A scientist weighs 30 fish. Their mean weight | vedclass

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(B) Let the original weights be $x_i$. The given mean is $\bar{x} = 30 	ext{ g}$ and the standard deviation is $\sigma = 2 	ext{ g}$.Since every wei

Vedclass · Accepted Answer

$32, 2$

Vedclass · Answer

(B) Let the original weights be $x_i$. The given mean is $\bar{x} = 30 	ext{ g}$ and the standard deviation is $\sigma = 2 	ext{ g}$.Since every weight was recorded $2 	ext{ g}$ less than the actual weight,the corrected weights are $y_i = x_i + 2$.The new mean is $\bar{y} = \bar{x} + 2 = 30 + 2 = 32 	ext{ g}$.Standard deviation is independent of the change of origin. Adding a constant to each observation does not change the dispersion of the data.Therefore,the corrected standard deviation remains $\sigma = 2 	ext{ g}$.The correct mean and standard deviation are $32 	ext{ g}$ and $2 	ext{ g}$ respectively.

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Group $A$	$9$	$17$	$32$	$33$	$40$	$10$	$9$
Group $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

Similar Questions

From the data given below,state which group is more variable,$A$ or $B$?

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Group $A$ $9$ $17$ $32$ $33$ $40$ $10$ $9$

Group $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$

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The mean of $5$ observations is $7$. If four of these observations are $6, 7, 8, 10$ and one is missing,then the variance of all the five observations is:

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