Question

The lengths of nails produced in a factory are normally distributed with a mean of 5.09 centimeters and a standard deviation of 0.04 centimeters. Find the two lengths that separate the top 7% and the bottom 7%

. 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:-

Given that,

mean = = 5.09 cm.

standard deviation = = 0.04 cm.

Using standard normal table,

P(Z > z) = 7%

= 1 - P(Z < z) = 0.07  

= P(Z < z) = 1 - 0.07

= P(Z < z ) = 0.93

= P(Z < 1.4758 ) = 0.93  

z = 1.4758

Using z-score formula,

x = z * +

x = 1.4758 * 0.04 + 5.09

x = 5.15 cm.

Top 7% = 5.15 cm.

Using standard normal table,

P(Z < z) = 7%

= P(Z < z ) = 0.07

= P(Z < -1.4758 ) = 0.07  

z = -1.4758

Using z-score formula,

x = z * +

x = -1.4758 * 0.04 + 5.09

x = 5.03 cm.

Bottom 7% = 5.03 cm.

length = ( 5.03, 5.15 )