Question

please show equations used & dont copy other solutions

A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100lb and 5 lb and standard deviation 1 lb and 0.5 lb, respectively. What percent of filled boxes weighing between 104 lb and 106 lb are to be expected?

Answer 1

We have the variable X+Y

Mean for X + Y :

Jx+y = 100 + 5 = 105

$\sigma_{X+Y} =\sqrt{1^2+0.5^2} = 1.118$

We need to find :

$P(104<X+Y<106) = P\left ( \frac{104-105}{1.118}< \frac{X+Y-105}{1.118}< \frac{106-105}{1.118} \right )$

= P(-0.894 < Z < 0.894)

$= P\left ( Z< 0.894 \right )-P\left (Z< -0.894 \right )$

We use Excel function "NORMSDIST()" as :

$=NORMSDIST(0.894)-NORMSDIST(-0.894)$

Hence,

62.9% of the filled boxes weighing between 104lb and 106 lb are to be expected.