Question

The weights of certain machine components are normally distributed with a mean of 5.19 ounces and a standard deviation of 0.05 ounces. Find the two weights that separate the top 8% and the bottom 8%. These weights could serve as limits used to identify which components should be rejected

Answer 1

Answer:

The  weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.

Step-by-step explanation:

We are given that

Mean,$\mu=5.19$

Standard deviation,$\sigma=0.05$

We have to find the two weights that separate the top 8% and the bottom 8%.

Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.

Z-value for p-value =0.08 =$-1.41$

For 8% bottom

$Z=\frac{x_1-\mu}{\sigma}=-1.41$

$\frac{x_1-5.19}{0.05}=-1.41$

$x_1-5.19=-1.41\times 0.05$

$x_1=-1.41\times 0.05+5.19$

$x_1=5.1195$

For 8% top

p-Value=1-0.08=0.92

Z- value=1.41

Now,

$\frac{x_2-5.19}{0.05}=1.41$

$x_2-5.19=1.41\times 0.05$

$x_2=1.41\times 0.05+5.19$

$x_2=5.2605$

(x1,x2)=(5.1195,5.2605)