CBSE Class 10 Mathematics Statistics Worksheet Set 04

STATISTICS

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Question 1. The empirical relationship between the three measures of central tendency is…………………
Answer: The mathematical relation connecting the three statistics is:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
In simple words: To find the mode of a dataset, multiply the median by three and subtract twice the mean value from it.

Exam Tip: Memorize this empirical relation thoroughly, as it is extremely useful for calculating any one of the three averages when the other two are given.

Question 2. …………………. Is called a positional average.
Answer: Median
In simple words: The median is called a positional average because its calculation is based on its position in the middle of a sorted list of numbers.

Exam Tip: Unlike the arithmetic mean, the median is called a positional average because it is unaffected by extreme values (outliers) in the data.

Question 3. The point of intersection of the less than ogive and the more than ogive gives us the ……………..
Answer: Median
In simple words: The exact spot where the less-than and more-than graph lines cross each other points directly down to the median on the bottom axis.

Exam Tip: On a graph where the two ogive curves cross at point \( (x, y) \), the horizontal value \( x \) is always the median, while the vertical value \( y \) is \( \frac{N}{2} \).

Question 4. The point of intersection of the less than ogive and the more than ogive is (36.5,15).The median is………
Answer: 36.5
In simple words: Since the horizontal coordinate of the crossing point represents the median, the median value is 36.5.

Exam Tip: Be careful not to swap the coordinates of the intersection point; the first number is always the median and the second number is half of the total frequency.

Question 5. The median and the mode of a data are 62 and 64 respectively. The mean is………..
Answer: Given values:
\( \text{Median} = 62 \)
\( \text{Mode} = 64 \)
Using the empirical relation:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
\( \implies 64 = 3(62) - 2 \ \text{Mean} \)
\( \implies 64 = 186 - 2 \ \text{Mean} \)
\( \implies 2 \ \text{Mean} = 186 - 64 = 122 \)
\( \implies \text{Mean} = 61 \)
In simple words: Substituting our numbers into the empirical formula shows that the mean value is 61.

Exam Tip: Carefully perform transposition and check algebraic signs to prevent simple calculation mistakes when rearranging the formula.

Question 6. The median and the mean of a data are 52 and 50 respectively. The mode is………..
Answer: Given values:
\( \text{Median} = 52 \)
\( \text{Mean} = 50 \)
Using the empirical formula:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
\( \implies \text{Mode} = 3(52) - 2(50) \)
\( \implies \text{Mode} = 156 - 100 = 56 \)
In simple words: Multiply the median by 3 and subtract twice the mean to get the mode, which equals 56.

Exam Tip: This is a direct substitution problem; simply multiply first and then subtract to find the mode.

Question 7. The mean and the mode of a data are 54 and 57 respectively. The median is………..
Answer: Given values:
\( \text{Mean} = 54 \)
\( \text{Mode} = 57 \)
Using the empirical formula:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
\( \implies 57 = 3 \ \text{Median} - 2(54) \)
\( \implies 57 = 3 \ \text{Median} - 108 \)
\( \implies 3 \ \text{Median} = 57 + 108 = 165 \)
\( \implies \text{Median} = \frac{165}{3} = 55 \)
In simple words: Putting the values into our formula gives three times the median as 165, which results in a median of 55.

Exam Tip: Remember to add twice the mean to the mode before dividing by 3 to find the correct median value.

Question 8. Change the following data to a frequency distribution table:

Answer: We convert this "less than" cumulative table to a grouped frequency distribution by finding the difference between successive cumulative frequencies:

In simple words: To make a normal table, we subtract each total from the next one to find the exact number of items in each 10-point group.

Exam Tip: The frequency of the first class interval is identical to its cumulative frequency, while the rest are calculated by subtracting the previous cumulative frequency from the current one.

Question 9. Change the following data to a frequency distribution table:

Answer: We convert this "more than / above" type cumulative frequency distribution into continuous class intervals by subtracting the next class cumulative frequency from the current class cumulative frequency:

In simple words: Subtract the cumulative value of the next class from the current one to find the frequency. The final class interval simply takes the last remaining cumulative value directly.

Exam Tip: For "more than" type tables, verify that the sum of all individual frequencies equals the first cumulative frequency value (which is 100 here).

Question 10. The mean of the following data is 38.2. Find the missing frequencies f1 and f2 if the total frequency is 50.

Answer: Let us prepare the calculation table:

Given that total frequency is 50:
\( \implies 31 + f_1 + f_2 = 50 \)
\( \implies f_1 + f_2 = 19 \) - (Equation 1)
Since Mean = 38.2:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 38.2 = \frac{1195 + 25f_1 + 45f_2}{50} \)
\( \implies 1195 + 25f_1 + 45f_2 = 38.2 \times 50 \)
\( \implies 1195 + 25f_1 + 45f_2 = 1910 \)
\( \implies 25f_1 + 45f_2 = 715 \)
Dividing by 5:
\( \implies 5f_1 + 9f_2 = 143 \) - (Equation 2)
Substituting \( f_1 = 19 - f_2 \) from Equation 1 into Equation 2:
\( \implies 5(19 - f_2) + 9f_2 = 143 \)
\( \implies 95 - 5f_2 + 9f_2 = 143 \)
\( \implies 4f_2 = 143 - 95 \)
\( \implies 4f_2 = 48 \implies f_2 = 12 \)
Substituting \( f_2 = 12 \) into Equation 1:
\( \implies f_1 + 12 = 19 \implies f_1 = 7 \)
Thus, the missing frequencies are \( f_1 = 7 \) and \( f_2 = 12 \).
In simple words: Add up the frequencies to write our first equation, and use the mean formula to establish the second equation. Solving these two together yields \( f_1 = 7 \) and \( f_2 = 12 \).

Exam Tip: Simplify the linear equations by dividing the coefficients by 5 to reduce complexity and perform faster manual calculations during the exam.

Question 11. The mean of the following frequency is 8. Find the value of p.

Answer: Let us prepare the calculation table:

Since Mean = 8:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 8 = \frac{303 + 9P}{41 + P} \)
\( \implies 8(41 + P) = 303 + 9P \)
\( \implies 328 + 8P = 303 + 9P \)
\( \implies 9P - 8P = 328 - 303 \)
\( \implies P = 25 \)
So, the value of P is 25.
In simple words: We multiply each variable value by its frequency, sum them up, and set up the mean formula equal to 8. Solving the linear equation gives us P = 25.

Exam Tip: Be careful not to add the variable \( P \) directly to the constant values when finding the sum of frequencies or the sum of \( f_i x_i \).

Question 12. Draw the less than ogive and decide the median.

Answer: Let us construct the less than cumulative frequency table:

The total frequency \( N = 35 \), so \( \frac{N}{2} = 17.5 \).
By plotting the cumulative frequencies on the y-axis against the upper class limits on the x-axis, we obtain the less-than ogive curve.
Locate \( y = 17.5 \) on the y-axis, move horizontally to intersect the curve, and then project vertically down to the x-axis to find the median.
By formula calculation:
The median class is 80-90 since cumulative frequency 29 is just greater than 17.5.
\( \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)
With \( l = 80, \ cf = 17, \ f = 12, \ h = 10 \):
\( \implies \text{Median} = 80 + \left(\frac{17.5 - 17}{12}\right) \times 10 = 80 + 0.417 \approx 80.42 \) (or approximately 82 when visually estimated on a hand-drawn graph paper).

Here is the graphical representation of the less than ogive: In simple words: Plot cumulative frequencies against upper limits to get the curve. A horizontal line from the middle point (17.5) down to the bottom axis reveals a median value of approximately 80.4.

Exam Tip: Less than ogives are always plotted using the upper class limits on the horizontal axis and their corresponding cumulative frequencies on the vertical axis.

Question 13. Find the unknown values a,b,c,d,e & f from the following table:

Answer: Let us calculate each of the unknown values step-by-step:
1. The cumulative frequency of the first class interval is equal to its frequency:
\( a = 12 \)
2. For the second class interval:
\( \implies 12 + b = 25 \implies b = 13 \)
3. For the third class interval:
\( c = 25 + 10 = 35 \)
4. For the fourth class interval:
\( \implies 35 + d = 43 \implies d = 8 \)
5. For the fifth class interval:
\( \implies 43 + e = 48 \implies e = 5 \)
6. For the sixth class interval:
\( f = 48 + 2 = 50 \)
Therefore, the calculated unknown values are:
\( a = 12, \ b = 13, \ c = 35, \ d = 8, \ e = 5, \ f = 50 \)
In simple words: Since cumulative frequency is the running total, we add and subtract from step to step to find the missing numbers: \( a = 12 \), \( b = 13 \), \( c = 35 \), \( d = 8 \), \( e = 5 \), and \( f = 50 \).

Exam Tip: Verify your calculations by ensuring that the final cumulative frequency \( f \) equals the stated total frequency of the data, which is 50.

Question 14. The median of the following data is 20.75 . Find x & y if the total frequency is 100.

Answer: Let us prepare the cumulative frequency table:

The total frequency is given as 100:
\( \implies 63 + x + y = 100 \)
\( \implies x + y = 37 \) - (Equation 1)
Since Median = 20.75, which lies in the class interval 20-25. Therefore, the median class is 20-25. We obtain:
Lower limit (\( l \)) = 20
Frequency of the median class (\( f \)) = y
Cumulative frequency of the preceding class (\( cf \)) = 30 + x
Class size (\( h \)) = 5
Total frequency (\( N \)) = 100
Applying the median formula:
\( \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)
\( \implies 20.75 = 20 + \left(\frac{50 - (30 + x)}{y}\right) \times 5 \)
\( \implies 0.75 = \frac{5(20 - x)}{y} \)
\( \implies 0.75y = 100 - 5x \)
\( \implies 5x + 0.75y = 100 \) - (Equation 2)
From Equation 1, substitute \( x = 37 - y \) into Equation 2:
\( \implies 5(37 - y) + 0.75y = 100 \)
\( \implies 185 - 5y + 0.75y = 100 \)
\( \implies -4.25y = -85 \implies y = \frac{85}{4.25} = 20 \)
Substitute \( y = 20 \) in Equation 1:
\( \implies x + 20 = 37 \implies x = 17 \)
Thus, the missing frequencies are \( x = 17 \) and \( y = 20 \).
In simple words: Since the median is 20.75, it belongs to the 20-25 class. Using the sum of frequencies and the median formula, we get two equations that solve to give \( x = 17 \) and \( y = 20 \).

Exam Tip: Pay special attention to decimals during algebra; expressing \( 4.25 \) as \( \frac{17}{4} \) is a useful shortcut to avoid division errors.

Question 15. The median of the following data is 525. Find f1 & f2 if the total frequency is 100.

Answer: Let us construct the cumulative frequency table:

The total frequency is given as 100:
\( \implies 76 + f_1 + f_2 = 100 \)
\( \implies f_1 + f_2 = 24 \) - (Equation 1)
Since Median = 525, which lies in the interval 500-600. Therefore, the median class is 500-600. We identify:
Lower limit (\( l \)) = 500
Frequency of the median class (\( f \)) = 20
Cumulative frequency of the preceding class (\( cf \)) = 36 + f1
Class size (\( h \)) = 100
Total frequency (\( N \)) = 100
Applying the median formula:
\( \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)
\( \implies 525 = 500 + \left(\frac{50 - (36 + f_1)}{20}\right) \times 100 \)
\( \implies 25 = (14 - f_1) \times 5 \)
\( \implies 5 = 14 - f_1 \implies f_1 = 9 \)
Substitute \( f_1 = 9 \) into Equation 1:
\( \implies 9 + f_2 = 24 \implies f_2 = 15 \)
Thus, the missing frequencies are \( f_1 = 9 \) and \( f_2 = 15 \).
In simple words: First we find the cumulative totals. Using the total frequency of 100 and the median value of 525, the formula helps us determine that \( f_1 = 9 \) and \( f_2 = 15 \).

Exam Tip: Be careful to apply the negative sign to both terms when simplifying \( 50 - (36 + f_1) \) so that it correctly becomes \( 14 - f_1 \).

Question 16. Find the mean, median and the mode of the following data.

Answer: Let us construct the joint calculation table:

1. Calculation of Mean:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{9320}{68} \approx 137.06 \) units

2. Calculation of Median:
Total frequency \( N = 68 \), so \( \frac{N}{2} = 34 \).
The cumulative frequency just greater than 34 is 42, which belongs to the class 125-145.
So, median class is 125-145.
We have: \( l = 125, \ cf = 22, \ f = 20, \ h = 20 \).
\( \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)
\( \text{Median} = 125 + \left(\frac{34 - 22}{20}\right) \times 20 = 125 + 12 = 137 \) units

3. Calculation of Mode:
The class with the highest frequency is 125-145 (frequency = 20). So modal class is 125-145.
We have: \( l = 125, \ f_1 = 20, \ f_0 = 13, \ f_2 = 14, \ h = 20 \).
\( \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \)
\( \text{Mode} = 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20 \)
\( \text{Mode} = 125 + \left(\frac{7}{40 - 27}\right) \times 20 = 125 + \frac{140}{13} \approx 135.76 \) units

Therefore, Mean = 137.06, Median = 137, and Mode = 135.76.
In simple words: We find the mean using our calculated totals to be 137.06, the median using cumulative sums to be 137, and the mode using the highest frequency class to be 135.76.

Exam Tip: Remember to state the units of measurement (such as units) alongside the calculated averages to ensure full marks.

Question 17. Find the median of the following data.

Answer: Since the class intervals are discontinuous, we make them continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit:

Total frequency \( N = 40 \), so \( \frac{N}{2} = 20 \).
The cumulative frequency just greater than 20 is 29, which corresponds to the class 144.5 - 153.5.
Therefore, the median class is 144.5 - 153.5.
We have:
Lower limit (\( l \)) = 144.5
Frequency (\( f \)) = 12
Cumulative frequency of the preceding class (\( cf \)) = 17
Class size (\( h \)) = 9
Applying the median formula:
\( \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \)
\( \implies \text{Median} = 144.5 + \left(\frac{20 - 17}{12}\right) \times 9 \)
\( \implies \text{Median} = 144.5 + \left(\frac{3}{12}\right) \times 9 \)
\( \implies \text{Median} = 144.5 + 2.25 = 146.75 \)
Thus, the median is 146.75.
In simple words: First we make the class intervals continuous. Finding the middle frequency point at 20 guides us to the 144.5-153.5 group, and the formula gives a median of 146.75.

Exam Tip: Always make the classes continuous before performing any calculations for median or mode. Check if the upper limit of one class matches the lower limit of the next class to see if this adjustment is needed.