Question

The lengths of nails produced in a factory are normally distributed with a mean of 4.77 centimeters and a standard deviation of 0.06 centimeters. Find the two lengths that separate the top 9% and the bottom 9%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.

Answer 1

Solution :

mean = = 4.77

standard deviation = = 0.06

Using standard normal table,

P(Z > z) = 9%

1 - P(Z < z) = 0.09

P(Z < z) = 1 - 0.09 = 0.91

P(Z < 1.341 ) = 0.91

z = 1.34

Using z-score formula,

x = z * +

x = 1.34 * 0.06 + 4.77

= 4.850

Minimum amount = 4.850

P(Z < z) = 9%

P(Z < z) = 0.09

P(Z < z) = 0.09

P(Z < -1.341 ) = 0.91

z = -1.34

Using z-score formula,

x = z * +

x = -1.34 * 0.06 + 4.77

= 4.690

Minimum amount = 4.690