GSEB Class 11 Statistics Solutions Chapter 4 Measures of Dispersion Solution Exercise 4.5

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures Of Dispersion Ex 4.5

Question 1. Price fluctuations of two shares A and B are given below, which type of share has more relative variation in its price?

Price (Rs.)	share A	321	322	325	322	324	320	323	316	319	318
Price (Rs.)	share B	141	146	130	146	142	145	132	134	132	152

Answer: To ascertain which share exhibits greater relative variation in its price, we need to compute the coefficient of variation for the prices of both share A and share B.

Share A				Share B
Price (Rs.)	\( (x-\bar{x}) \)	\( (x-\bar{x})^2 \)	X	Price (Rs.)	\( (x-\bar{x}) \)	\( (x-\bar{x})^2 \)	X
x	\( \bar{x} = 321 \)			x	\( \bar{x} = 140 \)
321	0	0		141	1	1
322	1	1		146	6	36
325	4	16		130	-10	100
322	1	1		146	6	36
324	3	9		142	2	4
320	-1	1		145	5	25
323	2	4		132	-8	64
316	-5	25		134	-6	36
319	-2	4		132	-8	64
318	-3	9		152	12	144
\( \Sigma x = 3210 \)	0	\( \Sigma(x-\bar{x})^2 = 70 \)		\( \Sigma x = 1400 \)	0	\( \Sigma(x-\bar{x})^2 = 510 \)

For Share A:
Mean:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{3210}{10} = 321 \)
Standard deviation:
\( s = \sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}} \)
\( \implies s = \sqrt{\frac{70}{10}} \)
\( \implies s = \sqrt{7} \)
\( \implies s = 2.65 \)
Coefficient of variation:
\( \text{Variation} = \frac{s}{\bar{x}} \times 100 \)
\( \implies \text{Variation} = \frac{2.65}{321} \times 100 \)
\( \implies \text{Variation} = 0.0083 \times 100 \)
\( \implies \text{Variation} = 0.83\% \)
For Share B:
Mean:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{1400}{10} = 140 \)
Standard deviation:
\( s = \sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}} \)
\( \implies s = \sqrt{\frac{510}{10}} \)
\( \implies s = \sqrt{51} \)
\( \implies s = 7.14 \)
Coefficient of variation:
\( \text{Variation} = \frac{s}{\bar{x}} \times 100 \)
\( \implies \text{Variation} = \frac{7.14}{140} \times 100 \)
\( \implies \text{Variation} = 0.051 \times 100 \)
\( \implies \text{Variation} = 5.1\% \)
Comparing the coefficients of variation, the price of share A has a variation of 0.83%, while share B's price shows a variation of 5.1%. Therefore, share B exhibits a greater relative measure of variation in its price.
In simple words: To find which share price varies more relative to its average price, we calculate the Coefficient of Variation (CV) for both. Share B has a higher CV (5.1%) compared to Share A (0.83%), indicating its price fluctuates more significantly.

🎯 Exam Tip: Remember that a higher coefficient of variation indicates greater relative variability or dispersion in data. For comparative analysis of two series, always use the coefficient of variation.

 

Question 2. The daily salary of administrative staff of two companies yielded the following results:

	Company A	Company B
Mean salary(Rs.)	600	2100
Standard Deviation (Rs.)	30	84

Which company has more stable salary?
Answer:
For Company A:
Mean salary \( \bar{x} \) = Rs. 600
Standard deviation \( s \) = Rs. 30
Coefficient of variation:
\( \text{Variation} = \frac{s}{\bar{x}} \times 100 \)
\( \implies \text{Variation} = \frac{30}{600} \times 100 \)
\( \implies \text{Variation} = 5\% \)
For Company B:
Mean salary \( \bar{x} \) = Rs. 2100
Standard deviation \( s \) = Rs. 84
Coefficient of variation:
\( \text{Variation} = \frac{s}{\bar{x}} \times 100 \)
\( \implies \text{Variation} = \frac{84}{2100} \times 100 \)
\( \implies \text{Variation} = 4\% \)
The coefficient of variation for the daily salary of employees in Company A is 5%, whereas for Company B, it is 4%. A lower coefficient of variation indicates greater stability. Consequently, the daily salaries in Company B are more stable.
In simple words: To determine which company has more stable salaries, we compare their Coefficients of Variation (CV). Company B has a lower CV (4%) than Company A (5%), meaning its salaries are more consistent and less variable.

🎯 Exam Tip: When comparing the consistency or stability of two datasets, the series with the smaller coefficient of variation is considered more stable. This is a common application of CV in real-world scenarios.

 

Question 3. The Coefficients of variation of two series are 30% and 25% and their standard deviations are 15 and 9 respectively. Find their means.
Answer:
For the First Series:
Coefficient of variation (CV): 30%
Standard deviation \( s \): 15
Mean \( \bar{x} \): ?
The formula for the coefficient of variation is:
\( \text{CV} = \frac{s}{\bar{x}} \times 100 \)
\( \implies 30 = \frac{15}{\bar{x}} \times 100 \)
\( \implies 30\bar{x} = 1500 \)
\( \implies \bar{x} = \frac{1500}{30} \)
\( \implies \bar{x} = 50 \)
For the Second Series:
Coefficient of variation (CV): 25%
Standard deviation \( s \): 9
Mean \( \bar{x} \): ?
The formula for the coefficient of variation is:
\( \text{CV} = \frac{s}{\bar{x}} \times 100 \)
\( \implies 25 = \frac{9}{\bar{x}} \times 100 \)
\( \implies 25\bar{x} = 900 \)
\( \implies \bar{x} = \frac{900}{25} \)
\( \implies \bar{x} = 36 \)
Therefore, the mean of the first series is 50, and the mean of the second series is 36.
In simple words: Given the Coefficient of Variation (CV) and standard deviation for two datasets, we can rearrange the CV formula to calculate the mean. For the first series, with a 30% CV and SD of 15, the mean is 50. For the second series, with a 25% CV and SD of 9, the mean is 36.

🎯 Exam Tip: This problem demonstrates how to use the Coefficient of Variation formula to find a missing component (mean or standard deviation) if the other two are provided. Always remember the formula: \( \text{CV} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100 \).