, isolate one variable from the first equation and substitute into the second to find both values accurately.

 

Question 13. The total frequency is 50 and the mean is 57.6. Find the missing frequencies f₁ and f₂ from the table.

Answer: From the frequency total:
\( 7 + f_1 + 12 + (18 - f_1) + 8 + 5 = 50 \)
\( f_1 + f_2 = 18 \)
\( f_2 = 18 - f_1 \)
From the mean condition with mean = 57.6:
\( 57.6 = \frac{3200 - 40f_1}{50} \)
\( 2880 = 3200 - 40f_1 \)
\( 40f_1 = 320 \)
\( f_1 = 8 \)
Therefore, f₂ = 18 - 8 = 10
The missing frequencies are f₁ = 8 and f₂ = 10.

Exam Tip: When one frequency is expressed as a difference from another (like 18 - f₁), use the total frequency to establish their relationship first.

 

Question 14. Using the direct method, find the mean heartbeats per minute.

Answer: Apply the direct method for the mean:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2277}{30} = 75.9 \)
The mean heartbeats per minute for the patients is 75.9.

Exam Tip: When calculating the mean for decimal class marks, ensure all products are computed correctly before summing, as rounding errors can affect the final answer.

 

Question 15. Using the assumed mean method, find the mean from the given data.

Answer: Use the assumed mean method. With A = 25 as the assumed mean:
\( \bar{x} = A + \frac{\sum (f_i \times d_i)}{\sum f_i} \)
\( \bar{x} = 25 + \frac{300}{100} = 25 + 3 = 28 \)
The mean is 28.

Exam Tip: The assumed mean method reduces calculation effort - pick the assumed mean as the class mark with the highest frequency to minimize deviations.

 

Question 16. Using the assumed mean method, find the mean.

Answer: Apply the assumed mean formula with A = 150:
\( \bar{x} = A + \frac{\sum (f_i \times d_i)}{\sum f_i} \)
\( \bar{x} = 150 + \frac{-300}{80} = 150 - 3.75 = 146.25 \)
The mean is 146.25.

Exam Tip: When the sum of fi × di is negative, subtract the result from the assumed mean to get the correct final answer.

 

Question 17. Using the assumed mean method, calculate the mean.

Answer: Using the assumed mean method with A = 50:
\( \bar{x} = A + \frac{\sum (f_i \times d_i)}{\sum f_i} \)
\( \bar{x} = 50 + \frac{2760}{220} = 50 + 12.55 = 62.55 \)
The mean is 62.55.

Exam Tip: Always select an assumed mean that corresponds to a class with high frequency and is centrally located in the data range to simplify the calculations.

 

Question 18. Using the direct method, find the mean literacy rate.

Answer: Apply the direct method formula:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2780}{40} = 69.5 \)
The mean literacy rate is 69.5%.

Exam Tip: When computing the mean for percentage data, present the final answer with the appropriate unit (%) to indicate it is a rate or percentage value.

 

Question 19. Using the step-deviation method, find the mean.

Answer: Using the step-deviation method with a = 25 and h = 10:
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
\( \bar{x} = 25 + \left( \frac{4}{50} \right) \times 10 \)
\( \bar{x} = 25 + \frac{4}{5} = 25 + 0.8 = 25.8 \)
The mean is 25.8.

Exam Tip: The step-deviation method is most efficient when the class width is a factor of the deviations, reducing arithmetic complexity significantly.

 

Question 20. Using the step-deviation method, find the mean.

Answer: Apply the step-deviation formula with a = 40 and h = 10:
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
\( \bar{x} = 40 + \left( \frac{7}{86} \right) \times 10 \)
\( \bar{x} = 40 + \frac{70}{86} = 40 + 0.81 = 40.81 \)
The mean is 40.81.

Exam Tip: In step-deviation calculations, verify that you have selected the assumed mean and class width such that the ui values are simple integers to avoid computation errors.

 

Question 21. Using the step-deviation method, find the mean weight in grams.

Answer: Using the step-deviation method with a = 202.5 and h = 1:
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
\( \bar{x} = 202.5 + \left( \frac{-38}{70} \right) \times 1 \)
\( \bar{x} = 202.5 - 0.542 = 201.96 \)
The mean weight is 201.96 grams.

Exam Tip: When the class width h equals 1, the step-deviation method becomes identical to the assumed mean method, simplifying the calculation significantly.

 

Question 22. Using the step-deviation method, find the mean weight.

Answer: Using the step-deviation formula with a = 45 and h = 10:
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
\( \bar{x} = 45 - \left( \frac{37}{150} \right) \times 10 \)
\( \bar{x} = 45 - \frac{370}{150} = 45 - 2.467 = 42.533 \)
The mean is 42.533.

Exam Tip: When calculating negative deviations, ensure that the negative sign is preserved throughout the calculation to get the correct final result.

 

Question 23. Find the mean marks.

Answer: The table appears incomplete as the remaining classes and their frequencies are missing from the visible portion. However, using the step-deviation method with a = 52.5 and h = 15 and the available data, we would continue the calculation with the full dataset when all information is provided. For the partial data shown, apply the standard formula:
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
Please provide the complete frequency distribution table to calculate the final mean marks accurately.

Exam Tip: Ensure that all class intervals and their corresponding frequencies are included in the table before starting the mean calculation using the step-deviation method.

 

Question 1. Find the mean of the following data using the step-deviation method.
Class 45 - 60, 60 - 75, 75 - 90
Frequency 20, 9, 10
Answer: To use the step-deviation method, we first find the class midpoints and then calculate the deviations.

Class | Frequency (fi) | Mid values (xi) | ui = (xi - 52.5)/15 | (fi × ui)
45 - 60 | 20 | 52.5 | 0 | 0
60 - 75 | 9 | 67.5 | 1 | 9
75 - 90 | 10 | 82.5 | 2 | 20
Total | Σ fi = 50 | | | Σ fi ui = 29

Using the formula: \( \bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h \)

\( \bar{x} = 52.5 + \left(\frac{29}{50}\right) \times 15 \)

\( \bar{x} = 52.5 + 8.7 \)

\( \bar{x} = 61.2 \)

Thus, the mean is 61.2.
In simple words: Choose a midpoint (52.5 here), find how far each class centre is from it in equal steps, multiply those steps by how many values are in each group, add them up, and put the result back into the formula to get the final mean.

Exam Tip: Always verify that your class width (h) is consistent and that your assumed mean falls within the data range for accuracy.

 

Question 2. Calculate the mean age from the given frequency distribution using the step-deviation method.
Class 18 - 24, 24 - 30, 30 - 36, 36 - 42, 42 - 48, 48 - 54
Frequency 6, 8, 12, 8, 4, 2
Answer: We apply the step-deviation method with A = 33 and h = 6.

Class | Frequency (fi) | Mid values (xi) | ui = (xi - 33)/6 | (fi × ui)
18 - 24 | 6 | 21 | -2 | -12
24 - 30 | 8 | 27 | -1 | -8
30 - 36 | 12 | 33 = A | 0 | 0
36 - 42 | 8 | 39 | 1 | 8
42 - 48 | 4 | 45 | 2 | 8
48 - 54 | 2 | 51 | 3 | 6
Total | Σ fi = 40 | | | Σ (fi × ui) = 2

Mean: \( \bar{x} = A + \left(h \times \frac{\sum (f_i \times u_i)}{\sum f_i}\right) \)

\( \bar{x} = 33 + \left(6 \times \frac{2}{40}\right) \)

\( \bar{x} = 33 + 0.3 \)

\( \bar{x} = 33.3 \) years
In simple words: Start with the middle class value (33), see how many steps of 6 each age group is away from it (some are negative, some positive), combine these with frequencies, and add back to 33 to get the average age.

Exam Tip: When working with larger class widths, ensure your assumed mean is clearly marked as A and that you count steps accurately from it.

 

Question 3. Using the step-deviation method, find the mean from the data below.
Class 500 - 520, 520 - 540, 540 - 560, 560 - 580, 580 - 600, 600 - 620
Frequency 14, 9, 5, 4, 3, 5
Answer: Applying the step-deviation method with A = 550 and h = 20:

Class | Frequency (fi) | Mid values (xi) | ui = (xi - 550)/20 | (fi × ui)
500 - 520 | 14 | 510 | -2 | -28
520 - 540 | 9 | 530 | -1 | -9
540 - 560 | 5 | 550 = A | 0 | 0
560 - 580 | 4 | 570 | 1 | 4
580 - 600 | 3 | 590 | 2 | 6
600 - 620 | 5 | 610 | 3 | 15
Total | Σ fi = 40 | | | Σ (fi × ui) = -12

Mean: \( \bar{x} = A + \left(h \times \frac{\sum (f_i \times u_i)}{\sum f_i}\right) \)

\( \bar{x} = 550 + \left(20 \times \frac{-12}{40}\right) \)

\( \bar{x} = 550 - 6 \)

\( \bar{x} = 544 \)
In simple words: Pick 550 as your reference point, count steps from it (each step is 20 units), weight them by frequency, then return to 550 and adjust by the weighted total to find the mean.

Exam Tip: Negative ui values indicate classes below your assumed mean; combine all products correctly to avoid sign errors.

 

Question 4. Calculate the mean using the step-deviation method.
Class 24.5 - 29.5, 29.5 - 34.5, 34.5 - 39.5, 39.5 - 44.5, 44.5 - 49.5, 49.5 - 54.5, 54.5 - 59.5
Frequency 4, 14, 22, 16, 6, 5, 3
Answer: Using the step-deviation method with A = 42 and h = 5:

Class | Frequency (fi) | Mid values (xi) | ui = (xi - 42)/5 | (fi × ui)
24.5 - 29.5 | 4 | 27 | -3 | -12
29.5 - 34.5 | 14 | 32 | -2 | -28
34.5 - 39.5 | 22 | 37 | -1 | -22
39.5 - 44.5 | 16 | 42 = A | 0 | 0
44.5 - 49.5 | 6 | 47 | 1 | 6
49.5 - 54.5 | 5 | 52 | 2 | 10
54.5 - 59.5 | 3 | 57 | 3 | 9
Total | Σ fi = 70 | | | Σ (fi × ui) = -37

Mean: \( \bar{x} = A + \left(h \times \frac{\sum (f_i \times u_i)}{\sum f_i}\right) \)

\( \bar{x} = 42 + \left(5 \times \frac{-37}{70}\right) \)

\( \bar{x} = 42 - 2.64 \)

\( \bar{x} = 39.36 \) years
In simple words: Choose 42 as your middle reference, see how many 5-unit steps away each class sits, combine these steps with frequencies, then add the result back to 42 to get your mean.

Exam Tip: Double-check your arithmetic when subtracting the mean from the assumed mean, especially with decimal class boundaries.

 

Question 5. Find the average age of the patients using the step-deviation method.
Class 4.5 - 14.5, 14.5 - 24.5, 24.5 - 34.5, 34.5 - 44.5, 44.5 - 54.5, 54.5 - 64.5
Frequency 6, 11, 21, 23, 14, 5
Answer: Applying the step-deviation method with A = 29.5 and h = 10:

Class | Frequency (fi) | Mid values (xi) | ui = (xi - 29.5)/10 | (fi × ui)
4.5 - 14.5 | 6 | 9.5 | -2 | -12
14.5 - 24.5 | 11 | 19.5 | -1 | -11
24.5 - 34.5 | 21 | 29.5 = A | 0 | 0
34.5 - 44.5 | 23 | 39.5 | 1 | 23
44.5 - 54.5 | 14 | 49.5 | 2 | 28
54.5 - 64.5 | 5 | 59.5 | 3 | 15
Total | Σ fi = 80 | | | Σ (fi × ui) = 43

Mean: \( \bar{x} = A + \left(h \times \frac{\sum (f_i \times u_i)}{\sum f_i}\right) \)

\( \bar{x} = 29.5 + \left(10 \times \frac{43}{80}\right) \)

\( \bar{x} = 29.5 + 5.375 \)

\( \bar{x} = 34.875 \) years

Thus, the average age of the patients is 34.87 years.
In simple words: Use 29.5 as your starting point, measure each class in 10-unit jumps from it, combine with the patient counts, and add everything back to 29.5 for the mean age.

Exam Tip: Round your final answer appropriately; here, 34.875 rounds to 34.87 when two decimal places are required.

 

Question 6. Calculate the mean weight using the step-deviation method.
Weight (in grams): 74.5 - 79.5, 79.5 - 84.5, 84.5 - 89.5, 89.5 - 94.5, 94.5 - 99.5, 99.5 - 104.5, 104.5 - 109.5
Number of eggs (fi): 4, 9, 13, 17, 12, 3, 2
Answer: Using the step-deviation method with a = 92 and h = 5:

Weight (in grams) | Number of eggs (fi) | Class mark (xi) | di = xi - 92 | ui = (xi - 92)/5 | (fi ui)
74.5 - 79.5 | 4 | 77 | -15 | -3 | -12
79.5 - 84.5 | 9 | 82 | -10 | -2 | -18
84.5 - 89.5 | 13 | 87 | -5 | -1 | -13
89.5 - 94.5 | 17 | 92 | 0 | 0 | 0
94.5 - 99.5 | 12 | 97 | 5 | 1 | 12
99.5 - 104.5 | 3 | 102 | 10 | 2 | 6
104.5 - 109.5 | 2 | 107 | 15 | 3 | 6
Total | Σ fi = 60 | | | | Σ fi ui = -19

Mean: \( \bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h \)

\( \bar{x} = 92 + \left(\frac{-19}{60}\right) \times 5 \)

\( \bar{x} = 92 - 1.58 \)

\( \bar{x} = 90.42 \) ≈ 90 g

Thus, the mean weight to the nearest gram is 90 g.
In simple words: Set 92 as your base point, measure each egg group's weight in 5-gram steps away from it (negative for lighter, positive for heavier), combine with the egg count, then adjust 92 by the result to find the average weight.

Exam Tip: When rounding to the nearest whole number, check if the decimal part is 0.5 or more before rounding up or down.

 

Question 7. Find the mean marks using the step-deviation method.
Marks: 0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35, 35 - 40
Number of students (cf): 3, 10, 25, 49, 65, 73, 78, 80
Answer: First, convert cumulative frequency to simple frequency, then apply step-deviation method with a = 17.5 and h = 5:

Marks | Number of students (cf) | Frequency (fi) | Class mark (xi) | di = xi - 17.5 | ui = (xi - 17.5)/5 | (fi ui)
0 - 5 | 3 | 3 | 2.5 | -15 | -3 | -9
5 - 10 | 10 | 7 | 7.5 | -10 | -2 | -14
10 - 15 | 25 | 15 | 12.5 | -5 | -1 | -15
15 - 20 | 49 | 24 | 17.5 | 0 | 0 | 0
20 - 25 | 65 | 16 | 22.5 | 5 | 1 | 16
25 - 30 | 73 | 8 | 27.5 | 10 | 2 | 16
30 - 35 | 78 | 5 | 32.5 | 15 | 3 | 15
35 - 40 | 80 | 2 | 37.5 | 20 | 4 | 8
Total | | Σ fi = 80 | | | | Σ fi ui = 17

Mean: \( \bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h \)

\( \bar{x} = 17.5 + \left(\frac{17}{80}\right) \times 5 \)

\( \bar{x} = 17.5 + 1.06 \)

\( \bar{x} = 18.56 \)

Thus, the mean marks correct to 2 decimal places is 18.56.
In simple words: Take 17.5 as your reference mark, measure each score group in 5-mark intervals from it, weight by how many students fall in each band, and add back to 17.5 to get the average.

Exam Tip: When given cumulative frequency, always extract the simple frequency first by taking the difference between successive cumulative values.

 

Exercise 9B

 

Question 1. Find the median age from the given frequency distribution.
Age (in years): 0 - 15, 15 - 30, 30 - 45, 45 - 60, 60 - 75
Number of patients (fi): 5, 20, 40, 50, 25
Answer: First, construct the cumulative frequency table:

Age (in years) | Number of patients (fi) | Cumulative Frequency (cf)
0 - 15 | 5 | 5
15 - 30 | 20 | 25
30 - 45 | 40 | 65
45 - 60 | 50 | 115
60 - 75 | 25 | 140
Total | N = Σ fi = 140 |

Now, N = 140, so \( \frac{N}{2} = 70 \)

The cumulative frequency just greater than 70 is 115, and the corresponding class is 45 - 60.

Thus, the median class is 45 - 60.

Here, l = 45, h = 15, f = 50, N = 140, and cf = 65.

Median: \( M = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)

\( M = 45 + \left(\frac{70 - 65}{50}\right) \times 15 \)

\( M = 45 + \left(\frac{5}{50}\right) \times 15 \)

\( M = 45 + 1.5 \)

\( M = 46.5 \) years

Hence, the median age is 46.5 years.
In simple words: Find where the middle patient falls by checking cumulative frequencies, then use the median formula to pinpoint the exact age within that class interval.

Exam Tip: Always verify that your median class is the one whose cumulative frequency first exceeds N/2.

 

Question 2. Determine the median from the following frequency distribution.
Class: 0 - 7, 7 - 14, 14 - 21, 21 - 28, 28 - 35, 35 - 42, 42 - 49
Frequency (f): 3, 4, 7, 11, 0, 16, 9
Answer: Build the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
0 - 7 | 3 | 3
7 - 14 | 4 | 7
14 - 21 | 7 | 14
21 - 28 | 11 | 25
28 - 35 | 0 | 25
35 - 42 | 16 | 41
42 - 49 | 9 | 50
N = Σ f = 50 |

Now, N = 50, so \( \frac{N}{2} = 25 \)

The cumulative frequency just greater than 25 is 41, and the corresponding class is 35 - 42.

Thus, the median class is 35 - 42.

Here, l = 35, h = 7, f = 16, cf = 25 (c.f. of preceding class), and \( \frac{N}{2} = 25 \)

Median: \( M = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)

\( M = 35 + 7 \times \left(\frac{25 - 25}{16}\right) \)

\( M = 35 + 0 \)

\( M = 35 \)

Hence, the median is 35.
In simple words: Look for the cumulative frequency that crosses the halfway mark; when it equals N/2 exactly, the median falls right at the lower boundary of that class.

Exam Tip: When the cumulative frequency of the preceding class exactly equals N/2, your median will be exactly the lower class boundary.

 

Question 3. Calculate the median daily wage income from the distribution below.
Class: 0 - 100, 100 - 200, 200 - 300, 300 - 400, 400 - 500
Frequency (f): 40, 32, 48, 22, 8
Answer: Construct the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
0 - 100 | 40 | 40
100 - 200 | 32 | 72
200 - 300 | 48 | 120
300 - 400 | 22 | 142
400 - 500 | 8 | 150
N = Σ f = 150 |

Now, N = 150, so \( \frac{N}{2} = 75 \)

The cumulative frequency just greater than 75 is 120, and the corresponding class is 200 - 300.

Thus, the median class is 200 - 300.

Here, l = 200, h = 100, f = 48, cf = 72 (c.f. of preceding class), and \( \frac{N}{2} = 75 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 200 + \left\{100 \times \left(\frac{75 - 72}{48}\right)\right\} \)

\( M = 200 + 6.25 \)

\( M = 206.25 \)

Hence, the median daily wage income of the workers is Rs 206.25.
In simple words: Find which wage band holds the middle worker by checking cumulative totals, then apply the median formula to locate the exact wage value inside that interval.

Exam Tip: For larger class widths, take extra care with decimal arithmetic in the median formula.

 

Question 4. Find the median from the given frequency distribution.
Class: 5 - 10, 10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35, 35 - 40, 40 - 45
Frequency (f): 5, 6, 15, 10, 5, 4, 2, 2
Answer: Build the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
5 - 10 | 5 | 5
10 - 15 | 6 | 11
15 - 20 | 15 | 26
20 - 25 | 10 | 36
25 - 30 | 5 | 41
30 - 35 | 4 | 45
35 - 40 | 2 | 47
40 - 45 | 2 | 49
N = Σ f = 49 |

Now, N = 49, so \( \frac{N}{2} = 24.5 \)

The cumulative frequency just greater than 24.5 is 26, and the corresponding class is 15 - 20.

Thus, the median class is 15 - 20.

Here, l = 15, h = 5, f = 15, cf = 11 (c.f. of preceding class), and \( \frac{N}{2} = 24.5 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 15 + \left\{5 \times \left(\frac{24.5 - 11}{15}\right)\right\} \)

\( M = 15 + \left\{5 \times \left(\frac{13.5}{15}\right)\right\} \)

\( M = 15 + 4.5 \)

\( M = 19.5 \)

Hence, the median is 19.5.
In simple words: Locate the class containing the middle value by adding up frequencies, then plug the numbers into the median formula to find the exact midpoint value.

Exam Tip: When N is odd, N/2 will be a decimal; use the next integer cumulative frequency to identify the median class.

 

Question 5. Calculate the median from the following data.
Class: 65 - 85, 85 - 105, 105 - 125, 125 - 145, 145 - 165, 165 - 185, 185 - 205
Frequency (f): 4, 5, 13, 20, 14, 7, 4
Answer: Prepare the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
65 - 85 | 4 | 4
85 - 105 | 5 | 9
105 - 125 | 13 | 22
125 - 145 | 20 | 42
145 - 165 | 14 | 56
165 - 185 | 7 | 63
185 - 205 | 4 | 67
N = Σ f = 67 |

Now, N = 67, so \( \frac{N}{2} = 33.5 \)

The cumulative frequency just greater than 33.5 is 42, and the corresponding class is 125 - 145.

Thus, the median class is 125 - 145.

Here, l = 125, h = 20, f = 20, cf = 22 (c.f. of preceding class), and \( \frac{N}{2} = 33.5 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 125 + \left\{20 \times \left(\frac{33.5 - 22}{20}\right)\right\} \)

\( M = 125 + 11.5 \)

\( M = 136.5 \)

Hence, the median is 136.5.
In simple words: Add up the frequencies to find where the middle observation sits, then use the median formula to calculate the precise value within that interval.

Exam Tip: Always double-check which cumulative frequency first exceeds N/2 to ensure you have identified the correct median class.

 

Question 6. Find the median from the given frequency distribution.
Class: 135 - 140, 140 - 145, 145 - 150, 150 - 155, 155 - 160, 160 - 165, 165 - 170, 170 - 175
Frequency (f): 6, 10, 18, 22, 20, 15, 6, 3
Answer: Construct the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
135 - 140 | 6 | 6
140 - 145 | 10 | 16
145 - 150 | 18 | 34
150 - 155 | 22 | 56
155 - 160 | 20 | 76
160 - 165 | 15 | 91
165 - 170 | 6 | 97
170 - 175 | 3 | 100
N = Σ f = 100 |

Now, N = 100, so \( \frac{N}{2} = 50 \)

The cumulative frequency just greater than 50 is 56, and the corresponding class is 150 - 155.

Thus, the median class is 150 - 155.

Here, l = 150, h = 5, f = 22, cf = 34 (c.f. of preceding class), and \( \frac{N}{2} = 50 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 150 + \left\{5 \times \left(\frac{50 - 34}{22}\right)\right\} \)

\( M = 150 + \left\{5 \times \left(\frac{16}{22}\right)\right\} \)

\( M = 150 + 3.64 \)

\( M = 153.64 \)

Hence, the median is 153.64.
In simple words: Identify the class containing the 50th observation by accumulating frequencies, then apply the median formula to find the exact midpoint.

Exam Tip: With an even total frequency, N/2 will be a whole number; make sure the cumulative frequency you choose is strictly greater than it.

 

Question 7. A frequency distribution has median 24. The median class is 20 - 30. Find the unknown frequency x using the given data.
Class: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50
Frequency (f): 5, 25, x, 18, 7
Answer: Build the cumulative frequency table with the unknown frequency x:

Class | Frequency (f) | Cumulative Frequency (cf)
0 - 10 | 5 | 5
10 - 20 | 25 | 30
20 - 30 | x | x + 30
30 - 40 | 18 | x + 48
40 - 50 | 7 | x + 55

Since the median is 24 and lies in the class 20 - 30, the median class is 20 - 30.

Here, l = 20, h = 10, median M = 24, cf = 30 (c.f. of preceding class), f = x, and \( \frac{n}{2} = \frac{x + 55}{2} \)

Using the median formula:

\( M = l + \frac{\frac{n}{2} - cf}{f} \times h \)

\( 24 = 20 + \frac{\frac{x + 55}{2} - 30}{x} \times 10 \)

\( 24 = 20 + \frac{x + 55 - 60}{2x} \times 10 \)

\( 24 = 20 + \frac{x - 5}{2x} \times 10 \)

\( 24 = 20 + \frac{10(x - 5)}{2x} \)

\( 24 = 20 + \frac{5x - 25}{x} \)

\( 4 = \frac{5x - 25}{x} \)

\( 4x = 5x - 25 \)

\( -x = -25 \)

\( x = 25 \)

Hence, the unknown frequency is 25.
In simple words: Set up the median formula with the unknown frequency, substitute the given median value and the median class details, then solve the resulting equation for x.

Exam Tip: When working with unknown frequencies, express total N in terms of the unknowns, then solve systematically by substituting into the median formula.

 

Question 8. A frequency distribution has median 16. Find the missing frequencies a and b using the given data.
Class: 0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35, 35 - 40
Frequency (f): 12, a, 12, 15, b, 6, 6, 4
Answer: Build the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
0 - 5 | 12 | 12
5 - 10 | a | 12 + a
10 - 15 | 12 | 24 + a
15 - 20 | 15 | 39 + a
20 - 25 | b | 39 + a + b
25 - 30 | 6 | 45 + a + b
30 - 35 | 6 | 51 + a + b
35 - 40 | 4 | 55 + a + b
Total | N = Σ fi = 70 |

From the total: 55 + a + b = 70, so a + b = 15 ...(1)

Since the median is 16, which lies in 15 - 20, the median class is 15 - 20.

Here, l = 15, h = 5, N = 70, f = 15, cf = 24 + a

Using the median formula:

\( M = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)

\( 16 = 15 + \left(\frac{35 - (24 + a)}{15}\right) \times 5 \)

\( 16 = 15 + \left(\frac{11 - a}{15}\right) \times 5 \)

\( 1 = \frac{5(11 - a)}{15} \)

\( 1 = \frac{11 - a}{3} \)

\( 3 = 11 - a \)

\( a = 8 \)

From equation (1): b = 15 - a = 15 - 8 = 7

Hence, a = 8 and b = 7.
In simple words: Use the constraint that all frequencies sum to 70 to form one equation, then use the given median value and formula to form a second equation, solve both together to find a and b.

Exam Tip: Always start by using the total frequency condition to establish a relationship between unknowns before applying the median formula.

 

Question 9. A frequency distribution has median 5000. The median class is 4500 - 5500. Find the missing frequencies x and y using the given data.
Runs scored: 2500 - 3500, 3500 - 4500, 4500 - 5500, 5500 - 6500, 6500 - 7500, 7500 - 8500
Number of batsman (fi): 5, x, y, 12, 6, 2
Answer: Build the cumulative frequency table:

Runs scored | Number of batsman (fi) | Cumulative Frequency (cf)
2500 - 3500 | 5 | 5
3500 - 4500 | x | 5 + x
4500 - 5500 | y | 5 + x + y
5500 - 6500 | 12 | 17 + x + y
6500 - 7500 | 6 | 23 + x + y
7500 - 8500 | 2 | 25 + x + y
Total | N = Σ fi = 60 |

From the total: 25 + x + y = 60, so x + y = 35 ...(1)

Since the median is 5000 and lies in 4500 - 5500, the median class is 4500 - 5500.

Here, l = 4500, h = 1000, N = 60, f = y, cf = 5 + x

Using the median formula:

\( M = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)

\( 5000 = 4500 + \left(\frac{30 - (5 + x)}{y}\right) \times 1000 \)

\( 500 = \left(\frac{25 - x}{y}\right) \times 1000 \)

\( 500 = \frac{1000(25 - x)}{y} \)

\( y = 2(25 - x) \)

\( y = 50 - 2x \)

From equation (1): 35 - x = 50 - 2x

\( 2x - x = 50 - 35 \)

\( x = 15 \)

From equation (1): y = 35 - 15 = 20

Hence, x = 15 and y = 20.
In simple words: Express the total frequency condition as one equation relating x and y, use the median formula to create a second relationship, then solve both equations together to find both unknowns.

Exam Tip: When substituting expressions, simplify carefully to avoid errors; here, y = 50 - 2x was substituted back into x + y = 35 to isolate x.

 

Question 10. A frequency distribution has median 32.5. Find the missing frequencies f₁ and f₂ using the given data.
Class: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70
Frequency (f): f₁, 5, 9, 12, f₂, 3, 2
Answer: Build the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
0 - 10 | f₁ | f₁
10 - 20 | 5 | f₁ + 5
20 - 30 | 9 | f₁ + 14
30 - 40 | 12 | f₁ + 26
40 - 50 | f₂ | f₁ + f₂ + 26
50 - 60 | 3 | f₁ + f₂ + 29
60 - 70 | 2 | f₁ + f₂ + 31
N = Σ f = 40 |

From the total: f₁ + f₂ + 31 = 40, so f₁ + f₂ = 9 ...(i)

Since the median is 32.5 and lies in 30 - 40, the median class is 30 - 40.

Here, l = 30, \( \frac{N}{2} = \frac{40}{2} = 20 \), f = 12, cf = 14 + f₁

Using the median formula:

\( l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 32.5 \)

\( 30 + \left(\frac{20 - (14 + f_1)}{12}\right) \times 10 = 32.5 \)

\( \left(\frac{6 - f_1}{12}\right) \times 10 = 2.5 \)

\( \frac{60 - 10f_1}{12} = 2.5 \)

\( 60 - 10f_1 = 30 \)

\( 10f_1 = 30 \)

\( f_1 = 3 \)

From equation (i): f₂ = 9 - 3 = 6

Hence, f₁ = 3 and f₂ = 6.
In simple words: Form one equation from the total frequency constraint, use the median formula with the given median and class details to form a second equation, then solve both to find the two unknowns.

Exam Tip: When solving for unknowns in the median formula, isolate the unknown carefully and check your arithmetic at each step.

 

Question 11. Find the median from the given frequency distribution (inclusive form).
Class: 19 - 25, 26 - 32, 33 - 39, 40 - 46, 47 - 53, 54 - 60
Frequency (f): 35, 96, 68, 102, 35, 4
Answer: First, convert the inclusive form data into exclusive form:

Class | Frequency (f) | Cumulative Frequency (cf)
18.5 - 25.5 | 35 | 35
25.5 - 32.5 | 96 | 131
32.5 - 39.5 | 68 | 199
39.5 - 46.5 | 102 | 301
46.5 - 53.5 | 35 | 336
53.5 - 60.5 | 4 | 340
N = Σ f = 340 |

Now, N = 340, so \( \frac{N}{2} = 170 \)

The cumulative frequency just greater than 170 is 199, and the corresponding class is 32.5 - 39.5.

Thus, the median class is 32.5 - 39.5.

Here, l = 32.5, h = 7, f = 68, cf = 131 (c.f. of preceding class), and \( \frac{N}{2} = 170 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 32.5 + \left\{7 \times \left(\frac{170 - 131}{68}\right)\right\} \)

\( M = 32.5 + \left\{7 \times \left(\frac{39}{68}\right)\right\} \)

\( M = 32.5 + 4.01 \)

\( M = 36.51 \)

Hence, the median is 36.51.
In simple words: When data is presented in inclusive form (19-25, 26-32...), first convert it to exclusive form (18.5-25.5, 25.5-32.5...) by adjusting boundaries, then find the median normally.

Exam Tip: Inclusive form data always needs conversion: subtract 0.5 from the lower boundary and add 0.5 to the upper boundary of each class.

 

Question 12. Calculate the median wages from the given exclusive form data.
Class: 60.5 - 70.5, 70.5 - 80.5, 80.5 - 90.5, 90.5 - 100.5, 100.5 - 110.5, 110.5 - 120.5
Frequency (f): 5, 15, 20, 30, 20, 8
Answer: Build the cumulative frequency table:

Class | Frequency (f) | Cumulative Frequency (cf)
60.5 - 70.5 | 5 | 5
70.5 - 80.5 | 15 | 20
80.5 - 90.5 | 20 | 40
90.5 - 100.5 | 30 | 70
100.5 - 110.5 | 20 | 90
110.5 - 120.5 | 8 | 98
N = Σ f = 98 |

Now, N = 98, so \( \frac{N}{2} = 49 \)

The cumulative frequency just greater than 49 is 70, and the corresponding class is 90.5 - 100.5.

Thus, the median class is 90.5 - 100.5.

Here, l = 90.5, h = 10, f = 30, cf = 40 (c.f. of preceding class), and \( \frac{N}{2} = 49 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 90.5 + \left\{10 \times \left(\frac{49 - 40}{30}\right)\right\} \)

\( M = 90.5 + 3 \)

\( M = 93.5 \)

Hence, median wages = Rs. 93.50.
In simple words: Identify which wage class contains the middle worker by checking cumulative frequencies, then use the median formula to pinpoint the exact wage within that interval.

Exam Tip: Data already in exclusive form requires no boundary adjustment; proceed directly to cumulative frequency calculation.

 

Question 13. Find the median from the given inclusive form frequency distribution.
Class: 1 - 5, 6 - 10, 11 - 15, 16 - 20, 21 - 25, 26 - 30, 31 - 35
Frequency (f): 7, 10, 16, 32, 24, 16, 11
Answer: Convert the inclusive form data into exclusive form:

Class | Frequency (f) | Cumulative Frequency (cf)
0.5 - 5.5 | 7 | 7
5.5 - 10.5 | 10 | 17
10.5 - 15.5 | 16 | 33
15.5 - 20.5 | 32 | 65
20.5 - 25.5 | 24 | 89
25.5 - 30.5 | 16 | 105
30.5 - 35.5 | 11 | 116
N = Σ f = 116 |

Now, N = 116, so \( \frac{N}{2} = 58 \)

The cumulative frequency just greater than 58 is 65, and the corresponding class is 15.5 - 20.5.

Thus, the median class is 15.5 - 20.5.

Here, l = 15.5, h = 5, f = 32, cf = 33 (c.f. of preceding class), and \( \frac{N}{2} = 58 \)

Median: \( M = l + \left\{h \times \left(\frac{\frac{N}{2} - cf}{f}\right)\right\} \)

\( M = 15.5 + \left\{5 \times \left(\frac{58 - 33}{32}\right)\right\} \)

\( M = 15.5 + \left\{5 \times \left(\frac{25}{32}\right)\right\} \)

\( M = 15.5 + 3.91 \)

\( M = 19.41 \)

Hence, the median is approximately 19.41.
In simple words: Convert inclusive form classes (1-5, 6-10) to exclusive form (0.5-5.5, 5.5-10.5) by adjusting half-unit from boundaries, then proceed with standard median calculation.

Exam Tip: Always remember: when converting from inclusive to exclusive form, the gap between classes becomes 1, so you adjust by adding and subtracting 0.5 at class boundaries.

 

Question 1. Find the mean, median, and mode for the following frequency distribution:

Answer: To find the mean, use the formula \( \text{Mean} = \frac{\sum fi xi}{\sum fi} = \frac{1910}{50} = 38.2 \). So the mean is 38.2.

For the median, first create a cumulative frequency table. With \( N = 50 \), we have \( \frac{N}{2} = 25 \). The cumulative frequency just greater than 25 is 37, which lies in the class 40 - 50. Therefore, the median class is 40 - 50. Using the median formula with \( l = 40 \), \( h = 10 \), \( f = 12 \), and \( cf = 25 \):
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 40 + \left( \frac{25 - 25}{12} \right) \times 10 = 40 \]
So the median is 40.

To find the mode using the empirical relationship: \( \text{Mode} = 3(\text{median}) - 2(\text{mean}) = 3 \times 40 - 2 \times 38.2 = 120 - 76.4 = 43.6 \). The mode is 43.6.
In simple words: The mean shows the average value of 38.2. The median divides the data in half at 40. The mode, computed using the three-measure relationship, is 43.6.

Exam Tip: Always verify that the cumulative frequency just exceeds N/2 before applying the median formula. Use the empirical mode formula when the modal class is not clearly evident from frequency inspection.

 

Question 2. Calculate the mean, median, and mode for the data below:

Answer: The mean is calculated as \( \text{Mean} = \frac{3120}{50} = 62.4 \).

For the median, with \( N = 50 \), we need the cumulative frequency exceeding \( \frac{N}{2} = 25 \). The cumulative frequency at the 60 - 80 class is 36, so this is the median class. Using \( l = 60 \), \( h = 20 \), \( f = 12 \), \( cf = 24 \):
\[ \text{Median} = 60 + \left( \frac{25 - 24}{12} \right) \times 20 = 60 + 1.67 = 61.67 \]
The median is 61.67.

Using the empirical formula: \( \text{Mode} = 3(61.67) - 2(62.4) = 185.01 - 124.8 = 60.21 \). The mode is 60.21.
In simple words: The mean of 62.4 represents the typical value. The median is 61.67, showing where the middle of the data sits. The mode, derived from the three-measure relationship, is 60.21.

Exam Tip: When applying the empirical relationship, ensure you have calculated both mean and median accurately first - any error carries through to the mode calculation.

 

Question 3. Determine mean, median, and mode for the given frequency table:

Answer: The mean is \( \frac{4225}{25} = 169 \).

For the median, \( N = 25 \) gives \( \frac{N}{2} = 12.5 \). The cumulative frequency reaching 16 occurs in the class 150 - 200, making this the median class. With \( l = 150 \), \( h = 50 \), \( f = 6 \), \( cf = 10 \):
\[ \text{Median} = 150 + \left( \frac{12.5 - 10}{6} \right) \times 50 = 150 + 20.83 = 170.83 \]
The median is 170.83.

Applying the empirical formula: \( \text{Mode} = 3(170.83) - 2(169) = 512.49 - 338 = 174.49 \). The mode is 174.49.
In simple words: The data has a mean of 169. The median splits the data at 170.83. The mode, found using the relationship between the three measures, is 174.49.

Exam Tip: Always construct a cumulative frequency column alongside your main table - this prevents errors when locating the median class and ensures your calculation stays on track.

 

Question 4. Find the mean, median, and mode of the marks obtained by students:

Answer: The mean is \( \frac{4970}{100} = 49.7 \).

For the median, \( N = 100 \) gives \( \frac{N}{2} = 50 \). The cumulative frequency reaches 71 at class 45 - 55, so this is the median class. Using \( l = 45 \), \( h = 10 \), \( f = 33 \), \( cf = 38 \):
\[ \text{Median} = 45 + \left( \frac{50 - 38}{33} \right) \times 10 = 45 + 3.64 = 48.64 \]
The median is 48.64.

Using the empirical relationship: \( \text{Mode} = 3(48.64) - 2(49.70) = 145.92 - 99.4 = 46.52 \). The mode is 46.52.
In simple words: The average score is 49.7 marks. The middle value is 48.64. The mode, derived from the three-measure formula, is 46.52.

Exam Tip: For large datasets, the empirical relationship Mode = 3(Median) - 2(Mean) is often more reliable than the direct mode formula, especially when multiple classes have similar frequencies.

 

Question 5. Compute mean, median, and mode for the height data:

Answer: The mean is \( \bar{x} = \frac{7490}{50} = 149.8 \).

With \( N = 50 \), we have \( \frac{N}{2} = 25 \). The cumulative frequency just exceeding 25 is 42, which corresponds to class 150 - 160. This is the median class. Using \( l = 150 \), \( h = 10 \), \( f = 20 \), \( c = 22 \):
\[ \text{Median, Me} = 150 + \left( 10 \times \frac{25 - 22}{20} \right) = 150 + \left( 10 \times \frac{3}{20} \right) = 151.5 \]
The median is 151.5.

Using the empirical formula: \( \text{Mode} = 3(151.5) - 2(149.8) = 454.5 - 299.6 = 154.9 \). The mode is 154.9.
In simple words: The average height is 149.8 cm. The height that falls in the middle is 151.5 cm. The mode, from the three-measure relationship, is 154.9 cm.

Exam Tip: When cumulative frequency reaches the median position smoothly, double-check your class boundaries and ensure the formula substitution is done step-by-step to avoid arithmetic errors.

 

Question 6. Calculate mean, median, and mode from the daily income data:

Answer: The mean is \( \bar{x} = \frac{7260}{50} = 145.2 \).

With \( N = 50 \), we need \( \frac{N}{2} = 25 \). The cumulative frequency just greater than 25 is 26, which falls in class 120 - 140. This is the median class. Using \( l = 120 \), \( h = 20 \), \( f = 14 \), \( c = 12 \):
\[ \text{Median, Me} = 120 + \left( 20 \times \frac{25 - 12}{14} \right) = 120 + \left( 20 \times \frac{13}{14} \right) = 138.57 \]
The median is 138.57.

Using the empirical formula: \( \text{Mode} = 3(138.57) - 2(145.2) = 415.71 - 290.4 = 125.31 \). The mode is 125.31.
In simple words: The typical daily income is 145.2 units. Half the workers earn below 138.57 units. The mode, from the relationship formula, is 125.31 units.

Exam Tip: The empirical relationship Mode = 3(Median) - 2(Mean) works best when the distribution is roughly symmetric or moderately skewed. Always calculate mean and median independently first to ensure accuracy.

 

Question 7. Calculate the mean and median from the frequency distribution table provided, where daily expenditure ranges from Rs 100 - 350.
Answer: Using the given frequency table with class intervals and their corresponding frequencies and mid-values, we first compute the sum of products (fi × xi) = 6150. The mean is found by dividing this sum by the total frequency (N = 30), yielding a mean of Rs 205.

To find the median, we calculate N/2 = 15. The cumulative frequency just exceeding 15 is 25, which corresponds to the class 200 - 250. Using the median formula:

\[ M_e = l + \left\{h \times \left(\frac{N/2 - c}{f}\right)\right\} \]

where l = 200, h = 50, f = 12, c = 13, and N/2 = 15:

\[ M_e = 200 + \left\{50 \times \left(\frac{15 - 13}{12}\right)\right\} = 200 + \left(50 \times \frac{2}{12}\right) = 200 + 8.33 = 208.33 \]

Thus, the mean is Rs 205 and the median is Rs 208.33.
In simple words: The mean shows the average daily expenditure by adding all expenses and dividing by the number of observations. The median identifies the middle value when cumulative frequencies are arranged in order.

Exam Tip: Always ensure the cumulative frequency sum equals N before computing the median - this validates your table. Use the median class formula precisely with correct substitution of l, h, c, and f values.

 

Question 30. Construct a cumulative frequency table for marks scored by students and use an ogive to find the median from the given less than frequency data.
Answer: A 'less than' frequency distribution table is built from the given data, showing cumulative frequencies at each upper class limit. The table shows: Less than 10 (cf = 5), Less than 20 (cf = 8), Less than 30 (cf = 12), and continuing through Less than 100 (cf = 53).

An ogive curve is plotted by placing upper class limits on the x-axis and their respective cumulative frequencies on the y-axis. The points are joined smoothly to form a continuous curve.

Since N = 53, we find N/2 = 26.5. A horizontal line is drawn from the point (0, 26.5) until it intersects the ogive curve at point A. From this intersection, a vertical line is drawn downward to meet the x-axis. The x-coordinate of this point is 66.4.

Therefore, the median of the data is 66.4.
In simple words: Plot the upper limits and cumulative frequencies on a graph to create an ogive. Find the middle position (N/2) on the y-axis, follow across to the curve, then down to the x-axis - the value there is your median.

Exam Tip: Ensure your ogive curve is smooth and continuous - never use straight line segments. Mark the half-way point (N/2) on the y-axis clearly to avoid reading errors from the graph.

 

Question 31. Use an ogive curve to determine the median from the given frequency distribution of upper class limits.
Answer: An ogive is plotted with upper class limits positioned on the x-axis and their respective cumulative frequencies on the y-axis. The points representing each upper class limit and its cumulative frequency are plotted and joined smoothly to create the ogive curve.

Given that N = 80, we calculate N/2 = 40. A horizontal reference line is drawn from the point (0, 40) on the y-axis. This line intersects the ogive curve at point A. From this intersection point, a perpendicular is dropped to the x-axis, meeting it at point M with coordinate 76.

Thus, the median of the data is 76.
In simple words: The ogive graph helps locate the median visually. Find where half of the total frequency falls by marking N/2 on the y-axis, trace to the curve, then drop down to read the median value on the x-axis.

Exam Tip: Double-check your calculation of N/2 and mark it accurately on the y-axis - small errors here lead to incorrect median readings from the graph.

 

Question 32. Construct a more than frequency distribution table and draw its ogive to find the median.
Answer: A 'more than' cumulative frequency distribution table is prepared from the raw data. This table lists each lower class limit alongside its corresponding cumulative frequency. The entries show: More than 0 (cf = 100), More than 10 (cf = 96), More than 20 (cf = 90), continuing through More than 70 (cf = 5).

The ogive is plotted by placing lower class limits on the x-axis and cumulative frequencies on the y-axis. Points are plotted and connected smoothly.

With N = 80, we calculate N/2 = 40. From point (0, 40) on the y-axis, a horizontal line is drawn to meet the ogive curve. A perpendicular is then drawn from this intersection to the x-axis, intersecting at the point x = 76.

Therefore, the median of the data is 76.
In simple words: A 'more than' ogive uses lower class limits instead of upper limits. The graphical method remains the same - find N/2 on the y-axis, trace to the curve, then read down to find the median on the x-axis.

Exam Tip: Remember the key difference - 'less than' ogives use upper class limits while 'more than' ogives use lower class limits. This affects how you plot and read the graph.

 

Question 33. Prepare a more than cumulative frequency table for height data and use an ogive to determine the median.
Answer: A 'more than' frequency distribution table is constructed showing: More than 135 cm (cf = 50), More than 140 cm (cf = 45), More than 145 cm (cf = 37), More than 150 cm (cf = 28), More than 155 cm (cf = 16), and More than 160 cm (cf = 2).

An ogive curve is drawn by plotting lower class limits (135, 140, 145, 150, 155, 160) on the x-axis and their cumulative frequencies on the y-axis. The plotted points are joined with a smooth curve.

Since the total frequency N = 50, half of this is N/2 = 25. However, following the 'more than' ogive approach, we locate the point where cumulative frequency equals 25. A horizontal line from y = 25 is drawn to intersect the curve, and from this intersection, a perpendicular meets the x-axis. The reading shows the median height is approximately between 145 and 150 cm.

Thus, the median height is found from the ogive graph.
In simple words: Plot height values on one axis and how many students have heights greater than each value on the other. The ogive curve shows where the middle observation falls, giving you the median height.

Exam Tip: When working with 'more than' frequency distributions, remember that cumulative frequencies decrease as the class limit increases - your ogive will slope downward, not upward.

 

Question 34. Construct a more than frequency table for height data and use an ogive to find the median.
Answer: A 'more than' cumulative frequency distribution table is organized showing height ranges and their cumulative frequencies: More than 140 cm (cf = 156), More than 160 cm (cf = 153), More than 180 cm (cf = 145), More than 200 cm (cf = 130), More than 220 cm (cf = 90), More than 240 cm (cf = 40), and More than 260 cm (cf = 10).

The ogive is constructed by plotting lower class limits on the x-axis (140, 160, 180, 200, 220, 240, 260) and cumulative frequencies on the y-axis. The points are connected smoothly to form the curve, which slopes downward as is typical for 'more than' ogives.

With N = 156, the median position corresponds to N/2 = 78. However, the exact median reading is obtained graphically by locating where the appropriate cumulative frequency intersects the ogive and reading the corresponding x-value.

The median is found from the graph by this method.
In simple words: The ogive shows a downward slope for 'more than' data. Find the point on the y-axis representing half the total observations, trace to where it meets the curve, then read down to find the median on the x-axis.

Exam Tip: Always check whether your ogive is increasing (less than) or decreasing (more than) - this helps verify you've plotted correctly before attempting to read the median.

 

Question 35. Create a more than cumulative frequency table for production yield data and use an ogive to determine the median.
Answer: A 'more than' frequency distribution table is established for production yield (kg/ha): More than 50 (cf = 100), More than 55 (cf = 98), More than 60 (cf = 90), More than 65 (cf = 78), More than 70 (cf = 54), and More than 75 (cf = 16).

An ogive curve is drawn by placing lower class limits on the x-axis and cumulative frequencies on the y-axis. The plotted points are connected smoothly to create the downward-sloping ogive characteristic of 'more than' frequency distributions.

Since N = 100, we have N/2 = 50. A horizontal line is drawn from (0, 50) to intersect the ogive at point A. From this intersection, a perpendicular is dropped to the x-axis, meeting it at x = 70.5.

Therefore, the median production yield is 70.5 kg/ha.
In simple words: Mark where half of all observations (50 out of 100) should fall on the y-axis. Trace horizontally to the ogive curve, then drop down vertically to read the median yield on the x-axis.

Exam Tip: For 'more than' ogives, the curve always descends from left to right - if yours doesn't, check your cumulative frequency calculations for errors.

 

Question 36. Construct both less than and more than frequency distribution tables for weekly expenditure and determine the median using ogive curves.
Answer: Two frequency distribution tables are prepared from the given weekly expenditure data. The 'less than' table shows cumulative frequencies increasing with upper class limits: Less than 200 (cf = 5) through Less than 900 (cf = 49).

The 'more than' table displays cumulative frequencies decreasing with lower class limits: More than 100 (cf = 49) through More than 800 (cf = 2).

Both ogive curves are drawn on the same graph. The 'less than' ogive is plotted with upper class limits on the x-axis, while the 'more than' ogive uses lower class limits. The cumulative frequencies form the y-axis for both.

The two ogive curves will intersect at a point. From this intersection point, a perpendicular is drawn to the x-axis. The x-coordinate where this perpendicular meets the axis represents the median.

By reading the graph, the median weekly expenditure can be determined from the intersection point.
In simple words: Plot both types of ogives on one graph - one rising (less than) and one falling (more than). Where they cross marks the median, so drop a line down to the x-axis to find it.

Exam Tip: The two ogive curves must intersect at exactly one point if calculations are correct - if they don't intersect or intersect at multiple points, recheck your frequency tables for errors.

 

Question 37. From the given frequency data of test scores, construct a more than frequency table and use its ogive to find the median.
Answer: A 'more than' frequency distribution table is prepared from the score data: More than 400 (cf = 230), More than 450 (cf = 210), More than 500 (cf = 175), More than 550 (cf = 135), More than 600 (cf = 103), More than 650 (cf = 79), More than 700 (cf = 52), and More than 750 (cf = 34).

The ogive is plotted by marking score values on the x-axis and their corresponding cumulative frequencies on the y-axis. Points representing A(750, 34), B(700, 52), C(650, 79), D(600, 103), E(550, 135), F(500, 175), G(450, 210), and H(400, 230) are plotted.

These points are connected with a smooth curve (not straight line segments) to form the ogive.

With N = 230, the median position is N/2 = 115. A horizontal line is drawn from point (0, 115) on the y-axis until it meets the ogive curve at point Q. From Q, a vertical line is drawn downward to meet the x-axis at M. The x-coordinate reading gives the median score.

From the graph, OM = 590 units. Therefore, the median score is 590 units.
In simple words: Plot the score data as a 'more than' ogive on a graph. Find the halfway point (115) on the frequency axis, trace to the curve, then read down to the score axis - that value is your median.

Exam Tip: Use a smooth, flowing curve for the ogive, not straight line segments between points - this gives more accurate median readings when using the graphical method.

 

Question 38. (i) Construct a less than frequency distribution table and draw its ogive. (ii) Construct a more than frequency distribution table and draw its ogive on the same graph.
Answer: (i) From the given mark data, a 'less than' cumulative frequency table is prepared: Less than 5 (cf = 2), Less than 10 (cf = 7), Less than 15 (cf = 13), continuing through Less than 50 (cf = 100). Points A(5, 2), B(10, 7), C(15, 13), D(20, 21), E(25, 31), F(30, 56), G(35, 76), H(40, 94), I(45, 98), and J(50, 100) are plotted and joined smoothly to form the 'less than' ogive curve.

(ii) The 'more than' table shows: More than 0 (cf = 100), More than 5 (cf = 98), More than 10 (cf = 93), continuing through More than 45 (cf = 2). On the same graph, points (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24), (40, 6), and (45, 2) are plotted and connected to form the 'more than' ogive.

The two ogives will intersect at a point L. A perpendicular LM is drawn from this intersection to the x-axis. The x-coordinate at M gives the median of the data.

From the graph, M = 29.5. Therefore, the median is 29.5.
In simple words: Plot both ogive curves on one graph - the less than curve rises from lower left to upper right, while the more than curve falls from upper left to lower right. They meet at the median point.

Exam Tip: Ensure both tables show all marks and their cumulative frequencies correctly - even one error will shift both curves and give an incorrect intersection point and median.

 

Question 39. (i) Construct a less than frequency distribution table and draw its ogive. (ii) Construct a more than frequency distribution table and draw its ogive on the same graph to find the median.
Answer: (i) From the given height data, a 'less than' cumulative frequency table is built: Less than 144 (cf = 3), Less than 148 (cf = 12), Less than 152 (cf = 36), Less than 156 (cf = 67), Less than 160 (cf = 109), Less than 164 (cf = 173), Less than 168 (cf = 248), Less than 172 (cf = 330), Less than 176 (cf = 416), and Less than 180 (cf = 450). These points are plotted and joined smoothly.

(ii) The 'more than' distribution table shows: More than 140 (cf = 450), More than 144 (cf = 447), More than 148 (cf = 438), More than 152 (cf = 414), More than 156 (cf = 383), More than 160 (cf = 341), More than 164 (cf = 277), More than 168 (cf = 202), More than 172 (cf = 120), and More than 176 (cf = 34). These values are plotted on the same graph and connected.

The two curves intersect at point L. A perpendicular LM is drawn from the intersection to the x-axis. The reading at M on the x-axis represents the median height.

From the graph intersection, the median height is 166 cm.
In simple words: Two ogives drawn together show where the data is equally split - one curves up (less than), the other curves down (more than). Their crossing point, when dropped to the horizontal axis, gives the median height.

Exam Tip: For height data spanning larger ranges (144-180 cm), use an appropriate graph scale to ensure accurate plotting and easy visual intersection identification.

 

Question 1. Find the median class from the given frequency distribution by calculating N/2 and identifying the cumulative frequency class.
Answer: To identify the median class, a frequency table is organized showing class intervals, frequencies, and cumulative frequencies. The classes range from 0 - 10 through 60 - 70, with a total frequency N = 50.

The value N/2 = 50/2 = 25. We examine the cumulative frequency column to find the smallest cumulative frequency greater than 25. This cumulative frequency is 26, which corresponds to the class interval 30 - 40.

Therefore, the median class is 30 - 40.
In simple words: Find half of the total number of observations. Then look at the running total of frequencies (cumulative frequency) to see which class interval contains this halfway mark - that's your median class.

Exam Tip: Always calculate N/2 first, then systematically check cumulative frequencies - the first cf value that equals or exceeds N/2 tells you the median class directly.

 

Question 2. Identify the modal class from the frequency distribution by finding the class with the maximum frequency.
Answer: The modal class is determined by identifying the class interval that has the highest frequency (maximum class frequency). Examining the frequency distribution, the class with frequency 27 is the highest frequency observed.

The class interval corresponding to this maximum frequency of 27 is 40 - 50. Therefore, the modal class is 40 - 50.

The lower limit (l) of this modal class is 40.
In simple words: Look through all the frequencies and find which class has the most observations. That class is the modal class, and its lower boundary value is what you're asked to find.

Exam Tip: The modal class always corresponds to the single highest frequency value - if two classes have equal frequencies, the distribution is bimodal, but typically one class dominates.

 

Question 3. Determine the modal class from the frequency distribution by locating the class interval with maximum frequency.
Answer: The modal class is found by identifying the class interval with the greatest frequency. Reviewing the given frequency distribution, the maximum class frequency is 30. The class interval that has this frequency of 30 is 150 - 200.

Therefore, the modal class is 150 - 200.
In simple words: Scan through all frequency values and pick the largest one. The class containing that largest frequency is your modal class.

Exam Tip: Modal class identification is straightforward - it's simply the class with the highest count or frequency, representing where most observations cluster.

 

Question 4. If the number of observations is odd, then the median is the \( \left(\frac{n+1}{2}\right)^{th} \) observation.
Answer: When you have an odd count of data points arranged in order, the middle value sits at position \( \left(\frac{n+1}{2}\right) \). For instance, with 25 observations, this gives you the 13th position, which represents your median.
In simple words: For an odd number of values, the median is simply the value in the exact middle when you line them all up in order.

Exam Tip: Always arrange data in ascending or descending order first, then count to the middle position using the formula - this ensures you identify the true median.

 

Question 5. Find the modal class and median class from the frequency distribution table.
Answer: To find the modal class, identify which class interval has the highest frequency. In this distribution, the maximum frequency is 25, making the 40-60 class the modal class. To find the median class, create a cumulative frequency column and find where \( \frac{N}{2} \) falls. With N = 50, we need \( \frac{N}{2} = 25 \). The cumulative frequency just greater than 25 is 35, corresponding to the 40-60 class. Therefore, both the modal class and median class are 40-60.
In simple words: The modal class is the group with the most items. The median class is the group where the middle value lands.

Exam Tip: Create a complete cumulative frequency table before answering - this prevents errors and makes both classes easy to spot.

 

Question 6. Calculate the class mark of the modal class.
Answer: The class mark (also called midpoint) is calculated by averaging the upper and lower boundaries. For the modal class 150-200, the class mark equals \( \frac{150+200}{2} = \frac{350}{2} = 175 \).
In simple words: Add the two ends of the class interval and divide by 2 to get the middle value of that class.

Exam Tip: The class mark represents the typical value in each class interval - always use it for calculations involving grouped data.

 

Question 7. Find the class marks for the intervals 10-25 and 35-55.
Answer: The class mark for interval 10-25 is calculated as \( \frac{10+25}{2} = \frac{35}{2} = 17.5 \). For interval 35-55, the class mark is \( \frac{35+55}{2} = \frac{90}{2} = 45 \).
In simple words: Find the average of the starting and ending numbers for each class to get its mark.

Exam Tip: Class marks are essential for computing the mean of grouped data - calculate them carefully for each interval.

 

Question 8. If the means of two groups are 4 and 3, with frequencies 36 and 64 respectively, find the combined mean.
Answer: For the first group: \( X = 4 \times 36 = 144 \). For the second group: \( Y = 3 \times 64 = 192 \). The combined sum is \( X + Y = 144 + 192 = 336 \). The total frequency is \( 36 + 64 = 100 \). Therefore, the combined mean equals \( \frac{336}{100} = 3.36 \).
In simple words: Multiply each group's mean by its count, add these products together, then divide by the total count.

Exam Tip: Always find the sum of all observations by multiplying mean × frequency for each group, then compute the overall mean.

 

Question 9. Find the upper class boundary of the highest class given the lowest class boundary is 8.1, class width is 2.5, and there are 12 classes.
Answer: Use the formula: Upper class boundary = Lowest class boundary + (width × number of classes). Substituting the values: \( 8.1 + (2.5 \times 12) = 8.1 + 30 = 38.1 \). Thus, the upper class boundary of the highest class is 38.1.
In simple words: Start from the lowest boundary and keep adding the class width once for each class interval until you reach the top.

Exam Tip: This formula helps you reconstruct the entire class structure when you know the starting point and interval width.

 

Question 10. If a dataset has 10 observations, and the values at positions 5 and 6 are x and (x+2), and the median is 63, find x.
Answer: For an even number of observations, the median is the average of the \( \left(\frac{n}{2}\right)^{th} \) and \( \left(\frac{n}{2}+1\right)^{th} \) values. With n = 10, these positions are 5 and 6. So: \( 63 = \frac{x+(x+2)}{2} \). Simplifying: \( 126 = 2x + 2 \), which gives \( 124 = 2x \), therefore \( x = 62 \).
In simple words: With an even count, the median is halfway between the two middle numbers. Use this to set up an equation and solve for x.

Exam Tip: Always identify which positions hold the "middle" values for even-sized datasets before calculating the median.

 

Question 11. If a median class is 20-30 with median 24, find the missing frequency x in the given distribution.
Answer: Given that the median is 24 and the median class is 20-30, we apply the median formula: \( \text{Median} = l + \left(\frac{\frac{N}{2}-cf}{f}\right) \times h \), where l = 20, h = 10, N = 55 + x, f = x, and cf = 30. Substituting: \( 24 = 20 + \left(\frac{\frac{55+x}{2}-30}{x}\right) \times 10 \). Simplifying: \( 4 = \left(\frac{\frac{55+x-60}{2x}}\right) \times 10 \), which leads to \( 8x = 10x - 50 \), giving \( x = 25 \).
In simple words: Use the median formula by substituting the known values and solve the equation step by step to find the missing frequency.

Exam Tip: Build the cumulative frequency table carefully and double-check your arithmetic when solving for missing values.

 

Question 12. If 8 is less than the median 30 and 32 is greater than the median 30, what is the median of 21 observations combined?
Answer: When combining two datasets of 10 and 11 observations, the middle value (the 11th observation overall) from the combined ordered list will be 30. Since 8 is below 30 and 32 is above 30, the value 30 remains unchanged as the median for all 21 observations taken together.
In simple words: When the middle position falls on the same value regardless of the datasets you're merging, that value is your median.

Exam Tip: Always arrange combined data in order and find the exact middle position to determine the true median.

 

Question 13. Arrange the values \( \frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x \) in ascending order and find x if the median is 8.
Answer: When arranged in ascending order, the sequence is \( \frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x \). With 5 values, the median is the 3rd term: \( \frac{x}{3} = 8 \). Solving: \( x = 3 \times 8 = 24 \).
In simple words: Arrange fractions in order, find which one sits in the middle, set it equal to the given median, and solve for x.

Exam Tip: Always verify that your answer produces the correct order - fractions with the same numerator get larger as the denominator gets smaller.

 

Question 14. Find the cumulative frequency of the modal class from the given distribution.
Answer: The modal class is identified by finding the class with the maximum frequency. In this distribution, the highest frequency is 23, which corresponds to class 12-15. To find the cumulative frequency of this class, sum all frequencies from the start up to and including this class. From the table, the cumulative frequency is 53.
In simple words: Spot the class with the highest count (modal class), then add up all frequencies from the first class through that modal class.

Exam Tip: Always double-check your cumulative frequency calculation by verifying that each cumulative value is at least as large as the previous one.

 

Question 15. Find the mode using the formula \( \text{Mode} = l + \left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right) \times h \) for the modal class 40-60.
Answer: Given: modal class = 40-60, lower limit (l) = 40, class size (h) = 20, frequency of modal class (\( f_1 \)) = 18, frequency of preceding class (\( f_0 \)) = 6, frequency of succeeding class (\( f_2 \)) = 10. Substituting into the formula: \( \text{Mode} = 40 + \left(\frac{18-6}{2(18)-6-10}\right) \times 20 = 40 + \left(\frac{12}{20}\right) \times 20 = 40 + 12 = 52 \).
In simple words: Plug all the given numbers into the mode formula and simplify step by step to get your final answer.

Exam Tip: Write out each parameter clearly before substituting - this prevents sign and value errors in the calculation.

 

Question 16. Create a 'less than type' cumulative frequency distribution table from the given data.
Answer: A 'less than type' cumulative frequency distribution shows the number of observations below each upper class limit:

In simple words: Start with the first frequency and keep adding each new frequency to the running total. Each entry shows how many observations fall below that upper limit.

Exam Tip: Always verify that the final cumulative frequency equals the total number of observations (N).

 

Question 17. Find the values of p and q, then identify the modal class and median class.
Answer: From the frequency distribution: \( p = 11 + 12 = 23 \) and \( q = 46 - 33 = 13 \). The maximum class frequency is 20, so the modal class is 500-600. With \( \Sigma f = N = 80 \), we have \( \frac{N}{2} = 40 \). The cumulative frequency just greater than 40 is 46, which corresponds to class 400-500. Therefore, the median class is 400-500.
In simple words: Find unknown frequencies using the given relationships, then identify the classes by checking which has the highest count and where the middle position falls.

Exam Tip: Always construct the complete frequency and cumulative frequency table before identifying modal and median classes.

 

Question 18. Create a 'more than type' cumulative frequency distribution table.
Answer: A 'more than type' cumulative frequency distribution shows the count of observations at or above each lower class limit:

In simple words: For a 'more than' table, start from the top and work downward by subtracting each class frequency. Each entry shows how many observations are at least that value.

Exam Tip: Check that 'more than' cumulative frequencies decrease as values increase - this confirms your calculation is correct.

 

Question 19. Convert the given data into a frequency distribution table.
Answer: The frequency distribution table is organized by grouping the life-time observations into class intervals:

In simple words: Group individual observations into class intervals and count how many fall within each range to create the frequency distribution.

Exam Tip: Ensure all class intervals are equal in width and that every observation falls into exactly one class.

 

Question 20. (a) Convert the discrete frequency distribution into continuous form. (b) Find the median class. (c) Calculate the class mark of the median class and identify the modal class.
Answer: (a) The continuous form adjusts boundaries by subtracting 0.5 from lower limits and adding 0.5 to upper limits:

(b) Building the cumulative frequency table: N = 2300, so \( \frac{N}{2} = 1150 \). The cumulative frequency just greater than 1150 is 1330, corresponding to class 40.5-50.5. Thus, the median class is 40.5-50.5. (c) The class mark of the median class is \( \frac{40.5+50.5}{2} = \frac{91}{2} = 45.5 \). The maximum class frequency is 529 (class 40.5-50.5), so the modal class is 40.5-50.5 with cumulative frequency 1330.In simple words: Change the boundaries to make the data continuous, then build a cumulative table to find where the middle falls, calculate its class mark as the average of the boundaries, and check which class has the most observations.

Exam Tip: When converting to continuous form, always subtract 0.5 from lower limits and add 0.5 to upper limits to remove gaps.

 

Question 21. Find the value of p using the mean formula for grouped data.
Answer: The mean for grouped data is \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \). From the table: \( \sum f_i = 43 + p \) and \( \sum f_i x_i = 1245 + 15p \). Given that the mean is 27: \[27 = \frac{1245+15p}{43+p}\] Cross-multiplying: \( 27(43+p) = 1245+15p \), which gives \( 1161+27p = 1245+15p \). Simplifying: \( 12p = 84 \), so \( p = 7 \).In simple words: Substitute the mean formula, set up an equation with the known mean, then solve for the missing frequency step by step.

Exam Tip: Always compute \( \sum f_i x_i \) carefully for each class before substituting into the mean formula.

 

Question 22. Find the missing frequency x using the median formula.
Answer: Given the median is 24, the median class is identified as 20-30. Using the median formula: \[\text{Median} = l + \left(\frac{\frac{N}{2}-cf}{f}\right) \times h\] With l = 20, h = 10, N = 55 + x, f = x, cf = 30: \[24 = 20 + \left(\frac{\frac{55+x}{2}-30}{x}\right) \times 10\] Simplifying: \( 4 = \left(\frac{x-5}{2x}\right) \times 10 \), which leads to \( 8x = 10x - 50 \). Solving: \( 2x = 50 \), so \( x = 25 \).In simple words: Set up the median equation with all known values and the variable x, then solve by clearing fractions and isolating x.

Exam Tip: Build a complete cumulative frequency column and verify that your median class is correct before solving for the missing frequency.

 

Question 1 (MCQ). Which of the following is a measure of dispersion?
(a) Mean
(b) Median
(c) Mode
(d) Standard Deviation
Answer: (d) Standard Deviation
In simple words: Standard deviation tells you how spread out your data is. The mean, median, and mode all describe the center of data, not how scattered it is.

Exam Tip: Remember that measures of central tendency (mean, median, mode) describe the "middle," while dispersion measures describe the "spread."

 

Question 2 (MCQ). Which measure of central tendency cannot be determined graphically?
(a) Mean
(b) Median
(c) Mode
(d) All of the options
Answer: (a) Mean
In simple words: The mean requires adding up all values, which can't be done by just looking at a graph. The median and mode can both be found from graphical displays.

Exam Tip: Median comes from ogive curves and mode from histograms, but mean needs actual calculations with all data values.

 

Question 3 (MCQ). Which measure of central tendency is influenced most by extreme values?
(a) Mean
(b) Median
(c) Mode
(d) All equally
Answer: (a) Mean
In simple words: One very large or very small number pulls the mean way up or down because you're averaging all values. The median and mode stay put because they depend on position or frequency, not the actual size.

Exam Tip: When a dataset has outliers, report both the mean and median - readers will then understand the data better.

 

Question 4 (MCQ). From which graphical tool can the mode of a frequency distribution be obtained directly?
(a) Ogive
(b) Frequency polygon
(c) Histogram
(d) Pie chart
Answer: (c) Histogram
In simple words: A histogram shows bars where the tallest bar represents the modal class. The mode is the value that appears most often, shown by the highest bar.

Exam Tip: The modal class (the class with the highest bar) corresponds to the mode of the data in a histogram.

 

Question 5 (MCQ). Which graphical representation is used to find the median of grouped data?
(a) Histogram
(b) Frequency polygon
(c) Bar graph
(d) Ogives
Answer: (d) Ogives
In simple words: An ogive is a curve showing cumulative frequencies. The median is found at the point where you reach half the total frequency on this curve.

Exam Tip: Ogives (both "less than" and "more than" types) meet at a point whose x-coordinate is the median.

 

Question 6 (MCQ). What is the purpose of a cumulative frequency table?
(a) To find the mean
(b) To find the median
(c) To find the mode
(d) To find standard deviation
Answer: (b) To find the median
In simple words: By adding up frequencies as you go down the table, you can spot exactly where the middle falls, which gives you the median.

Exam Tip: Always build a cumulative frequency table as your first step when finding the median of grouped data.

 

Question 7 (MCQ). What does the x-coordinate of the intersection point of 'less than' and 'more than' ogive curves represent?
(a) Mode
(b) Median
(c) Mean
(d) Range
Answer: (b) Median
In simple words: The 'less than' ogive climbs upward while 'more than' descends downward. They meet at the point representing the median because it's where exactly half the data sits below and half above.

Exam Tip: Plotting both ogive curves and finding their intersection point gives you a graphical method to determine the median quickly.

 

Question 8 (MCQ). Find the value of \( \sum f_i (x_i - \bar{x}) \).
(a) 1
(b) 0
(c) \( \bar{x} \)
(d) \( \sum f_i \)
Answer: (b) 0
In simple words: This sum always equals zero because the positive deviations (values above the mean) and negative deviations (values below the mean) cancel each other out perfectly.

Exam Tip: This property is fundamental in statistics - the deviations from the mean always balance to zero, which helps verify calculations.

 

Question 9 (MCQ). What is the formula for the deviation \( u_i \) from an assumed mean A?
(a) \( u_i = x_i - A \)
(b) \( u_i = \frac{x_i - A}{h} \)
(c) \( u_i = \frac{x_i}{A} \)
(d) \( u_i = x_i + A \)
Answer: (b) \( u_i = \frac{x_i - A}{h} \)
In simple words: First subtract the assumed mean A from each class mark, then divide by the class width h to get a scaled deviation. This simplifies calculations.

Exam Tip: Use this formula in the step-deviation method to reduce large numbers and make arithmetic easier.

 

Question 10 (MCQ). What do the deviations \( d_i \) represent in the step-deviation method?
(a) Deviations from the lower limit of the modal class
(b) Deviations from the mean
(c) Deviations from A (assumed mean) of midpoints of classes
(d) Deviations from the median
Answer: (c) Deviations from A (assumed mean) of midpoints of classes
In simple words: Choose any class mark as your assumed mean A, then calculate how far each class mark is from A. These differences are your \( d_i \) values.

Exam Tip: Pick an assumed mean near the center of your data - this keeps the \( d_i \) values small and calculations quick.

 

Question 11 (MCQ). When computing the mean of grouped data, what assumption is made about frequencies?
(a) They are spread evenly across each class
(b) They are centered at the class marks of classes
(c) They are concentrated at the upper limit
(d) They are concentrated at the lower limit
Answer: (b) They are centered at the class marks of classes
In simple words: We pretend all values in a class are located at the class mark (the middle), then use these marks to calculate the mean.

Exam Tip: This assumption is why finding class marks is the first step in computing the mean for grouped data.

 

Question 12 (MCQ). Which formula relates mode, median, and mean?
(a) Mode = Median + Mean
(b) Mode = (3 × Median) - (2 × Mean)
(c) Mode = 3 × Median
(d) Mode = Mean + Median
Answer: (b) Mode = (3 × Median) - (2 × Mean)
In simple words: This empirical relationship shows how the three measures connect - triple the median, subtract twice the mean, and you get the mode.

Exam Tip: Use this relationship to find one measure if you know the other two - it's a powerful shortcut in grouped data problems.

 

Question 13 (MCQ). A graph shows the intersection of two ogive curves at x = 20.5. What does this represent?
(a) Mode is 20.5
(b) Mean is 20.5
(c) Median is 20.5
(d) Range is 20.5
Answer: (c) Median is 20.5
In simple words: The x-value where 'less than' and 'more than' ogive curves cross shows the median - exactly the point where half the data is below and half is above.

Exam Tip: Reading the x-coordinate of the intersection point directly from the graph gives you the median without calculation.

 

Question 14 (MCQ). From a frequency distribution with modal class 150-155 and cumulative frequency data, find the sum of the modal class lower limit and the upper limit of the median class.
(a) 310
(b) 315
(c) 320
(d) 325
Answer: (b) 315
In simple words: The modal class 150-155 has a lower limit of 150. From cumulative frequencies, the median class has an upper limit of 165. Their sum is 150 + 165 = 315.

Exam Tip: Always identify both the modal and median classes separately before adding their boundaries together.

 

Question 15 (MCQ). Which class interval is the modal class in a frequency distribution where class 30-40 has maximum frequency 30?
(a) 20-30
(b) 40-50
(c) 30-40
(d) 50-60
Answer: (c) 30-40
In simple words: The modal class is simply the one with the highest frequency count - in this case, 30-40 with 30 observations.

Exam Tip: Scan the frequency column first to spot the maximum - no calculation needed for this step.

 

Question 16 (MCQ). Which formula calculates mode using the frequency formula?
(a) \( \text{Mode} = x_k + h \left(\frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}}\right) \)
(b) \( \text{Mode} = x_k + h \left(\frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}}\right) \)
(c) \( \text{Mode} = \frac{f_k}{2f_k} \)
(d) \( \text{Mode} = h + f_k \)
Answer: (b) \( \text{Mode} = x_k + h \left(\frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}}\right) \)
In simple words: This formula uses the lower limit of the modal class, the class width, and frequencies of the modal class plus its neighbors to calculate a precise mode.

Exam Tip: Always identify \( f_k \), \( f_{k-1} \), and \( f_{k+1} \) correctly - these are the frequencies of the modal class and its neighbors.

 

Question 17 (MCQ). What is the correct median formula for grouped data?
(a) \( \text{Median} = l + \left(h \times \frac{\frac{N}{2} - cf}{f}\right) \)
(b) \( \text{Median} = h + \left(l \times \frac{N}{2}\right) \)
(c) \( \text{Median} = \frac{N}{2} + f \)
(d) \( \text{Median} = l - h \)
Answer: (a) \( \text{Median} = l + \left(h \times \frac{\frac{N}{2} - cf}{f}\right) \)
In simple words: Start at the lower limit l of the median class. Adjust by a fraction of the class width h, where the fraction depends on how far \( \frac{N}{2} \) is from the cumulative frequency cf.

Exam Tip: Double-check that cf is the cumulative frequency UP TO (but not including) the median class - a common mistake.

 

Question 18 (MCQ). If mean = 8.9 and median = 9, find the mode using the empirical formula.
(a) 9.0
(b) 9.1
(c) 9.2
(d) 9.3
Answer: (c) 9.2
In simple words: Apply the formula: Mode = (3 × Median) - (2 × Mean) = (3 × 9) - (2 × 8.9) = 27 - 17.8 = 9.2.

Exam Tip: Keep three decimal places in intermediate steps to avoid rounding errors in the final answer.

 

Question 19 (MCQ). Find the median from the grouped data: class intervals 35-45, 45-55, 55-65, 65-75 with frequencies 8, 12, 20, 10 respectively.
(a) 56.5
(b) 57.5
(c) 58.5
(d) 59.5
Answer: (b) 57.5
In simple words: With N = 50, we need \( \frac{N}{2} = 25 \). This falls in class 55-65 (cumulative frequency 40). Using the median formula with l = 55, h = 10, cf = 20, f = 20: Median = \( 55 + \frac{10 \times (25-20)}{20} = 55 + 2.5 = 57.5 \).

Exam Tip: Always build the cumulative frequency column first to identify which class contains the median position.

 

Question 20 (MCQ). If the maximum frequency is 25 in class 22-26 with \( f_{k-1} = 16 \) and \( f_{k+1} = 19 \), find the mode.
(a) 23.4
(b) 24.0
(c) 24.4
(d) 25.0
Answer: (c) 24.4
In simple words: Using the mode formula with \( x_k = 22 \), \( f_k = 25 \), \( f_{k-1} = 16 \), \( f_{k+1} = 19 \), and h = 4: Mode = \( 22 + 4 \times \frac{25-16}{2(25)-16-19} = 22 + 4 \times \frac{9}{15} = 22 + 2.4 = 24.4 \).

Exam Tip: Verify that the denominator \( 2f_k - f_{k-1} - f_{k+1} \) is positive - if negative, double-check your frequency values.

 

Question 21 (MCQ). If mode = 16 and mean = 28, find the median using the empirical relationship.
(a) 20
(b) 22
(c) 24
(d) 26
Answer: (c) 24
In simple words: From Mode = (3 × Median) - (2 × Mean), we get: 16 = (3 × Median) - 56. Solving: 3 × Median = 72, so Median = 24.

Exam Tip: Rearrange the empirical formula to solve for the unknown measure - this is much faster than calculating from raw data.

 

Question 22 (MCQ). If mode = 29 and median = 26, find the mean using the empirical formula.
(a) 23.5
(b) 24.5
(c) 25.5
(d) 26.5
Answer: (b) 24.5
In simple words: From Mode = (3 × Median) - (2 × Mean), rearranging: 2 × Mean = (3 × 26) - 29 = 78 - 29 = 49. Therefore, Mean = 24.5.

Exam Tip: Always verify your answer by substituting back into the original formula to check accuracy.

 

Question 23. In a symmetric distribution, which of the following is true?
(a) mean > median > mode
(b) mean < median < mode
(c) mean = mode = median
(d) None of the options
Answer: (c) mean = mode = median
In simple words: When a distribution is symmetric, the three measures of central tendency - mean, median, and mode - all have the same value. The data is evenly balanced on both sides around this central point.

Exam Tip: Recognize that symmetry means the left and right sides mirror each other, so all three measures coincide at the center.

 

Question 24. How many families have a monthly income in the range of Rs. 20,000 - Rs. 25,000?
Answer: (c) 13

To find the number of families in a specific income bracket, we convert the given cumulative data into a frequency table by calculating the difference between consecutive cumulative frequencies.

The frequency for the income range Rs. 20,000 - Rs. 25,000 is obtained by subtracting the cumulative frequency of the previous class from the current class: 50 - 37 = 13.
In simple words: When you have cumulative data (running totals), you find how many people are in one group by subtracting the previous total from the new total. In this case, 50 - 37 = 13 families.

Exam Tip: Always remember that frequency = current cumulative frequency - previous cumulative frequency. Check your subtraction carefully to avoid calculation errors.

 

Question 25. What is the median of the first 8 prime numbers?
Answer: (b) 9

The first 8 prime numbers are: 2, 3, 5, 7, 11, 13, 17, and 19.

For an even number of values, the median is found by taking the average of the two middle terms. With 8 numbers, the two middle positions are the 4th and 5th.

4th term = 7
5th term = 11

Median = \( \frac{7 + 11}{2} = \frac{18}{2} = 9 \)

In simple words: List the numbers in order, find the two middle numbers (the 4th and 5th in a list of 8), and then find their average. That average is your median.

Exam Tip: For an even number of observations, the median always lies between the two middle values. Make sure you identify the correct positions and compute their average correctly.

 

Question 26. If the mean of 20 numbers is zero, what is the maximum number of positive numbers that can exist in this set?
Answer: (d) 19

We are told that the mean of 20 numbers equals zero, which means the sum of all 20 numbers must equal zero (since mean = sum / count, and 0 = sum / 20 only when sum = 0).

For the sum to be zero while having as many positive numbers as possible, we can have 19 positive numbers whose sum equals some value x, and then the 20th number must be - x (a negative number) to balance out to zero.

For example: if the sum of 19 positive numbers is 100, the 20th number would be - 100, making the total sum = 100 + (- 100) = 0.

We cannot have 20 positive numbers because then the sum would be positive, not zero. Therefore, at most 19 numbers can be positive.

In simple words: If all 20 numbers were positive, their sum would be positive, not zero. So at least one number must be negative to make the sum equal zero. This means you can have at most 19 positive numbers.

Exam Tip: Remember that mean = 0 automatically tells you the sum = 0. Use this constraint to figure out how many numbers can have a particular sign.

 

Question 27. The median of 6 numbers is 13. If two of the numbers are (x - 1) and (x - 3), and they are the 3rd and 4th terms respectively, find the value of x.
Answer: (c) 15

For 6 numbers arranged in order, the median is the average of the 3rd and 4th terms.

Given: Median = 13, 3rd term = (x - 1), 4th term = (x - 3)

\[ 13 = \frac{(x-1) + (x-3)}{2} \]

Multiplying both sides by 2:
\[ 26 = (x-1) + (x-3) \]
\[ 26 = 2x - 4 \]
\[ 2x = 30 \]
\[ x = 15 \]

In simple words: The median of 6 numbers is the average of the middle two numbers (positions 3 and 4). Set up an equation using the given expressions, solve for x, and you get 15.

Exam Tip: Always recall that for an even count of data points, the median is the average of the two middle values. Set up the equation carefully and solve step by step.

 

Question 28. Two sets of observations are given. In the first set of 4 observations {2, 7, 6, x}, the mean is 15. In the second set of 5 observations {18, 1, 6, x, y}, the mean is 10. Find the value of y.
Answer: (c) - 20

For the first set of 4 observations, using the mean formula:

\[ \text{Mean} = \frac{\text{sum of observations}}{\text{number of observations}} \]

\[ 15 = \frac{2 + 7 + 6 + x}{4} \]
\[ 60 = 15 + x \]
\[ x = 45 \quad \text{...(1)} \]

Now for the second set of 5 observations:

\[ 10 = \frac{18 + 1 + 6 + x + y}{5} \]
\[ 50 = 25 + x + y \]
\[ x + y = 25 \]
\[ y = 25 - x \]

Substituting x = 45 from equation (1):

\[ y = 25 - 45 = -20 \]

In simple words: Use the mean formula to find x from the first set, then use that same formula on the second set to find y. Substitute the value of x into the equation for y to get your answer.

Exam Tip: When working with multiple sets, find unknowns step by step. Always check that you substitute correctly from one equation into the next.

 

Question 29. Match each statement in Column I with the correct definition or property in Column II.
Answer:

In simple words: These matching questions test your knowledge of statistical terms. The mode is the most frequent value; the mean cannot be read from a graph; an ogive helps find the median; and standard deviation measures spread, not central tendency.

Exam Tip: Remember the key definitions: mode (most frequent), median (middle value), mean (average), and standard deviation (spread). Know which can or cannot be determined graphically.

 

Question 30(a). Assertion (A): The mean of a distribution is 148. Reason (R): If 3 × median - mode = 2 × mean, and the median is 150 and mode is 154, then the mean is 148. Are both the assertion and reason true? If so, is the reason a correct explanation of the assertion?
Answer: Both Assertion (A) and Reason (R) are true, and Reason (R) is a correct explanation of Assertion (A).

The relationship given in the reason is: \( 2 \times \text{mean} = 3 \times \text{median} - \text{mode} \)

Substituting the given values (median = 150, mode = 154):

\[ 2 \times \text{mean} = (3 \times 150) - 154 = 450 - 154 = 296 \]
\[ \text{mean} = 148 \]

This confirms that the assertion is true. The reason provides the exact formula that allows us to calculate and verify the mean value, making it a direct and correct explanation of the assertion.

In simple words: The reason gives you a formula that connects mean, median, and mode. When you plug in the median and mode values, you get the mean of 148, which proves the assertion is correct.

Exam Tip: For assertion-reason questions, verify both statements independently first, then check whether the reason logically supports the assertion. A true reason must explain why the assertion is true.

 

Question 30(b). Assertion (A): The mode of a grouped frequency distribution with modal class 12 - 15 and maximum frequency 23 is 12.4. Reason (R): The mode of grouped data cannot be determined graphically. Are both the assertion and reason true? If so, is the reason a correct explanation of the assertion?
Answer: Both Assertion (A) and Reason (R) are true, but Reason (R) is NOT a correct explanation of Assertion (A).

To find the mode of grouped data, we use the mode formula for a frequency distribution. Given: xk = 12, fk = 23 (maximum frequency), fk-1 = 21, fk+1 = 23, and h = 3 (class width).

\[ \text{Mode} = x_k + h \times \frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}} \]

\[ \text{Mode} = 12 + 3 \times \frac{23 - 21}{(2 \times 23) - 21 - 23} \]

\[ = 12 + 3 \times \frac{2}{46 - 21 - 23} \]

\[ = 12 + 3 \times \frac{2}{2} \]

\[ = 12 + 3 = 12.4 \]

The assertion is therefore true. The reason is also true - the mode cannot be read directly from a graph. However, the reason does not explain how we calculated the mode or why the assertion is true. The reason is an independent fact that does not logically support the assertion.

In simple words: Both statements are correct facts, but they are about different things. One tells you the mode value; the other tells you that mode requires calculation, not a graph. One doesn't explain the other.

Exam Tip: In assertion-reason questions, being true is not enough - you must determine whether the reason actually explains the assertion. Two true statements are not necessarily linked.