Assume the weight (in ounces) of a major league baseball is a random variable, a carton contains 144 baseballs. Assume now that the weights of individual baseballs are independent and normally distributed with mean $\mu=5$ and standard deviation $\sigma = 2/5$, and let $T$ represent the total weight of all baseballs in the carton. Find the probability that the total weight of baseballs in a carton is at most 725 ounces.
Let $B_1,B_2,…,B_{144}$ be the random variables for the weight of each baseball in the carton. Consider the change of variable $T = B_1+B_2+…+B_{144}$. The Jacobian is 1. The density function $f(T)$ is
$$\left(\frac1{\sqrt{2\pi}\sigma}\right)^{144}\int_0^{\infty}\int_0^{\infty}…\int_0^\infty\exp\left(\frac{-1}{2\sigma^2}[(B_2-\mu)^2+…+(B_{144}-\mu)^2+(T-B_2-…-B_{144})^2]\right)dB_2dB_3…dB_{144}$$
and the answer will be $\int_0^{725} f(T)dT$. Is my set up correct? Is there an efficient way to evaluate this? Thank you!
(The book's answer is 0.8508)
Best Answer
The sum of $n$ i.i.d. normal random variables with mean $\mu$ and variance $\sigma^2$ is normal with mean $n\mu$ and variance $n\sigma^2$. In your case, $T=n\mu+\sqrt{n\sigma^2}Z$ for some standard normal $Z$, $n=144$, $\mu=5$, $\sigma=\frac25$.
Hence $P[T\leqslant t]=P[Z\leqslant(t-n\mu)/\sqrt{n\sigma^2}]=\Phi((t-n\mu)/(\sigma\sqrt{n}))$.
Numerically, for $t=725$, one gets $\Phi(\frac{25}{24})\approx85\%$.