GSEB Class 11 Statistics Solutions Chapter 3 Measures of Central Tendency Exercise 3.2

GSEB Solutions Class 11 Statistics Chapter 3 Measures Of Central Tendency Ex 3.2

 

Question 1. The mean daily wage paid to 75 skilled workers of a factory was Rs. 280 whereas the mean daily wage paid to 125 unskilled workers was Rs. 150. Find the mean wage of all the workers.

Answer: To determine the combined mean wage for all workers, we first identify the given values:
\(n_1\) = Number of skilled workers = 75
\( \bar{x}_1 \) = Mean daily wage of skilled workers = Rs. 280
\(n_2\) = Number of unskilled workers = 125
\( \bar{x}_2 \) = Mean daily wage of unskilled workers = Rs. 150

The formula for the combined mean \( \bar{x}_c \) of daily wages is:
\( \bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} \)

Substituting the given values into the formula:
\( \bar{x}_c = \frac{(75 \times 280) + (125 \times 150)}{75 + 125} \)
\( \implies \bar{x}_c = \frac{21000 + 18750}{200} \)
\( \implies \bar{x}_c = \frac{39750}{200} \)
\( \implies \bar{x}_c = 198.75 \)
Therefore, the combined mean daily wage for all workers is Rs. 198.75.
In simple words: We calculated the average wage by adding the total earnings of skilled and unskilled workers and dividing by the total number of workers.

🎯 Exam Tip: Remember to clearly define all variables \(n_1, \bar{x}_1, n_2, \bar{x}_2\) and the combined mean formula to ensure full marks for numerical problems involving combined averages.

 

Question 2. Find the weighted mean of the percentage change in prices from the following data:

Food item	Rice	Wheat	Tea	Sugar	Pulses
Weight	7	10	5	8	2
Percentage change in price	134	125	115	97	120

Answer: Given the weights corresponding to the percentage change in price, we prepare the following table to calculate the weighted mean:

Food items	Weight \(w\)	Percentage change of price \(x\)	\(wx\)
Rice	7	134	938
Wheat	10	125	1250
Tea	5	115	575
Sugar	8	97	776
Pulses	2	120	240
Total	\( \Sigma w = 32 \)	-	\( \Sigma wx = 3779 \)

The weighted mean of the percentage change in price, \( \bar{x}_w \), is calculated using the formula:
\( \bar{x}_w = \frac{\Sigma wx}{\Sigma w} \)

Substituting \( \Sigma wx = 3779 \) and \( \Sigma w = 32 \) into the formula:
\( \bar{x}_w = \frac{3779}{32} \)
\( \implies \bar{x}_w = 118.09\% \)
Therefore, the weighted mean of the percentage change in price is 118.09%.
In simple words: We find the weighted average by multiplying each price change by its importance (weight), summing these products, and then dividing by the total weight.

🎯 Exam Tip: When calculating a weighted mean, always construct a clear table to organize \(w\), \(x\), and \(wx\) values. This helps prevent errors in summation and application of the formula.

 

Question 3. 2 officers, 10 clerks and 3 peons contributed for a staff picnic. The contribution collected per person is shown in the following table:

Officer	Clerk	Peon
Rs. 1000	Rs. 500	Rs. 200

Calculate the mean contribution per person using weighted mean.

Answer: For calculating the weighted mean contribution, the number of staff members for each category is taken as the weight. The calculation is organized in the following table:

Staff	No. of staff \(w\)	Contribution for picnic \(x\) (Rs.)	\(wx\)
Officer	2	1000	2000
Clerk	10	500	5000
Peon	3	200	600
Total	\( \Sigma w = 15 \)	-	\( \Sigma wx = 7600 \)

The mean of individual contribution \( \bar{x}_w \) is calculated using the formula:
\( \bar{x}_w = \frac{\Sigma wx}{\Sigma w} \)

Substituting \( \Sigma wx = 7600 \) and \( \Sigma w = 15 \) into the formula:
\( \bar{x}_w = \frac{7600}{15} \)
\( \implies \bar{x}_w = 506.67 \)
Therefore, the mean individual contribution is Rs. 506.67.
In simple words: We calculate the average contribution per person by summing the total amount contributed by all staff categories and dividing it by the total number of staff.

🎯 Exam Tip: Clearly define the "weight" in weighted mean problems. In this case, it's the number of people. Ensure all currency symbols are converted to the standard abbreviation (Rs.).

 

Question 4. The mean marks of a student in 7 theory papers is 62. What should be the mean marks in 3 practical examinations so that his mean marks for the entire examination is 68? (The marks of each theory paper and practical examination are the same.)

Answer: We are given the following information:
\(n_1\) = Number of theory papers = 7
\( \bar{x}_1 \) = Mean marks of theory papers = 62
\(n_2\) = Number of papers for practical exam = 3
\( \bar{x}_2 \) = Mean marks of practical exam = ? (This is what we need to find)
\( \bar{x}_c \) = Mean marks of theory and practical combined = 68

The combined mean formula is:
\( \bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} \)

Substituting the known values:
\( 68 = \frac{(7 \times 62) + (3 \times \bar{x}_2)}{7 + 3} \)
\( \implies 68 = \frac{434 + 3\bar{x}_2}{10} \)
\( \implies 68 \times 10 = 434 + 3\bar{x}_2 \)
\( \implies 680 = 434 + 3\bar{x}_2 \)
\( \implies 3\bar{x}_2 = 680 - 434 \)
\( \implies 3\bar{x}_2 = 246 \)
\( \implies \bar{x}_2 = \frac{246}{3} \)
\( \implies \bar{x}_2 = 82 \) marks
Thus, the mean marks for the practical examination should be 82 marks.
In simple words: To achieve an overall average of 68 marks, the student needs to score an average of 82 marks in the practical examinations, given their average theory score.

🎯 Exam Tip: When solving for an unknown mean in a combined average problem, set up the combined mean formula and algebraically solve for the missing variable. Show each step clearly.