Maharashtra Board Class 11 Maths Part 2 Chapter 2 Measures of Dispersion 2.3 Solutions

Question 1. The mean and standard deviation of two distributions of 100 and 150 items are 50, 5, and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Answer:
Here, \( n_1 = 100, n_2=150, \bar{x}_1=50, \bar{x}_2 = 40, \sigma_1 = 5, \sigma_2 = 6 \)
Combined mean is given by,
\( \bar{x}_c = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} \)
\( = \frac{(100 \times 50)+(150 \times 40)}{100+150} \)
\( = \frac{5000+6000}{250} \)
\( = \frac{11000}{250} = 44 \)
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 \( d_1 = \bar{x}_1-\bar{x}_c, d_2 = \bar{x}_2-\bar{x}_c \)
\( d_1 = 50-44 = 6 \)
and \( d_2 = 40-44 = -4 \)
\[ \sigma_c = \sqrt{\frac{100(5^2+6^2)+150[6^2+(-4)^2]}{100+150}} \] \[ \sigma_c = \sqrt{\frac{100(25+36)+150(36+16)}{250}} \] \[ \sigma_c = \sqrt{\frac{(100 \times 61)+(150 \times 52)}{250}} \] \[ \sigma_c = \sqrt{\frac{6100+7800}{250}} \] \[ \sigma_c = \sqrt{\frac{13900}{250}} \] \[ \sigma_c = \sqrt{55.6} \]
The mean and standard deviation of all 250 items taken together are 44 and \( \sqrt{55.6} \) respectively.
In simple words: This problem involves calculating the combined mean and standard deviation for two separate datasets. We first find the combined mean using a weighted average formula, then calculate the individual deviations from this combined mean to use in the combined standard deviation formula.

🎯 Exam Tip: Remember to correctly apply the formulas for combined mean and combined standard deviation, paying close attention to the deviations from the combined mean (\(d_1\) and \(d_2\)) as these are crucial for accuracy.

Question 2. For certain bivariate data, the following information is available.

Obtain the combined standard deviation.
Answer:
Here, \( \bar{x} = 13, \bar{y} = 17, \sigma_x = 3, \sigma_y = 2, n_x = 10, n_y = 10 \)
Combined mean is given by \( \bar{x}_c = \frac{n_x\bar{x}+n_y\bar{y}}{n_x+n_y} \)
\( = \frac{10(13)+10(17)}{10+10} \)
\( = \frac{130+170}{20} \)
\( = \frac{300}{20} = 15 \)
Combined standard deviation is given by,
\[ \sigma_c = \sqrt{\frac{n_x(\sigma_x^2+d_1^2)+n_y(\sigma_y^2+d_2^2)}{n_x+n_y}} \] where \( d_1 = \bar{x}-\bar{x}_c, d_2 = \bar{y}-\bar{x}_c \)
\( d_1 = 13-15 = -2 \)
and \( d_2 = 17-15 = 2 \)
\[ \sigma_c = \sqrt{\frac{10[3^2+(-2)^2]+10(2^2+2^2)}{10+10}} \] \[ \sigma_c = \sqrt{\frac{10(9+4)+10(4+4)}{20}} \] \[ \sigma_c = \sqrt{\frac{10(13)+10(8)}{20}} \] \[ \sigma_c = \sqrt{\frac{130+80}{20}} \] \[ \sigma_c = \sqrt{\frac{210}{20}} \] \[ \sigma_c = \sqrt{10.5} \]
In simple words: To find the combined standard deviation, we first calculate the combined mean for the two variables (X and Y). Then, we use this combined mean to find the deviations for each variable and apply the formula for combined standard deviation.

🎯 Exam Tip: When dealing with bivariate data for combined standard deviation, carefully distinguish between individual means and standard deviations of X and Y, and remember that deviations \(d_1\) and \(d_2\) are calculated from the *combined* mean.

Question 3. Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are: 2, 4, 6, 8, 10. (Given: \( \sqrt{8} = 2.8284 \))
Answer:

Here, \( n = 5 \),
\( \bar{x} = \frac{\Sigma x_i}{n} = \frac{30}{5} = 6 \)
\( Var(X) = \sigma_x^2 = \frac{1}{n}\Sigma (x_i-\bar{x})^2 = \frac{40}{5} = 8 \)
\( S.D. = \sigma_x = \sqrt{Var(X)} = \sqrt{8} = 2.8284 \)
Now, \( C.V. = 100 \times \frac{\sigma_x}{\bar{x}} = 100 \times \frac{2.8284}{6} = 47.14\% \)
In simple words: The coefficient of variation (C.V.) measures the relative variability of data. To calculate it, we first find the mean and standard deviation of the given marks, then divide the standard deviation by the mean and multiply by 100.

🎯 Exam Tip: Remember the formula for Coefficient of Variation (\(C.V. = \frac{\sigma}{\bar{x}} \times 100\)). Ensure you correctly calculate the mean (\(\bar{x}\)) and standard deviation (\(\sigma\)) before applying the formula.

Question 4. Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Answer:
Given, \( \bar{x} = 25, \sigma = 5 \)
\( C.V. = 100 \times \frac{\sigma}{\bar{x}} \)
\( = 100 \times \frac{5}{25} \)
\( = 20\% \)
In simple words: The coefficient of variation is a percentage that shows how much variation there is relative to the mean. It's calculated by dividing the standard deviation by the mean and multiplying by 100.

🎯 Exam Tip: This is a direct application of the C.V. formula. Ensure you correctly substitute the given mean and standard deviation values.

Question 5. A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variation of 2.5%. What is the standard deviation of their height?
Answer:
Given, \( n = 65, \bar{x} = 150.4, C.V. = 2.5\% \)
\( C.V. = 100 \times \frac{\sigma}{\bar{x}} \)
\( \therefore 2.5 = 100 \times \frac{\sigma}{150.4} \)
\( \therefore \sigma = \frac{2.5 \times 150.4}{100} \)
\( \therefore \sigma = 3.76 \)
The standard deviation of students' height is 3.76.
In simple words: If you know the average (mean) and the coefficient of variation, you can work backward to find the standard deviation. This measures the typical spread of heights around the average height for the students.

🎯 Exam Tip: When given C.V. and mean, rearrange the C.V. formula to solve for standard deviation (\(\sigma = \frac{C.V. \times \bar{x}}{100}\)). Be careful with decimal places in calculations.

Question 6. Two workers on the same job show the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Answer:
Here, \( \bar{p}=33, \bar{q} = 21, \sigma_p = 9, \sigma_q = 7 \)
\( C.V. (P) = 100 \times \frac{\sigma_p}{\bar{p}} = 100 \times \frac{9}{33} = 27.27\% \)
\( C.V. (Q) = 100 \times \frac{\sigma_q}{\bar{q}} = 100 \times \frac{7}{21} = 33.33\% \)
(i) Since, \( C.V. (P) < C.V.(Q) \)
Worker P is more consistent regarding the time required to complete the job.
(ii) Since, \( \bar{p} > \bar{q} \)
i.e., the expected time for completing the job is less for worker Q.
Worker Q seems to be faster in completing the job.
In simple words: We use the Coefficient of Variation to compare consistency; a lower C.V. means more consistency. To determine who is faster, we simply compare their average completion times, with a lower average indicating faster work.

🎯 Exam Tip: For consistency comparisons, always use the Coefficient of Variation (C.V.); the lower C.V. indicates higher consistency. For comparing speed or efficiency, directly compare the means (average values).

Question 7. A company has two departments with 42 and 60 employees respectively. Their average weekly wages are Rs. 750 and Rs. 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Answer:
Let \( n_1 = 42, n_2 = 60, \bar{x}_1=750, \bar{x}_2 = 400, \sigma_1 = 8, \sigma_2 = 10 \)
\( C.V. (1) = 100 \times \frac{\sigma_1}{\bar{x}_1} = 100 \times \frac{8}{750} = 1.07\% \)
\( C.V. (2) = 100 \times \frac{\sigma_2}{\bar{x}_2} = 100 \times \frac{10}{400} = 2.5\% \)
(i) Since, \( \bar{x}_1 > \bar{x}_2 \)
i.e., average weekly wages are more for the first department.
The first department has a larger bill.
(ii) Since, \( C.V. (1) < C.V. (2) \)
The second department is less consistent.
The second department has larger variability in wages.
In simple words: To find which department has a larger total wage bill, we compare their average wages (mean). To assess variability, we compare their Coefficients of Variation (C.V.); a higher C.V. indicates greater variability in wages.

🎯 Exam Tip: For comparing the "bill" or total expenditure, focus on the average wages (mean). For comparing "variability" or "consistency", always calculate and compare the Coefficient of Variation (C.V.) for each department.

Question 8. The following table gives the weights of the students of class A. Calculate the coefficient of variation (Given \( \sqrt{8} = 0.8944 \))

Answer:

\( \bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{800}{20} = 40 \)
\( Var(X) = \sigma_x^2 = \frac{\Sigma f_i x_i^2}{N} - (\bar{x})^2 \)
\( = \frac{33600}{20} - (40)^2 \)
\( = 1680 - 1600 \)
\( = 80 \)
\( S.D. = \sigma_x = \sqrt{80} = \sqrt{100 \times 8} \)
\( = \sqrt{100} \times \sqrt{8} = 10 \times 0.8944 = 8.944 \)
\( C.V.(X) = 100 \times \frac{\sigma_x}{\bar{x}} = 100 \times \frac{8.944}{40} = 22.36\% \)
In simple words: To find the Coefficient of Variation for grouped data, first calculate the mean (\(\bar{x}\)) and standard deviation (\(\sigma\)) using the mid-values of the classes. Then, apply the standard C.V. formula.

🎯 Exam Tip: For grouped data, remember to use the mid-point of each class interval for \(x_i\) and the formulas for calculating mean and variance that incorporate frequencies (\(f_i\)).

Question 9. Compute coefficient of variation for team A and team B. (Given: \( \sqrt{2.5162} = 1.5863, \sqrt{2.244} = 1.4980 \)) Which team is more consistent?
Answer:

Let \( f_1 \) denote no. of goals of team A and \( f_2 \) denote no. of goals of team B.

For Team A:
\( \bar{x}_1 = \frac{\Sigma f_1x_i}{N_1} = \frac{121}{60} = 2.0167 \)
\( \sigma_1^2 = \frac{\Sigma f_1x_i^2}{N_1} - (\bar{x}_1)^2 \)
\( = \frac{395}{60} - (2.0167)^2 \)
\( = 6.5833 - 4.0671 = 2.5162 \)
\( \sigma_1 = \sqrt{2.5162} = 1.5863 \)
\( C.V. \text{ of team A} = C.V.(x_1) = 100 \times \frac{\sigma_1}{\bar{x}_1} = 100 \times \frac{1.5863}{2.0167} = 78.66\% \)
For Team B:
\( \bar{x}_2 = \frac{\Sigma f_2x_i}{N_2} = \frac{148}{70} = 2.1143 \)
\( \sigma_2^2 = \frac{\Sigma f_2x_i^2}{N_2} - (\bar{x}_2)^2 \)
\( = \frac{470}{70} - (2.1143)^2 \)
\( = 6.7143 - 4.4703 = 2.244 \)
\( \sigma_2 = \sqrt{2.244} = 1.4980 \)
\( C.V. \text{ of team B} = C.V.(x_2) = 100 \times \frac{\sigma_2}{\bar{x}_2} = 100 \times \frac{1.4980}{2.1143} = 70.85\% \)
Since C.V. of team A > C.V. of team B.
Team B is more consistent.
In simple words: To compare consistency between two teams' performances (goals), we calculate the Coefficient of Variation for each. The team with the lower C.V. is considered more consistent because their scores vary less relative to their average performance.

🎯 Exam Tip: When comparing consistency between two datasets, always compute the Coefficient of Variation (C.V.) for each. A smaller C.V. indicates greater consistency, as it represents less relative variability.

Question 10. Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Answer:
Here, \( \bar{x}_m = 20, \bar{x}_s = 25, \sigma_m = 2, \sigma_s = 3 \)
\( C.V. (M) = 100 \times \frac{\sigma_m}{\bar{x}_m} \)
\( = 100 \times \frac{2}{20} = 10\% \)
\( C.V. (S) = 100 \times \frac{\sigma_s}{\bar{x}_s} = 100 \times \frac{3}{25} = 12\% \)
Since C.V. (S) > C.V. (M)
The subject statistics show higher variability in marks.
In simple words: To determine which subject has more variable marks, we calculate the Coefficient of Variation for both Mathematics and Statistics. The subject with the higher C.V. indicates greater variability in its scores.

🎯 Exam Tip: Variability is best compared using the Coefficient of Variation (C.V.), especially when means are different. The subject with the higher C.V. exhibits greater relative variability in its marks.