Question

The weights of certain machine components are normally distributed with a mean of 8.81g and a standard deviation of 0.1g. Find the two weights that separate the top​ 3% and the bottom​ 3%. These weights could serve as limits used to identify which components should be rejected. Round to the nearest hundredth of a gram.

Answer 1

Solution:-

Given that,

mean = = 8.81g

standard deviation = = 0.1g

Using standard normal table,

P(Z > z) = 3%

= 1 - P(Z < z) = 0.03  

= P(Z < z) = 1 - 0.03

= P(Z < z ) = 0.97

= P(Z < 1.88) = 0.97  

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 0.1 + 8.81

x = 9.00

Using standard normal table,

P(Z < z) = 3%

= P(Z < z ) = 0.03

= P(Z < - 1.88) = 0.03  

z = - 1.88

Using z-score formula,

x = z * +

x = - 1.88 * 0.1 +8.81

x = 8.62