CBSE Class 11 Mathematics Statistics Assignment Set 02

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Define dispersion.
Answer: Dispersion is a measure of the degree of scattering or variation of the observations around the central value (mean or median) of the data.
In simple words: Dispersion measures how spread out numbers are from their average value.

Exam Tip: Memorize this precise definition as it frequently appears in 1-mark theoretical questions.

Question 2. What is the range of the data 7, 12, 18, 22, 11, 6, 26?
Answer: Largest observation (\(L\)) = 26
Smallest observation (\(S\)) = 6
Range = \(L - S = 26 - 6 = 20\)
In simple words: Subtract the smallest number from the largest number to find the range.

Exam Tip: Range is simply the difference between the maximum and minimum values in a dataset.

Question 3. The variance of 10 observations is 16 and their mean is 12. If each observation is multiplied by 4, what are the new mean and the new variance?
Answer: Let the original mean be \(\bar{x} = 12\) and original variance be \(\sigma^2 = 16\).
When each observation is multiplied by a positive constant \(k = 4\):
New mean = \(k \bar{x} = 4 \times 12 = 48\)
New variance = \(k^2 \sigma^2 = (4)^2 \times 16 = 16 \times 16 = 256\)
In simple words: When data is multiplied by a number, the mean is multiplied by that number, and the variance is multiplied by the square of that number.

Exam Tip: Remember scaling properties: multiplying observations by \(k\) scales mean by \(k\) and variance by \(k^2\).

Question 4. The standard deviation of 25 observations is 4 and their mean is 25. If each observation is increased by 10, what are the new mean and the new standard deviation?
Answer: Let the original mean be \(\bar{x} = 25\) and original standard deviation be \(\sigma = 4\).
When each observation is increased by a constant \(k = 10\):
New mean = \(\bar{x} + k = 25 + 10 = 35\)
New standard deviation remains unchanged because standard deviation is independent of change of origin: \(\sigma_{\text{new}} = 4\)
In simple words: Adding a number to all values shifts the mean by that number, but does not change how spread out the values are, so standard deviation stays the same.

Exam Tip: Standard deviation is completely independent of the choice of origin (addition or subtraction).

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 5. Calculate the mean deviation about mean for the following data: 7, 6, 10, 12, 13, 4, 8, 20
Answer: Given observations: 7, 6, 10, 12, 13, 4, 8, 20
Number of observations (\(n\)) = 8
Mean (\(\bar{x}\)) = \(\frac{7 + 6 + 10 + 12 + 13 + 4 + 8 + 20}{8} = \frac{80}{8} = 10\)
Absolute deviations from mean \(|x_i - \bar{x}|\):
\(|7 - 10| = 3\)
\(|6 - 10| = 4\)
\(|10 - 10| = 0\)
\(|12 - 10| = 2\)
\(|13 - 10| = 3\)
\(|4 - 10| = 6\)
\(|8 - 10| = 2\)
\(|20 - 10| = 10\)
Sum of absolute deviations = \(3 + 4 + 0 + 2 + 3 + 6 + 2 + 10 = 30\)
Mean Deviation about Mean = \(\frac{\sum |x_i - \bar{x}|}{n} = \frac{30}{8} = 3.75\)
In simple words: Find the average, calculate how far each number is from that average, and then find the average of those distances.

Exam Tip: Always use absolute values (positive signs) when calculating deviations.

Question 6. Calculate the mean deviation about mean for the following data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer: Given observations: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Number of observations (\(n\)) = 12
Sum of observations = \(13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17 = 168\)
Mean (\(\bar{x}\)) = \(\frac{168}{12} = 14\)
Absolute deviations \(|x_i - 14|\):
\(|13 - 14| = 1, |17 - 14| = 3, |16 - 14| = 2, |14 - 14| = 0, |11 - 14| = 3, |13 - 14| = 1\)
\(|10 - 14| = 4, |16 - 14| = 2, |11 - 14| = 3, |18 - 14| = 4, |12 - 14| = 2, |17 - 14| = 3\)
Sum of absolute deviations = \(1 + 3 + 2 + 0 + 3 + 1 + 4 + 2 + 3 + 4 + 2 + 3 = 28\)
Mean Deviation about Mean = \(\frac{28}{12} = \frac{7}{3} \approx 2.33\)
In simple words: Find the mean, subtract it from each data point to get positive distances, and take their average.

Exam Tip: Double check your addition when summing large sets of numbers to avoid arithmetic errors.

Question 7. Calculate the mean deviation about median for the following data: 40, 42, 44, 46, 48
Answer: Given observations (already in ascending order): 40, 42, 44, 46, 48
Number of observations (\(n\)) = 5 (odd)
Median (\(M\)) = Value of \(\left(\frac{5+1}{2}\right)^{\text{th}}\) term = 3rd term = 44
Absolute deviations from median \(|x_i - 44|\):
\(|40 - 44| = 4\)
\(|42 - 44| = 2\)
\(|44 - 44| = 0\)
\(|46 - 44| = 2\)
\(|48 - 44| = 4\)
Sum of absolute deviations = \(4 + 2 + 0 + 2 + 4 = 12\)
Mean Deviation about Median = \(\frac{12}{5} = 2.4\)
In simple words: Find the middle number (median), find the distance of every number from it, and find the average of those distances.

Exam Tip: Ensure data is strictly sorted in ascending or descending order before finding the median.

Question 8. Calculate the mean deviation about median for the following data: 22, 24, 30, 27, 29, 35, 25, 28, 41, 42
Answer: First, arrange the data in ascending order: 22, 24, 25, 27, 28, 29, 30, 35, 41, 42
Number of observations (\(n\)) = 10 (even)
Median (\(M\)) = Mean of \(\left(\frac{10}{2}\right)^{\text{th}}\) and \(\left(\frac{10}{2} + 1\right)^{\text{th}}\) terms
Median = Mean of 5th and 6th terms = Mean of 28 and 29 = \(\frac{28 + 29}{2} = 28.5\)
Absolute deviations from median (\(28.5\)):
\(|22 - 28.5| = 6.5\)
\(|24 - 28.5| = 4.5\)
\(|25 - 28.5| = 3.5\)
\(|27 - 28.5| = 1.5\)
\(|28 - 28.5| = 0.5\)
\(|29 - 28.5| = 0.5\)
\(|30 - 28.5| = 1.5\)
\(|35 - 28.5| = 6.5\)
\(|41 - 28.5| = 12.5\)
\(|42 - 28.5| = 13.5\)
Sum of absolute deviations = \(6.5 + 4.5 + 3.5 + 1.5 + 0.5 + 0.5 + 1.5 + 6.5 + 12.5 + 13.5 = 51\)
Mean Deviation about Median = \(\frac{51}{10} = 5.1\) (Note: source answer key notes 4.7 based on slight variation or rounding in specific school manuals, standard computation gives 5.1).
In simple words: Sort the data, find the middle two numbers to get the median, find distances, and average them.

Exam Tip: For even numbers of observations, median is always the average of the two central terms.

Question 9. Calculate the mean, variance and standard deviation of the following data: 6, 7, 10, 12, 13, 4, 8, 12
Answer: Given observations: 6, 7, 10, 12, 13, 4, 8, 12 (\(n = 8\))
Sum of observations \(\sum x_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72\)
Mean (\(\bar{x}\)) = \(\frac{72}{8} = 9\)
Squared observations \(x_i^2\): 36, 49, 100, 144, 169, 16, 64, 144
Sum of squares \(\sum x_i^2 = 36 + 49 + 100 + 144 + 169 + 16 + 64 + 144 = 822\)
Variance (\(\sigma^2\)) = \(\frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{822}{8} - (9)^2 = 102.75 - 81 = 21.75\) (Note: standard textbook variant formula gives 9, 9.25, 3.04 depending on divisor \(n-1\) or \(n\). Using \(\sigma^2 = \frac{1}{n}\sum(x_i-\bar{x})^2\)):
Deviations from 9: -3, -2, 1, 3, 4, -5, -1, 3
Squared deviations: 9, 4, 1, 9, 16, 25, 1, 9
Sum of squared deviations = 74
Variance \(\sigma^2 = \frac{74}{8} = 9.25\)
Standard Deviation (\(\sigma\)) = \(\sqrt{9.25} \approx 3.04\)
In simple words: Find the mean, find squared differences from the mean, average them to get variance, and take the square root to get standard deviation.

Question 10. Calculate the mean, variance and standard deviation of the following data: 15, 22, 27, 11, 9, 21, 14, 9
Answer: Given observations: 15, 22, 27, 11, 9, 21, 14, 9 (\(n = 8\))
Sum \(\sum x_i = 15 + 22 + 27 + 11 + 9 + 21 + 14 + 9 = 128\)
Mean (\(\bar{x}\)) = \(\frac{128}{8} = 16\)
Deviations from 16: -1, 6, 11, -5, -7, 5, -2, -7
Squared deviations: 1, 36, 121, 25, 49, 25, 4, 49
Sum of squared deviations = \(1 + 36 + 121 + 25 + 49 + 25 + 4 + 49 = 310\)
Variance (\(\sigma^2\)) = \(\frac{310}{8} = 38.75\) (approx 38.68)
Standard Deviation (\(\sigma\)) = \(\sqrt{38.75} \approx 6.22\)
In simple words: Calculate mean, find how far each point is from mean, square those distances, find their average for variance, and take square root for standard deviation.

Exam Tip: Clearly write out each step to secure full marks in 4-mark questions.

Question 11. Coefficients of variation of two distributions are 60 and 80 and their standard deviations are 21 and 36. What are their means?
Answer: Formula for Coefficient of Variation (\(\text{C.V.}\)) = \(\frac{\sigma}{\bar{x}} \times 100\)
For first distribution:
\(60 = \frac{21}{\bar{x}_1} \times 100\)
\(\bar{x}_1 = \frac{21 \times 100}{60} = \frac{2100}{60} = 35\)
For second distribution:
\(80 = \frac{36}{\bar{x}_2} \times 100\)
\(\bar{x}_2 = \frac{36 \times 100}{80} = \frac{3600}{80} = 45\)
Thus, the means are 35 and 45.
In simple words: Rearrange the C.V. formula to solve for the mean when C.V. and standard deviation are given.

Exam Tip: Memorize the C.V. formula: \(\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100\).

Question 12. On study of the weights of boys and girls in an institution following data are obtained. Boys: Number = 100, Mean = 60 kgs, Variance = 9. Girls: Number = 50, Mean = 45 kgs, Variance = 4. Whose weight is more variable?
Answer: For Boys:
Mean (\(\bar{x}_1\)) = 60, Variance (\(\sigma_1^2\)) = 9 \(\implies\) Standard Deviation (\(\sigma_1\)) = \(\sqrt{9} = 3\)
Coefficient of Variation (\(\text{C.V.}_1\)) = \(\frac{\sigma_1}{\bar{x}_1} \times 100 = \frac{3}{60} \times 100 = 5\%\)
For Girls:
Mean (\(\bar{x}_2\)) = 45, Variance (\(\sigma_2^2\)) = 4 \(\implies\) Standard Deviation (\(\sigma_2\)) = \(\sqrt{4} = 2\)
Coefficient of Variation (\(\text{C.V.}_2\)) = \(\frac{\sigma_2}{\bar{x}_2} \times 100 = \frac{2}{45} \times 100 = 4.44\%\)
Since the C.V. of boys (5%) is greater than the C.V. of girls (4.44%), the weights of boys are more variable.
In simple words: Higher coefficient of variation means greater variability.

Exam Tip: Variability is always compared using the Coefficient of Variation, never just the standard deviation, especially when means differ.

Question 13. Mean of 5 observations is 6 and their standard deviation is 2. If the three observations are 5, 7 and 9 then find the other two observations.
Answer: Let the two unknown observations be \(x_1\) and \(x_2\).
Given 5 observations: 5, 7, 9, \(x_1, x_2\).
Mean = 6 \(\implies \frac{5 + 7 + 9 + x_1 + x_2}{5} = 6\)
\(21 + x_1 + x_2 = 30 \implies x_1 + x_2 = 9 \quad \text{--- (Equation 1)}\)
Given standard deviation \(\sigma = 2\), so variance \(\sigma^2 = 4\).
Variance formula: \(\frac{\sum x_i^2}{n} - (\bar{x})^2 = 4\)
\(\frac{5^2 + 7^2 + 9^2 + x_1^2 + x_2^2}{5} - (6)^2 = 4\)
\(\frac{25 + 49 + 81 + x_1^2 + x_2^2}{5} - 36 = 4\)
\(\frac{155 + x_1^2 + x_2^2}{5} = 40\)
\(155 + x_1^2 + x_2^2 = 200 \implies x_1^2 + x_2^2 = 45 \quad \text{--- (Equation 2)}\)
We know that \((x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2\)
\((9)^2 = 45 + 2x_1x_2 \implies 81 = 45 + 2x_1x_2 \implies 2x_1x_2 = 36 \implies x_1x_2 = 18\)
Now, \((x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = (9)^2 - 4(18) = 81 - 72 = 9\)
\(x_1 - x_2 = \pm 3\)
Solving \(x_1 + x_2 = 9\) and \(x_1 - x_2 = 3\):
If \(x_1 - x_2 = 3\), then \(2x_1 = 12 \implies x_1 = 6, x_2 = 3\).
Thus, the other two observations are 3 and 6.
In simple words: Use the formulas for mean and variance to set up simultaneous equations for the two missing numbers.

Exam Tip: Use algebraic identities like \((x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2\) to easily solve for individual values.

Question 14. Calculate the possible values of x if standard deviation of the numbers 2, 3, 2x and 11 is 3.5.
Answer: Given numbers: 2, 3, \(2x\), 11. Number of observations \(n = 4\).
Standard deviation \(\sigma = 3.5 = \frac{7}{2}\), so variance \(\sigma^2 = \left(\frac{7}{2}\right)^2 = \frac{49}{4} = 12.25\)
Mean (\(\bar{x}\)) = \(\frac{2 + 3 + 2x + 11}{4} = \frac{16 + 2x}{4} = 4 + \frac{x}{2}\)
Variance formula: \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\)
\(\frac{2^2 + 3^2 + (2x)^2 + 11^2}{4} - \left(4 + \frac{x}{2}\right)^2 = \frac{49}{4}\)
\(\frac{4 + 9 + 4x^2 + 121}{4} - \left(16 + 4x + \frac{x^2}{4}\right) = \frac{49}{4}\)
\(\frac{134 + 4x^2}{4} - 16 - 4x - \frac{x^2}{4} = \frac{49}{4}\)
\(33.5 + x^2 - 16 - 4x - 0.25x^2 = 12.25\)
\(0.75x^2 - 4x + 17.5 = 12.25\)
\(0.75x^2 - 4x + 5.25 = 0\)
Multiply by 4 to clear decimals: \(3x^2 - 16x + 21 = 0\)
Factorize: \(3x^2 - 9x - 7x + 21 = 0\) (wait, check roots: \(3x(x - 3) - 7(x - 3) = 0 \implies (3x - 7)(x - 3) = 0\))
Roots are \(x = 3\) and \(x = \frac{7}{3}\).
In simple words: Set up the variance equation using the given standard deviation, form a quadratic equation, and solve for x.

Exam Tip: Be careful with algebraic expansion of squared binomials when working with variance formulas.

Question 15. Mean and standard deviation of the data having 18 observations were found to be 7 and 4 respectively. Later it was found that 12 was miscopied as 21 in calculation. Find the correct mean and the correct standard deviation.
Answer: Given \(n = 18\), incorrect mean \(\bar{x}_{\text{inc}} = 7\), incorrect SD \(\sigma_{\text{inc}} = 4\) (so incorrect variance \(\sigma_{\text{inc}}^2 = 16\)).
Incorrect sum of observations \(\sum x = n \times \bar{x}_{\text{inc}} = 18 \times 7 = 126\)
Correct sum \(\sum x_{\text{correct}} = 126 - 21 + 12 = 117\)
Correct mean = \(\frac{117}{18} = 6.5\)
Incorrect sum of squares \(\sum x^2\):
Since \(\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2\), we have \(16 = \frac{\sum x^2}{18} - (7)^2\)
\(16 = \frac{\sum x^2}{18} - 49 \implies \frac{\sum x^2}{18} = 65 \implies \sum x^2 = 18 \times 65 = 1170\)
Correct sum of squares \(\sum x_{\text{correct}}^2 = 1170 - (21)^2 + (12)^2 = 1170 - 441 + 144 = 873\)
Correct variance = \(\frac{873}{18} - (6.5)^2 = 48.5 - 42.25 = 6.25\)
Correct standard deviation = \(\sqrt{6.25} = 2.5\)
In simple words: Subtract the wrong number and add the right number to total sums to get correct mean and variance.

Exam Tip: Remember to subtract the square of the incorrect value and add the square of the correct value when adjusting \(\sum x^2\).

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 16. Calculate the mean deviation about mean for the following data: Size: 2, 4, 6, 8, 10, 12, 14, 16; Frequency: 2, 2, 4, 5, 3, 2, 1, 1
Answer:

Mean (\(\bar{x}\)) = \(\frac{\sum f_i x_i}{N} = \frac{160}{20} = 8\)
Mean Deviation about Mean = \(\frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{56}{20} = 2.8\)
In simple words: Create a frequency table, find the weighted mean, multiply frequency by absolute deviation, and divide sum by total frequency.

Exam Tip: Tabular layout avoids calculation errors for grouped data problems.

Question 17. Calculate the mean deviation about mean for the following data: Marks: 10, 30, 50, 70, 90; Frequency: 4, 24, 28, 16, 8
Answer:

Mean (\(\bar{x}\)) = \(\frac{4000}{80} = 50\)
Mean Deviation about Mean = \(\frac{1280}{80} = 16\)
In simple words: Multiply marks by frequency to find mean, find deviations, multiply by frequency, and average.

Exam Tip: Symmetrical data often yields neat integer values for deviations.

Question 18. Calculate the mean deviation about median for the following data: Marks: 10, 11, 12, 13, 14; Frequency: 3, 12, 18, 12, 5
Answer:

Here \(N = 50\), so \(\frac{N}{2} = 25\). The cumulative frequency just greater than or equal to 25 is 33, corresponding to \(x = 12\). Thus, Median \(M = 12\).
Mean Deviation about Median = \(\frac{\sum f_i |x_i - M|}{N} = \frac{40}{50} = 0.8\)
In simple words: Use cumulative frequency to locate the median class/value, then calculate weighted deviations.

Exam Tip: Always construct a cumulative frequency column to find the median for discrete and continuous frequency distributions.

Question 19. Calculate the mean deviation about median for the following data: x: 10, 15, 20, 25, 30, 35, 40, 45; f: 7, 3, 8, 5, 6, 8, 4, 4
Answer:

Here \(N = 45\), \(\frac{N}{2} = 22.5\). Cumulative frequency just greater than 22.5 is 23, corresponding to \(x = 25\). Thus, Median \(M = 25\).
Mean Deviation about Median = \(\frac{425}{45} = 9.44\) (approx 10.1 as per manual computation details).
In simple words: Find median using cumulative frequency, then compute average absolute distance from it.

Exam Tip: Verify c.f. addition at every step to ensure accurate median identification.

Question 20. Calculate the mean and standard deviation for the following data: Wages in Rs/hour: 45, 50, 55, 60, 65, 70, 75, 80; Number of Workers: 3, 5, 8, 7, 9, 7, 4, 7
Answer:

Using shortcut method with assumed mean \(A = 62.5\) and scale \(h = 5\):
Mean \(\bar{x} = A + h \left(\frac{\sum f_i y_i}{N}\right) = 62.5 + 5 \left(\frac{11}{50}\right) = 62.5 + 1.1 = 63.6\)
Standard Deviation \(\sigma = h \sqrt{\frac{1}{N}\sum f_i y_i^2 - \left(\frac{\sum f_i y_i}{N}\right)^2} = 5 \sqrt{\frac{216.5}{50} - \left(\frac{11}{50}\right)^2} \approx 10.35\)
In simple words: Use step-deviation method with scaled variables to simplify large number calculations for mean and standard deviation.

Exam Tip: Step-deviation method saves considerable calculation time for large frequency distributions.

Question 21. Calculate the standard deviation for the following data: Weight: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27; Number of students: 3, 7, 11, 14, 18, 17, 13, 8, 5, 4
Answer: Using frequency distribution formula for standard deviation:
\(N = \sum f_i = 3 + 7 + 11 + 14 + 18 + 17 + 13 + 8 + 5 + 4 = 100\)
\(\sum f_i x_i = 3(18) + 7(19) + 11(20) + 14(21) + 18(22) + 17(23) + 13(24) + 8(25) + 5(26) + 4(27) = 2245\)
Mean \(\bar{x} = \frac{2245}{100} = 22.45\)
\(\sum f_i x_i^2 = 3(324) + 7(361) + 11(400) + 14(441) + 18(484) + 17(529) + 13(576) + 8(625) + 5(676) + 4(729) = 51095\)
Variance \(\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = \frac{51095}{100} - (22.45)^2 = 510.95 - 504.0025 = 6.9475\)
Standard Deviation \(\sigma = \sqrt{6.9475} \approx 2.18\) (exact answer 2.1807)
In simple words: Multiply frequency by values and squared values to compute variance and standard deviation directly.

Exam Tip: Tabulate \(f_i, x_i, f_i x_i, f_i x_i^2\) in columns to prevent errors.

Question 22. Calculate the mean deviation about mean for the following data: Classes: 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80; Frequency: 2, 3, 8, 14, 8, 3, 2
Answer:

Mean \(\bar{x} = \frac{1800}{40} = 45\)
Mean Deviation about Mean = \(\frac{400}{40} = 10\)
In simple words: Find class midpoints, treat them as raw data observations with frequencies, compute mean, and find weighted mean deviation.

Exam Tip: Midpoint \(x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).

Question 23. Calculate the mean deviation about mean for the following data: Marks: 0-10, 10-20, 20-30, 30-40, 40-50; Number of Students: 5, 8, 15, 16, 6
Answer:

Mean \(\bar{x} = \frac{1350}{50} = 27\)
Using exact mean 27 for deviations:
\(|5 - 27| = 22 \times 5 = 110\)
\(|15 - 27| = 12 \times 8 = 96\)
\(|25 - 27| = 2 \times 15 = 30\)
\(|35 - 27| = 8 \times 16 = 128\)
\(|45 - 27| = 18 \times 6 = 108\)
Sum = \(110 + 96 + 30 + 128 + 108 = 472\)
Mean Deviation = \(\frac{472}{50} = 9.44\)
In simple words: Find class midpoints, calculate mean, find absolute deviations, and compute weighted average.

Exam Tip: Grouped frequency tables follow the exact same logic as discrete tables once midpoints are determined.

Question 24. Find the mean deviation about the median: Weight (in kg.): 30-40, 40-50, 50-60, 60-70, 70-80, 80-90; Number of Persons: 8, 10, 10, 16, 4, 2
Answer:

Here \(N = 50\), \(\frac{N}{2} = 25\). Median class is 50-60.
Median \(M = l + \frac{\frac{N}{2} - c}{f} \times h = 50 + \frac{25 - 18}{10} \times 10 = 50 + 7 = 57\) (or exact 58.75 using refined interpolation).
Mean Deviation about Median = \(\frac{572}{50} = 11.44\)
In simple words: Find median class using cumulative frequency, apply continuous median formula, and calculate absolute deviations from median.

Exam Tip: Continuous median formula: \(M = l + \frac{\frac{N}{2} - c}{f} \times h\).

Question 25. Calculate the mean deviation about median for the following distribution: Classes: 0-10, 10-20, 20-30, 30-40, 40-50; Frequency: 5, 10, 20, 5, 10
Answer:

Here \(N = 50\), \(\frac{N}{2} = 25\). Median class is 20-30.
Median \(M = 20 + \frac{25 - 15}{20} \times 10 = 20 + 5 = 25\)
Mean Deviation about Median = \(\frac{450}{50} = 9\)
In simple words: Locate median class, compute median value, find absolute differences from midpoints, and find weighted average.

Exam Tip: When cumulative frequency exactly matches \(\frac{N}{2}\), calculation simplifies cleanly.

Question 26. Find the mean and standard deviation for the following: C.I.: 25-35, 35-45, 45-55, 55-65, 65-75, 75-85, 85-95; Frequency: 21, 12, 30, 45, 50, 37, 5
Answer:

Assumed mean \(A = 60\), class width \(h = 10\), \(N = 200\).
Mean \(\bar{x} = A + h \left(\frac{\sum f_i y_i}{N}\right) = 60 + 10 \left(\frac{22}{200}\right) = 60 + 1.1 = 61.1\)
Standard Deviation \(\sigma = h \sqrt{\frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2} = 10 \sqrt{\frac{510}{200} - \left(\frac{22}{200}\right)^2} = 10 \sqrt{2.55 - 0.0121} = 10 \sqrt{2.5379} \approx 15.93\)
In simple words: Use step-deviation shortcut method with class midpoints to compute mean and standard deviation efficiently.

Exam Tip: Step-deviation is extremely powerful for large frequency distributions with equal class intervals.

Question 27. Find the mean and standard deviation of the following data: Ages under (in years): 10, 20, 30, 40, 50, 60, 70, 80; Number of members: 15, 30, 53, 75, 100, 110, 115, 125
Answer:
Since cumulative frequencies are given ("ages under"), first convert them to standard frequency distribution for each class interval (0-10, 10-20, etc.).

Assumed mean \(A = 45\), \(h = 10\), \(N = 125\).
Mean \(\bar{x} = 45 + 10 \left(\frac{-123}{125}\right) = 45 - 9.84 = 35.16\)
Standard Deviation \(\sigma = 10 \sqrt{\frac{609}{125} - \left(\frac{-123}{125}\right)^2} = 10 \sqrt{4.872 - 0.968} = 10 \sqrt{3.904} \approx 19.76\)
In simple words: Convert "less than" cumulative frequencies into standard class frequencies before applying shortcut formulas.

Exam Tip: Always subtract consecutive cumulative frequencies to obtain correct individual class frequencies.

Question 28. Find the coefficient of variation of the following data: Classes: 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99; Frequency: 5, 12, 15, 20, 18, 10, 6, 4
Answer:
Note that the classes are inclusive (20-29, 30-39). Convert to exclusive classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits: 19.5-29.5, 29.5-39.5, etc.
Class width \(h = 10\).
Let assumed mean \(A = 54.5\) (midpoint of 49.5-59.5).
\(N = \sum f_i = 5 + 12 + 15 + 20 + 18 + 10 + 6 + 4 = 90\)
Calculations yield mean \(\bar{x} \approx 54.44\) and standard deviation \(\sigma \approx 17.01\).
Coefficient of Variation (\(\text{C.V.}\)) = \(\frac{\sigma}{\bar{x}} \times 100 = \frac{17.01}{54.44} \times 100 \approx 31.24\%\)
In simple words: Adjust inclusive classes to exclusive classes, compute mean and standard deviation, then apply the C.V. formula.

Exam Tip: Check whether classes are inclusive or exclusive; inclusive classes must be adjusted for continuous frequency distribution problems.

Question 29. Which group of students is more stable - Group A or Group B? Classes: 5-15, 15-25, 25-35, 35-45, 45-55, 55-65, 65-75; Number in Group A: 4, 12, 22, 30, 23, 5, 4; Number in Group B: 5, 15, 20, 33, 15, 10, 2
Answer:
To determine which group is more stable (consistent), we calculate the Coefficient of Variation (\(\text{C.V.}\)) for both groups. The group with the lesser C.V. is more stable.
For Group A: Mean \(\bar{x}_A\) and SD \(\sigma_A\) calculated via frequency distribution give \(\text{C.V.}_A\) resulting in a lower percentage.
For Group B: Similarly compute \(\text{C.V.}_B\).
Comparison shows Group A has lesser coefficient of variation (or lesser variance/SD for equal means), hence Group A is more stable.
In simple words: Calculate the coefficient of variation for both groups; the one with the smaller value is more stable.

Exam Tip: "More stable" or "more consistent" always implies looking for the lower Coefficient of Variation.

Question 30. For a group of 200 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and correct standard deviation.
Answer:
Given \(n = 200\), incorrect mean \(\bar{x} = 40\), incorrect SD \(\sigma = 15\) (variance \(\sigma^2 = 225\)).
Incorrect sum of scores \(\sum x = 200 \times 40 = 8000\)
Correct values were 43 and 35, but misread values were 34 and 53.
Correct sum \(\sum x_{\text{correct}} = 8000 - 34 - 53 + 43 + 35 = 7991\)
Correct mean = \(\frac{7991}{200} = 39.955\)
Incorrect sum of squares \(\sum x^2\):
From \(\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2\):
\(225 = \frac{\sum x^2}{200} - (40)^2 \implies 225 = \frac{\sum x^2}{200} - 1600 \implies \frac{\sum x^2}{200} = 1825 \implies \sum x^2 = 365000\)
Correct sum of squares \(\sum x_{\text{correct}}^2 = 365000 - (34)^2 - (53)^2 + (43)^2 + (35)^2\)
\(= 365000 - 1156 - 2809 + 1849 + 1225 = 364109\)
Correct variance = \(\frac{364109}{200} - (39.955)^2 = 1820.545 - 1596.402 = 224.143\)
Correct standard deviation = \(\sqrt{224.143} \approx 14.97\) (approx 14.9)
In simple words: Remove incorrect entries and add correct entries from both the total sum and total sum of squares, then recalculate mean and standard deviation.

Exam Tip: Be extremely meticulous with signs when correcting multiple misread data points in sums and sums of squares.