Question

We load on a plane 100 packages whose weights are independent random variables between 20 and 120 pounds. The mean weight is 70 pounds. The distribution is unknown. What is the probability that the total weight will exceed 7500 pounds?

Answer 1

We know that if $X\sim U(a,b)$ , then $E(X)=\frac{a+b}{2}$

Let X denote the weight of a package (in pounds).

Now, $Mean=70=\frac{20+120}{2}=\frac{a+b}{2}$

$\Rightarrow X\sim U(20,120)$

So, $V(X)=\frac{(b-a)^2}{12}=\frac{(120-20)^2}{12}=\frac{10000}{12}=\frac{2500}{3}$

Using Central Limit theorem, we know,

$\sum X_i\sim N(100(70),100(\frac{2500}{3}))\sim N(7000,\frac{250000}{3})$

Required probability =

$\\P(\sum X>7500)=P(Z>\frac{7500-7000}{\sqrt{\frac{250000}{3}}})=P(Z>1.73) \\ \\=1-P(Z<1.73)=1-0.95818=0.04182$