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If the coefficient of variation of two d...

If the coefficient of variation of two distribution are 50 ,60 and their arithmetic means are 30 and 25 respectively then the difference of their standard deviation is

A

0

B

1

C

1.5

D

2.5

Text Solution

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The correct Answer is:
To find the difference of the standard deviations of two distributions given their coefficients of variation and arithmetic means, we can follow these steps: ### Step 1: Understand the Coefficient of Variation The coefficient of variation (Cv) is defined as: \[ Cv = \frac{\sigma}{\mu} \times 100 \] where \(\sigma\) is the standard deviation and \(\mu\) is the mean. ### Step 2: Set Up the Equations For the first distribution: - Coefficient of Variation \(Cv_1 = 50\) - Mean \( \mu_1 = 30\) Using the formula: \[ Cv_1 = \frac{\sigma_1}{\mu_1} \times 100 \] we can rearrange it to find \(\sigma_1\): \[ \sigma_1 = \frac{Cv_1 \times \mu_1}{100} = \frac{50 \times 30}{100} \] For the second distribution: - Coefficient of Variation \(Cv_2 = 60\) - Mean \( \mu_2 = 25\) Using the same formula: \[ Cv_2 = \frac{\sigma_2}{\mu_2} \times 100 \] we can rearrange it to find \(\sigma_2\): \[ \sigma_2 = \frac{Cv_2 \times \mu_2}{100} = \frac{60 \times 25}{100} \] ### Step 3: Calculate \(\sigma_1\) and \(\sigma_2\) Now we can calculate the values: \[ \sigma_1 = \frac{50 \times 30}{100} = \frac{1500}{100} = 15 \] \[ \sigma_2 = \frac{60 \times 25}{100} = \frac{1500}{100} = 15 \] ### Step 4: Find the Difference of Standard Deviations Now, we can find the difference: \[ \text{Difference} = \sigma_1 - \sigma_2 = 15 - 15 = 0 \] ### Final Answer The difference of their standard deviations is: \[ \boxed{0} \]

To find the difference of the standard deviations of two distributions given their coefficients of variation and arithmetic means, we can follow these steps: ### Step 1: Understand the Coefficient of Variation The coefficient of variation (Cv) is defined as: \[ Cv = \frac{\sigma}{\mu} \times 100 \] where \(\sigma\) is the standard deviation and \(\mu\) is the mean. ...
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