Question

A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 90 finished production and computes the sample mean product weights bar. If results over a long period of time show that 5% of bar values are more than 8.5 pounds and 5% are less than 7.5 pounds, what are the mean and standard deviation for the population of production produced within the process? It may help to sketch the standard normal curve.

Answer 1

Solution:- Given that n = 90

μ = (8.5+7.5)/2 = 8

the Z-score corresponing with a probability of 5% or 0.05 in Z = -1.645 or Z = 1.645

THe Z-score is the value decreased by the mean, divided by the standard deviation

Z = (X-μx)/σx

so, the standard error = σx = (X-μx)/Z = (7.5-8)/-1.645 = 0.3040

The standard deviation is the product of the standard error and the square root of tghe sample size

σ = σx*sqrt(90) = 0.3040*sqrt(90) = 2.8840

=> μ = 8, σ = 2.8840