TS EAMCET · Maths · Statistics
Let \(\sigma_1, \sigma_2\) be the standard deviations of two distributions \(D_1\) and \(D_2\) respectively and \(D_1\) be more consistent than \(D_2\). If the means of \(D_1\) and \(D_2\) are same, then the percentage increase in the standard deviation of \(D_2\) over the standard deviation of \(D_1\) is
- A \(\frac{\sigma_1-\sigma_2}{\sigma_2} \times 100\)
- B \(\frac{\sigma_1-\sigma_2}{\sigma_1} \times 100\)
- C \(\frac{\sigma_2-\sigma_1}{\sigma_2} \times 100\)
- D \(\frac{\sigma_2-\sigma_1}{\sigma_1} \times 100\)
Answer & Solution
Correct Answer
(D) \(\frac{\sigma_2-\sigma_1}{\sigma_1} \times 100\)
Step-by-step Solution
Detailed explanation
Since mean of both the distribution are same, therefore Percentage increase in the standard deviation of \(D_2\) over the standard deviation of \(D_1\) \( =\frac{\sigma_2-\sigma_1}{\sigma_1} \times 100 \)
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