I came across a problem today in which it indicated the weight of a baseball $X$ has mean 5 and variance $4\over25$. It then indicated a carton contains 144 baseballs, and we are to obtain $T$ the total weight of the baseballs in one carton, assuming the weights of all the baseballs in the carton are independent. I thought of two approaches to this - one is "correct" (the Correct Method below) and the other is not (Incorrect Method). I'm trying to where the fault is in my reasoning. Why is it not the cases that both answers should be right?

Correct Method

Let $T=\sum_{i=1}^{144}X_i$, where $X_i$ is the weight of the $i$-th baseball, for $i=1, 2, ..., 144$. Then:

$Var(T) = Var[\sum_{i=1}^{144}X_i]=\sum_{i=1}^{144}Var(X_i)=144(\frac{4}{25})=23.04$

Incorrect Method

Let $T = 144X_1$ since each $X_i$ are iid. Then:


I realize the incorrect method is obviously not correct, but why aren't these two quantities the same? What am I missing?

  • 1
    $\begingroup$ because $T\neq 144X_1$ $\endgroup$
    – gunes
    Mar 23, 2019 at 18:41
  • 1
    $\begingroup$ @StatsStudent, sure. I just thought that the OP doesn't know that $T\neq 144X_1$, but knows that the true answer is $23.04$. $\endgroup$
    – gunes
    Mar 23, 2019 at 20:10
  • $\begingroup$ While $X_1$ and $X_2$ are identically distributed, $X_1+X_2$ has a different distribution to $X_1+X_1\,(=2X_1)$ (consider that the variance of the first expression is half the variance of the second expression) $\endgroup$
    – Glen_b
    Mar 24, 2019 at 1:52

1 Answer 1


As I've also pointed out in my comment, $T$, i.e. the total weight, is not equal to $144X_1$, i.e. $144$ times the weight of the first ball. Being iid doesn't produce the equality $\sum X_i=144 X_1$. It just means that the weights are drawn from the same population, so they may be quiet similar, but not constrained to be the same. So, if the two expressions are not equivalent, we shouldn't expect their variances to be equal (even if they could have been since the variance is a summary statistic).

Also, intuitively, $144X_1$ shouldn't have the same variance with $\sum_{i=1}^{144} X_i=144\bar{X}$ because it just depends on one sample drawn from the population, but the sum (or the mean) is more robust. When you draw $144$ samples from the population, you'd get samples like $5,4.7,5.1,...$ and the mean will be typically around $5$. So, since since we know that $\bar{X}$ is more certain than $X_1$, $144\times\bar{X}$ will have lower variance than $144\times X_1$.


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