GSEB Class 11 Statistics Solutions Chapter 4 Measures of Dispersion Solution Exercise 4.6

Official GSEB Solutions for Class 11 Statistics: Chapter 04 Measures of Dispersion

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Chapter-wise Solutions for Statistics: Chapter 04 Measures of Dispersion

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Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures Of Dispersion Ex 4.6

Question 1. The information regarding marks of the students of two classes of a school is given below. Find the combined standard deviation of the marks obtained by the students.

Division ADivision B
No. of Students5060
Mean marks6048
Standard deviation1012

Answer:For Class A: \(n_1\) (Number of students) = 50 \(\bar{X}_1\) (Average marks) = 60 \(s_1\) (Standard deviation) = 10 For Class B: \(n_2\) (Number of students) = 60 \(\bar{X}_2\) (Average marks) = 48 \(s_2\) (Standard deviation) = 12 First, calculate the combined mean of the marks for students from both Class A and Class B:
\[ \bar{X}_c = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2} \]
\[ = \frac{(50 \times 60) + (60 \times 48)}{50 + 60} \]
\[ = \frac{3000 + 2880}{110} \]
\[ = \frac{5880}{110} \]
\[ = 53.45 \text{ marks} \] Next, compute the combined standard deviation of the marks for students from Class A and Class B:
Calculate the squared differences from the combined mean:
\(d_1 = \bar{X}_1 - \bar{X}_c = 60 - 53.45 = 6.55\)
\(d_1^2 = (6.55)^2 = 42.9025\)
\(d_2 = \bar{X}_2 - \bar{X}_c = 48 - 53.45 = -5.45\) (As per source, \(d_2\) is \(5.45\), but the calculation is \(48-53.45\). The square will be positive regardless.)
\(d_2^2 = (-5.45)^2 = 29.7025\)
Also, we need the variances:
\(s_1^2 = (10)^2 = 100\)
\(s_2^2 = (12)^2 = 144\)
Now, apply the formula for combined standard deviation:
\[ s_c = \sqrt{\frac{n_1 (s_1^2 + d_1^2) + n_2 (s_2^2 + d_2^2)}{n_1 + n_2}} \]
\[ = \sqrt{\frac{50 (100 + 42.9025) + 60 (144 + 29.7025)}{50 + 60}} \]
\[ = \sqrt{\frac{50 (142.9025) + 60 (173.7025)}{110}} \]
\[ = \sqrt{\frac{7145.125 + 10422.150}{110}} \]
\[ = \sqrt{\frac{17567.275}{110}} \]
\[ = \sqrt{159.7025} \]
\[ = 12.64 \text{ marks} \]In simple words: This problem involves combining data from two groups (classes) to find an overall measure of spread (standard deviation) and average (mean). We first calculate the combined average mark for all students, and then use a specific formula that accounts for each class's individual spread and how far its average is from the combined average to determine the combined standard deviation.

🎯 Exam Tip: Remember to calculate the combined mean accurately before proceeding with the combined standard deviation. Mistakes in the combined mean will propagate through the 'd' values and lead to an incorrect final standard deviation.

 

Question 2. The following information is available for two sections of a factory. Obtain the combined standard deviation of the production time.

Section ASection B
No. of workers1040
Average production time per unit (minute)2520
Variance1625

Answer:For Section A: \(n_1\) (Number of workers) = 10 \(\bar{X}_1\) (Average production time per unit) = 25 minutes \(s_1^2\) (Variance) = 16 For Section B: \(n_2\) (Number of workers) = 40 \(\bar{X}_2\) (Average production time per unit) = 20 minutes \(s_2^2\) (Variance) = 25 First, determine the combined mean of the production time per unit for both Section A and Section B:
\[ \bar{X}_c = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2} \]
\[ = \frac{(10 \times 25) + (40 \times 20)}{10 + 40} \]
\[ = \frac{250 + 800}{50} \]
\[ = \frac{1050}{50} \]
\[ = 21 \text{ minutes} \] Next, compute the combined standard deviation of the production time per unit for both sections:
Calculate the squared differences from the combined mean:
\(d_1 = \bar{X}_1 - \bar{X}_c = 25 - 21 = 4\)
\(d_1^2 = (4)^2 = 16\)
\(d_2 = \bar{X}_2 - \bar{X}_c = 20 - 21 = -1\)
\(d_2^2 = (-1)^2 = 1\)
Now, use the formula for combined standard deviation:
\[ s_c = \sqrt{\frac{n_1 (s_1^2 + d_1^2) + n_2 (s_2^2 + d_2^2)}{n_1 + n_2}} \]
\[ = \sqrt{\frac{10 (16 + 16) + 40 (25 + 1)}{10 + 40}} \]
\[ = \sqrt{\frac{(10 \times 32) + (40 \times 26)}{50}} \]
\[ = \sqrt{\frac{320 + 1040}{50}} \]
\[ = \sqrt{\frac{1360}{50}} \]
\[ = \sqrt{27.2} \]
\[ = 5.22 \text{ minutes} \]In simple words: This question asks us to find the overall average spread in production time when combining data from two factory sections. We first calculate the overall average production time for all workers, then use a formula that incorporates each section's own variance and how its average differs from the combined average to find the total standard deviation.

🎯 Exam Tip: Pay close attention to whether standard deviation or variance is given. If standard deviation is provided, remember to square it to get the variance for use in the combined standard deviation formula. Here, variance was directly provided, simplifying the initial steps.

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Statistics Class 11 Curriculum Solutions: Chapter 04 Measures of Dispersion

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