Maharashtra Board Class 11 Maths Part 2 Chapter 2 Measures of Dispersion Miscellaneous Solutions

Question 1. Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Answer: Solution:
Here, largest value (L) = 203, smallest value (S) = 114
\(\therefore\) Range = L - S
= 203 - 114
= 89
In simple words: The range is the difference between the highest and lowest values in a dataset, providing a simple measure of data spread. Here, subtracting the smallest value (114) from the largest (203) gives a range of 89.

🎯 Exam Tip: Always identify the maximum and minimum values accurately to calculate the range correctly, as it's a basic measure of dispersion.

Question 2. Given below the frequency distribution of weekly wages of 400 workers. Find the range.
Answer: Solution:

Here, largest value (L) = 40, smallest value (S) = 10
\(\therefore\) Range = L - S
= 40 - 10
= 30
In simple words: For a frequency distribution, the range is found by taking the difference between the largest and smallest observed values. In this case, the highest wage is 40 and the lowest is 10, resulting in a range of 30.

🎯 Exam Tip: When given a frequency distribution for discrete data, the largest and smallest values are directly observable from the data categories, not the frequencies themselves.

Question 3. Find the range of the following data.
Answer: Solution:

Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115
\(\therefore\) Range = L - S
= 175 - 115
= 60
In simple words: For grouped data, the range is calculated by subtracting the lower limit of the first class from the upper limit of the last class. Here, it's \(175 - 115 = 60\).

🎯 Exam Tip: For continuous frequency distributions, the range is always determined by the extreme values of the class intervals, specifically the upper limit of the highest class and the lower limit of the lowest class.

Question 4. The city traffic police issued challans for not observing the traffic rules:
Answer: Solution:

Find Q.D.
The given data can be arranged in ascending order as follows:
24, 36, 40, 58, 62, 80
Here, n = 6
\[Q_1 = \text{value of } \left( \frac{n+1}{4} \right)^\text{th} \text{observation}\]
\[ = \text{value of } \left( \frac{6+1}{4} \right)^\text{th} \text{observation}\]
\[ = \text{value of } (1.75)^\text{th} \text{observation}\]
\[ = \text{value of } 1^\text{st} \text{observation} + 0.75(\text{value of } 2^\text{nd} \text{observation} - \text{value of } 1^\text{st} \text{observation}) \]
\[ = 24 + 0.75(36 - 24) \]
\[ = 24 + 0.75(12) \]
\[ = 24 + 9 \]
\(\therefore\) \(Q_1 = 33\)
\[Q_3 = \text{value of } 3 \left( \frac{n+1}{4} \right)^\text{th} \text{observation}\]
\[ = \text{value of } 3 \left( \frac{6+1}{4} \right)^\text{th} \text{observation}\]
\[ = \text{value of } (3 \times 1.75)^\text{th} \text{observation}\]
\[ = \text{value of } (5.25)^\text{th} \text{observation}\]
\[ = \text{value of } 5^\text{th} \text{observation} + 0.25(\text{value of } 6^\text{th} \text{observation} - \text{value of } 5^\text{th} \text{observation}) \]
\[ = 62 + 0.25(80 - 62) \]
\[ = 62 + 0.25(18) \]
\[ = 62 + 4.5 \]
\[ = 66.5 \]
\(\therefore\) Q.D. \( = \frac{Q_3 - Q_1}{2} = \frac{66.5 - 33}{2} = \frac{33.5}{2} = 16.75\)
In simple words: Quartile Deviation (Q.D.) measures the spread of the middle 50% of the data. First, the data is ordered, then the first quartile (Q1) and third quartile (Q3) are calculated. Q.D. is half the difference between Q3 and Q1.

🎯 Exam Tip: Remember to arrange data in ascending order before calculating quartiles. For fractional positions, interpolate between observations to find accurate Q1 and Q3 values.

Question 5. Calculate Q.D. from the following data.
Answer: Solution:

We construct the less than cumulative frequency table as follows:

Here, N = 35
\(Q_1 \text{ class} = \text{class containing } \left( \frac{N}{4} \right)^\text{th} \text{observation}\)
\(\therefore \frac{N}{4} = \frac{35}{4} = 8.75\)
Cumulative frequency which is just greater than (or equal to) 8.75 is 15.
\(\therefore Q_1 \text{ lies in the class 20-30.}\)
\(\therefore L = 20, c.f. = 8, f = 7, h = 10\)
\[Q_1 = L + \frac{h}{f} \left( \frac{N}{4} - c.f. \right) \]
\[ = 20 + \frac{10}{7} (8.75 - 8) \]
\[ = 20 + \frac{10}{7} \times 0.75 = 20 + 1.07 \]
\(\therefore Q_1 = 21.07\)
\[Q_3 \text{ class} = \text{class containing } \left( \frac{3N}{4} \right)^\text{th} \text{observation}\]
\(\therefore \frac{3N}{4} = \frac{3 \times 35}{4} = 26.25\)
Cumulative frequency which is just greater than (or equal to) 26.25 is 30.
\(\therefore Q_3 \text{ lies in the class 40-50.}\)
\(L = 40, c.f. = 20, f = 10, h = 10\)
\[Q_3 = L + \frac{h}{f} \left( \frac{3N}{4} - c.f. \right) \]
\[ = 40 + \frac{10}{10} (26.25 - 20) = 40 + 6.25 \]
\(\therefore Q_3 = 46.25\)
\[Q.D. = \frac{Q_3 - Q_1}{2} = \frac{46.25 - 21.07}{2} = \frac{25.18}{2} = 12.59 \]
In simple words: For grouped frequency data, we first construct a cumulative frequency table to find the classes containing Q1 and Q3. Then, specific formulas involving the class limits, frequencies, and cumulative frequencies are used to calculate the precise values of Q1 and Q3, and finally, the Quartile Deviation.

🎯 Exam Tip: Ensure correct identification of the Q1 and Q3 classes based on cumulative frequency. Pay close attention to the formula for quartiles in grouped data, especially `L`, `h`, `f`, and `c.f.` values for the respective classes.

Question 6. Calculate the appropriate measure of dispersion for the following data.
Answer: Solution:

Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.
We construct the less than cumulative frequency table as follows:

Here N = 250
\(Q_1 \text{ class class containing } \left( \frac{N}{4} \right)^\text{th} \text{observation}\)
\(\therefore \frac{N}{4} = \frac{250}{4} = 62.5\)
Cumulative frequency which is just greater than (or equal to) 62.5 is 65.
\(\therefore Q_1 \text{ lies in the class 35-40.}\)
\(\therefore L = 35, c.f. = 15, f = 50, h = 5\)
\[Q_1 = L + \frac{h}{f} \left( \frac{N}{4} - c.f. \right) \]
\[ = 35 + \frac{5}{50} (62.5 - 15) \]
\[ = 35 + \frac{1}{10} \times 47.5 \]
\[ = 35 + 4.75 \]
\(\therefore Q_1 = 39.75\)
\[Q_3 \text{ class} = \text{class containing } \left( \frac{3N}{4} \right)^\text{th} \text{observation}\]
\(\therefore \frac{3N}{4} = \frac{3 \times 250}{4} = 187.5\)
The cumulative frequency which is just greater than (or equal to) 187.5 is 190.
\(\therefore Q_3 \text{ lies in the class 45-50.}\)
\(\therefore L = 45, c.f. = 150, f = 40, h = 5\)
\[Q_3 = L + \frac{h}{f} \left( \frac{3N}{4} - c.f. \right) \]
\[ = 45 + \frac{5}{40} (187.5 - 150) \]
\[ = 45 + \frac{1}{8} \times 37.5 \]
\[ = 45 + 4.6875 \]
\(\therefore Q_3 = 49.6875\)
\[Q.D. = \frac{Q_3 - Q_1}{2} = \frac{49.6875 - 39.75}{2} = \frac{9.9375}{2} \]
\(\therefore Q.D. = 4.9688\)
In simple words: When data contains open-ended classes, Quartile Deviation (Q.D.) is the preferred measure of dispersion because it only depends on the values of the first and third quartiles, which can be calculated without knowing the exact boundaries of the open-ended classes. It involves building a cumulative frequency table, finding Q1 and Q3 using interpolation, and then applying the Q.D. formula.

🎯 Exam Tip: For datasets with open-ended classes (e.g., "Less than X" or "Greater than Y"), Quartile Deviation is the most suitable measure of dispersion as it does not require knowledge of the exact limits of these extreme classes.

Question 7. Calculate Q.D. of the following data.
Answer: Solution:

We construct the less than cumulative frequency table as follows:

Here, N = 120
\(Q_1 \text{ class} = \text{class containing } \left( \frac{N}{4} \right)^\text{th} \text{observation}\)
\(\therefore \frac{N}{4} = \frac{120}{4} = 30\)
Cumulative frequency which is just greater than (or equal to) 30 is 35.
\(\therefore Q_1 \text{ lies in the class 4-6.}\)
\(\therefore L = 4, c.f. = 15, f = 20, h = 2\)
\[Q_1 = L + \frac{h}{f} \left( \frac{N}{4} - c.f. \right) \]
\[ = 4 + \frac{2}{20} (30 - 15) \]
\[ = 4 + \frac{1}{10} \times 15 \]
\[ = 4 + 1.5 \]
\[ = 5.5 \]
\[Q_3 \text{ class} = \text{class containing } \left( \frac{3N}{4} \right)^\text{th} \text{observation}\]
\(\therefore \frac{3N}{4} = \frac{3 \times 120}{4} = 90\)
Cumulative frequency which is just greater than (or equal to) 90 is 90.
\(\therefore Q_3 \text{ lies in the class 10-12.}\)
\(\therefore L = 10, c.f. = 72, f = 18, h = 2\)
\[Q_3 = L + \frac{h}{f} \left( \frac{3N}{4} - c.f. \right) \]
\[ = 10 + \frac{2}{18} (90 - 72) \]
\[ = 10 + \frac{2}{18} \times 18 \]
\[ = 10 + 2 \]
\(\therefore Q_3 = 12\)
\[Q.D. = \frac{Q_3 - Q_1}{2} = \frac{12 - 5.5}{2} = \frac{6.5}{2} = 3.25 \]
In simple words: To calculate Quartile Deviation for grouped data, you first create a cumulative frequency table. Then, identify the class intervals containing the first (Q1) and third (Q3) quartiles using their respective position formulas. Finally, apply the interpolation formula for grouped data to find Q1 and Q3, and then use the Q.D. formula.

🎯 Exam Tip: Pay careful attention to the class boundaries and corresponding cumulative frequencies when determining the quartile classes for grouped data. A small error in reading the table can lead to incorrect quartile values.

Question 8. Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 27, 25 (Given \(\sqrt{6.6} = 2.57\))
Answer: Solution:
We prepare the following table for the calculation of variance and S.D.:

Here, n = 10
\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{250}{10} = 25\]
\[\text{Var}(X) = \sigma_x^2 = \frac{1}{n}\Sigma(x_i - \bar{x})^2 = \frac{1}{10} \times 66 = 6.6\]
\[\text{S.D.} = \sigma_x = \sqrt{\text{Var}(X)} = \sqrt{6.6} = 2.57\]
In simple words: Variance measures the average squared difference from the mean, while Standard Deviation (S.D.) is the square root of variance, providing a measure of spread in the original units. To calculate, first find the mean, then the squared deviations from the mean for each value, sum them, divide by the count for variance, and finally take the square root for S.D.

🎯 Exam Tip: When calculating variance and standard deviation, accurately computing the mean (\(\bar{x}\)) is critical, as all subsequent calculations depend on it. Ensure the sum of squared deviations is divided by 'n' for population variance or 'n-1' for sample variance (here, we assume population for simplicity as 'n' is used).

Question 9. Following data gives no. of goals scored by a team in 90 matches.
Answer: Solution:

Compute the variance and standard deviation for the above data.
We prepare the following table for the calculation of variance and S.D:

\[\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{220}{90} = 2.44\]
\[\text{Var}(X) = \sigma_x^2 = \frac{\Sigma f_i x_i^2}{N} - (\bar{x})^2 = \frac{700}{90} - (2.44)^2 = 7.78 - 5.9536 = 1.83\]
\[\text{S.D.} = \sigma_x = \sqrt{\text{Var}(X)} = \sqrt{1.83}\]
In simple words: For a frequency distribution, variance and standard deviation are calculated by first finding the mean using \( \Sigma f_i x_i / N \). Then, compute the sum of \(f_i x_i^2\). Variance is found by subtracting the square of the mean from \( (\Sigma f_i x_i^2) / N \), and S.D. is the square root of this variance.

🎯 Exam Tip: When dealing with frequency distributions, ensure to correctly use \(f_i\) in the summation for both \(x_i\) and \(x_i^2\). Mistakes often occur in calculating \(f_i x_i^2\) or applying the formula for mean and variance.

Question 10. Compute the arithmetic mean and S.D. and C.V. (Given \(\sqrt{296} = 17.20\))
Answer: Solution:

We prepare the following table for calculation of arithmetic mean and S.D.:

\[\text{Arithmetic mean} = \bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{1950}{25} = 78\]
\[\text{Var}(X) = \sigma_x^2 = \frac{\Sigma f_i x_i^2}{N} - (\bar{x})^2 = \frac{159500}{25} - (78)^2 = 6380 - 6084 = 296\]
\[\text{S.D.} = \sigma_x = \sqrt{\text{Var}(X)} = \sqrt{296} = 17.20\]
\[\text{C.V.} = 100 \times \frac{\sigma_x}{\bar{x}} = 100 \times \frac{17.20}{78} = 22.05\%\]
In simple words: To find the arithmetic mean, standard deviation (S.D.), and coefficient of variation (C.V.) for grouped data, calculate midpoints (\(x_i\)) for each class. Then compute \(f_i x_i\) and \(f_i x_i^2\). The mean is \( \Sigma f_i x_i / N \), variance is \( (\Sigma f_i x_i^2 / N) - \bar{x}^2 \), S.D. is the square root of variance, and C.V. is \( (\text{S.D.} / \text{Mean}) \times 100 \).

🎯 Exam Tip: Always calculate the mid-value (\(x_i\)) correctly for each class interval in grouped data. The Coefficient of Variation (C.V.) is a relative measure of dispersion, useful for comparing variability between datasets with different means or units.

Question 11. The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.
Answer: Solution:
Here, n = 200, \(\bar{x}\) = Mean = 60, S.D. = 20
Wrongly taken items are 3 and 67.
Correct items are 13 and 17.
Now, \(\bar{x} = 60\)
\(\therefore \frac{1}{n} \Sigma_{i=1}^{n} x_i = 60\)
\(\therefore \frac{1}{200} \Sigma_{i=1}^{n} x_i = 60\)
\(\therefore \Sigma_{i=1}^{n} x_i = 200 \times 60\)
\(\therefore \Sigma_{i=1}^{n} x_i = 12000\)
Correct value of \( \Sigma_{i=1}^{n} x_i = \Sigma_{i=1}^{n} x_i \) (sum of wrongly taken items) + (sum of correct items)
= 12000 - (3 + 67) + (13 + 17)
= 12000 - 70 + 30
= 11960
Correct value of mean \( = \frac{1}{n} \times \) correct value of \( \Sigma_{i=1}^{n} x_i \)
\( = \frac{1}{200} \times 11960 \)
= 59.8
Now, S.D. = 20
Variance = (S.D.)\(^2\) = \(20^2\)
\(\therefore\) Variance = 400
\(\therefore \frac{1}{n} \Sigma_{i=1}^{n} x_i^2 - (\bar{x})^2 = 400\)
\(\therefore \frac{1}{200} \Sigma_{i=1}^{n} x_i^2 - (60)^2 = 400\)
\(\therefore \frac{1}{200} \Sigma_{i=1}^{n} x_i^2 = 400 + 3600\)
\(\therefore \Sigma_{i=1}^{n} x_i^2 = 4000 \times 200\)
\(\therefore \Sigma_{i=1}^{n} x_i^2 = 800000\)
Correct value of \( \Sigma_{i=1}^{n} x_i^2 = \Sigma_{i=1}^{n} x_i^2 \) - (Sum of squares of wrongly taken items) + (Sum of squares of correct items)
= 800000 - (\(3^2 + 67^2\)) + (\(13^2 + 17^2\))
= 800000 - (9 + 4489) + (169 + 289)
= 800000 - 4498 + 458
= 795960
\(\therefore\) Correct value of Variance \( = \left( \frac{1}{n} \times \Sigma_{i=1}^{n} x_i^2 \right) - (\text{correct value of } \bar{x})^2\)
\( = \frac{1}{200} \times 795960 - (59.8)^2\)
= 3979.8 - 3576.04
= 403.76
\(\therefore\) The correct mean is 59.8 and correct variance is 403.76.
In simple words: To correct mean and variance after wrong observations, first calculate the original sum of observations and sum of squares using the given incorrect mean and S.D. Then, subtract the wrongly recorded values (and their squares) and add the correct values (and their squares) to get the true sums. Finally, re-calculate the mean and variance using these corrected sums.

🎯 Exam Tip: When correcting mean and variance, be careful to subtract the *square* of the incorrect observations and add the *square* of the correct observations when adjusting \( \Sigma x_i^2 \). Also, remember to use the *corrected mean* for the final variance calculation.

Question 12. The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.
Answer: Solution:
n = 48, \(\bar{x}\) = 40, \(\sigma_x\) = 8 ...[Given]
\[\bar{x} = \frac{1}{n} \Sigma x_i \]
\(\therefore \Sigma x_i = n \bar{x} = 48 \times 40 = 1920\)
New \(\Sigma x_i = \Sigma x_i + 60 + 65 = 1920 + 60 + 65 = 2045\)
\[\text{New Mean} = \frac{2045}{50} = 40.9\]
Now, \(\sigma_x = 8\)
\(\therefore \sigma_x^2 = 64\)
\[\text{Since, } \sigma_x^2 = \frac{1}{n} (\Sigma x_i^2) - (\bar{x})^2 \]
\[64 = \frac{1}{48} (\Sigma x_i^2) - (40)^2 \]
\[64 = \frac{1}{48} (\Sigma x_i^2) - 1600 \]
\[\frac{1}{48} \Sigma x_i^2 = 64 + 1600 = 1664 \]
\(\therefore \Sigma x_i^2 = 48 \times 1664 = 79872\)
\[\text{New } \Sigma x_i^2 = \Sigma x_i^2 + (60)^2 + (65)^2 = 79872 + 3600 + 4225 = 87697 \]
\[\text{New S.D.} = \sqrt{\frac{\text{New } \Sigma x_i^2}{n} - (\text{New mean})^2} \]
\[ = \sqrt{\frac{87697}{50} - (40.9)^2 = \sqrt{1753.94 - 1672.81} = \sqrt{81.13}} \]
In simple words: To find the new mean and S.D. after adding observations, first calculate the original sum of observations (\(\Sigma x\)) and sum of squares (\(\Sigma x^2\)). Add the new observations to these sums to get the new \(\Sigma x\) and \(\Sigma x^2\), and update 'n'. Then, use these new values to compute the updated mean and standard deviation.

🎯 Exam Tip: Remember to update both the total number of observations (n) and the sum of observations (\(\Sigma x\)) and sum of squares (\(\Sigma x^2\)) when new data points are added. Use the new mean when calculating the new standard deviation.

Question 13. The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Answer: Solution:
Let \(n_1\) and \(n_2\) be the number of boys and girls respectively.
Let \(n = 200, \bar{X}_c = 65, \bar{X}_1 = 70, \bar{X}_2 = 62, \sigma_1 = 8, \sigma_2 = 10\)
Here, \(n_1 + n_2 = n\)
\(\therefore n_1 + n_2 = 200\) .......(i)
Combined mean is given by
\[\bar{X}_c = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2}\]
\[65 = \frac{n_1 (70) + n_2 (62)}{200}\]
\(\therefore 70n_1 + 62n_2 = 13000\)
\(\therefore 35n_1 + 31n_2 = 6500\) ........(ii)
Solving (i) and (ii), we get
\(n_1 = 75, n_2 = 125\)
Combined standard deviation is given by,
\[\sigma_c = \sqrt{\frac{n_1 (\sigma_1^2 + d_1^2) + n_2 (\sigma_2^2 + d_2^2)}{n_1 + n_2}}\]
where \(d_1 = \bar{X}_1 - \bar{X}_c, d_2 = \bar{X}_2 - \bar{X}_c\)
\(\therefore d_1 = 70 - 65 = 5\) and \(d_2 = 62 - 65 = -3\)
\[\sigma_c = \sqrt{\frac{75(64 + 25) + 125(100 + 9)}{200}}\]
\[ = \sqrt{\frac{6675 + 13625}{200}} \]
\[ = \sqrt{\frac{20300}{200}} \]
\[ = \sqrt{101.5} = 10.07 \]
In simple words: To find the number of boys and combined S.D., first use the combined mean formula and the total number of students to set up a system of equations, solving for \(n_1\) and \(n_2\). Then, calculate the deviations (\(d_1, d_2\)) of each group's mean from the combined mean. Finally, use the combined standard deviation formula, incorporating each group's individual S.D. and the calculated deviations.

🎯 Exam Tip: Accurately solve the system of linear equations to find \(n_1\) and \(n_2\). Remember that the deviation \(d_i\) is the difference between the group mean and the combined mean, and its square is always positive in the formula for combined S.D.

Question 14. From the following data available for 5 pairs of observations of two variables x and y, obtain the combined S.D. for all 10 observations,
where \( \Sigma_{i=1}^{n} x_i = 30, \Sigma_{i=1}^{n} y_i = 40, \Sigma_{i=1}^{n} x_i^2 = 225, \Sigma_{i=1}^{n} Y_i^2 = 340 \)
Answer: Solution:
Here, \( \Sigma_{i=1}^{n} x_i = 30, \Sigma_{i=1}^{n} y_i = 40, \Sigma_{i=1}^{n} x_i^2 = 225, \Sigma_{i=1}^{n} y_i^2 = 340, n_x = 5, n_y = 5\)
\[\bar{x} = \frac{\Sigma x_i}{n_x} = \frac{30}{5} = 6\]
\[\bar{y} = \frac{\Sigma y_i}{n_y} = \frac{40}{5} = 8\]
Combined mean is given by
\[\bar{x}_c = \frac{n_x \bar{x} + n_y \bar{y}}{n_x + n_y} = \frac{5(6) + 5(8)}{5+5} = \frac{30 + 40}{10} = \frac{70}{10} = 7\]
Combined standard deviation is given by,
\[\sigma_c = \sqrt{\frac{n_x (\sigma_x^2 + d_x^2) + n_y (\sigma_y^2 + d_y^2)}{n_x + n_y}}\]
where \(d_x = \bar{x} - \bar{x}_c, d_y = \bar{y} - \bar{x}_c\)
\[\sigma_x^2 = \frac{1}{n_x} \Sigma x_i^2 - (\bar{x})^2 = \frac{1}{5} (225) - (6)^2 = 45 - 36 = 9\]
\[\sigma_y^2 = \frac{1}{n_y} \Sigma y_i^2 - (\bar{y})^2 = \frac{1}{5} (340) - (8)^2 = 68 - 64 = 4\]
\(d_x = 6 - 7 = -1\) and \(d_y = 8 - 7 = 1\)
\[\therefore \sigma_c = \sqrt{\frac{5[9+(-1)^2] + 5[4+(1)^2]}{5+5}}\]
\[ = \sqrt{\frac{5(9+1) + 5(4+1)}{10}} \]
\[ = \sqrt{\frac{5(10) + 5(5)}{10}} \]
\[ = \sqrt{\frac{50 + 25}{10}} \]
\[ = \sqrt{\frac{75}{10}} = \sqrt{7.5} \]
In simple words: To find the combined standard deviation of two groups, calculate the mean (\(\bar{x}\), \(\bar{y}\)) and variance (\(\sigma_x^2\), \(\sigma_y^2\)) for each group first. Then determine the combined mean (\(\bar{x}_c\)) and the deviations (\(d_x\), \(d_y\)) of each group's mean from the combined mean. Finally, substitute these values into the combined standard deviation formula.

🎯 Exam Tip: Carefully calculate individual means and variances for each variable before combining. Pay close attention to the definition of \(d_x\) and \(d_y\), which are the differences between each group's mean and the overall combined mean.

Question 15. The mean and standard deviations of two brands of watches are given below:
Answer: Solution:

Calculate the coefficient of variation of the two brands and interpret the results.
Here, \(\bar{x}_I=36, \bar{X}_{II} = 48, \sigma_I = 8, \sigma_{II} = 10\)
\[\text{C.V. (I)} = 100 \times \frac{\sigma_I}{\bar{x}_I} = 100 \times \frac{8}{36} = 22.22\%\]
\[\text{C.V. (II)} = 100 \times \frac{\sigma_{II}}{\bar{X}_{II}} = 100 \times \frac{10}{48} = 20.83\%\]
Since C.V. (I) > C.V. (II)
\(\therefore\) the brand I is more variable.
In simple words: The Coefficient of Variation (C.V.) compares the relative variability of two datasets, especially when their means are different. It is calculated as (Standard Deviation / Mean) * 100. A higher C.V. indicates greater relative variability or inconsistency.

🎯 Exam Tip: The Coefficient of Variation (C.V.) is a critical measure for comparing the consistency or variability of two or more datasets. A smaller C.V. indicates greater consistency or uniformity in the data, while a larger C.V. implies higher variability.

Question 16. Calculate the coefficient of variation for the data given below. [Given \(\sqrt{3.3} = 1.8166\)]
Answer: Solution:

\[\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{4600}{100} = 46\]
\[\text{Var}(X) = \sigma_x^2 = \frac{1}{N} \Sigma f_i x_i^2 - (\bar{x})^2 = \frac{1}{100} \times 244600 - (46)^2 = 2446 - 2116 = 330\]
\[\text{S.D.} = \sigma_x = \sqrt{330} = \sqrt{3.3 \times 100} = 10\sqrt{3.3} = 10 \times 1.8166 = 18.166\]
\[\text{C.V.} = 100 \times \frac{\sigma_x}{\bar{x}} = 100 \times \frac{18.166}{46} = 39.49\%\]
In simple words: To calculate the Coefficient of Variation for grouped data, first determine the midpoints of each class interval. Then, compute the mean and standard deviation from these midpoints and their frequencies. Finally, divide the standard deviation by the mean and multiply by 100 to get the C.V., which indicates relative variability.

🎯 Exam Tip: When calculating C.V. for grouped data, ensure the calculation of mid-values for classes is correct, as all subsequent steps depend on them. Also, properly apply the grouped data formulas for mean and standard deviation before determining the C.V.

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