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If both the mean and the standard deviation of 50 observations `x_(1), x_(2),…, x_(50)` are equal to 16, then the mean of `(x_(1) -4)^(2), (x_(2) -4)^(2), …., (x_(50) - 4)^(2)` is k. The value of `(k)/(80)` is _______.

Text Solution

Verified by Experts

The correct Answer is:
5
For given numbers mean `(mu) = 16`
Standard deviation `(sigma ) = 4`
Mean of `(x_(i) -4) = 16-4 = 12`
Standard deviation of `(x_(i) -4) = 16` (as standard deviation is not affected by shifting)
`(sum (x_(i) -4)^(2))/(5) - (12)^(2) = (16)^(2)`
` (sum (x_(i) -4)^(2))/(50) = 256 + 144 = 400`
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