Spread of a parcel's total weight from four independent items

Question

A shipping box holds four items whose weights vary independently. Each item's weight has variance $2.25$ (standard deviation $1.5$ kilograms). The total weight is

$S = X_1 + X_2 + X_3 + X_4.$

What is the standard deviation of the total weight?

Accepted Answer

Because the four items are independent, the variance of their total is just the sum of the four variances. They all have the same variance, so this is four copies of it: \mathrm{Var}(S) = 2.25 + 2.25 + 2.25 + 2.25 = 4 	imes 2.25 = 9. In words: independent wobbles add up as variances, so n identical independent items give n times the single-item variance. Then take the square root: \mathrm{SD}(S) = \sqrt{9} = 3 	ext{ kilograms}. Notice the spread grew from 1.5 to 3, a factor of 2, not a factor of 4. That is because the standard deviation of a sum of n identical independent items grows like \…

Spread of a parcel's total weight from four independent items

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