The tensile strength of an aluminum rod is normally distribu

The tensile strength of an aluminum rod is normally distributed with mean of 40 Kilograms and standard deviation of 5 Kilograms. If 50,000 parts are produced, how many would you expect to:

A. Fail to meet a minimum specification limit of 35 Kilograms tensile strength?

B. Have a tensile strength in excess of 48 Kilograms?

Solution

The tensile strength of an aluminum rod is normally distributed with mean of 40 Kilograms and standard deviation of 5 Kilograms. If 50,000 parts are produced, how many would you expect to:

A. Fail to meet a minimum specification limit of 35 Kilograms tensile strength?

Z value for 35, z = (35-40)/5 = -1

P( x < 35) = P( z < -1) = 0.1587

50000* 0.1587 = 7935

7935 rods fail to meet a minimum specification limit of 35 Kilograms tensile strength

B. Have a tensile strength in excess of 48 Kilograms?

Z value for 48, z = (48-40)/5 = 1.6

P( x >48) = P( z > 1.6) = 0.0548

50000* 0.0548 = 2740

2740 rods have a tensile strength in excess of 48 Kilograms.