Question

The weights of certain machine components are normally distributed with a mean of 5.195.19 ounces and a standard deviation of 0.050.05 ounces. Find the two weights that separate the top 8%8% and the bottom 8%8%. These weights could serve as limits used to identify which components should be rejected.

________ ounces and ________ ounces

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 5.19 ounces

Standard deviation = 0.05 ounces

Let the weight that separates top 8% be T and bottom 8% be B

P(X > T) = 0.08

P(X < T) = 1 - 0.08 = 0.92

P(Z < (T - 5.19)/0.05) = 0.92

Take value corresponding to 0.92 from standard normal distribution table

(T - 5.19)/0.05 = 1.405

T = 5.26 ounces

Similarly,

(B - 5.19)/0.05 = 5.12

5.12 ounces and 5.26 ounces are the values that separates bottom and top 8% respectively