Question

A manufacturing process produces a critical part of average length 90 millimeters, with a standard deviation 2 of millimeters. All parts deviating by more than 5 millimeters from the mean must be rejected. What percentage of the parts must be rejected, on average? Assume a normal distribution.

Answer 1

We have that

`X\sim N(\mu=90,\sigma^2=4^{})`

The parts the will be rejected when it's above 95 or when it's under 85, if we plot the normal distribution it would be

Then, the percentage of the parts that will be rejected corresponds to the area in blue, then, we must calculate the area under the normal distribution for

`P(X<85)+P(X>95)`

The normal distribution is symmetrical, then calculate P(X < 85) is the same as P(X > 95), then we write it as

`2\cdot P(X>95)`

Calculate that integral is very hard, then, we must transform that in a standard normal X ~ N(0, 1) and use a table to find the result, to do that we must write a value z, it's a transformation to take a value on our normal and leads it to the standard normal, it's

`Z=(X-\mu)/(\sigma)`

We have X = 95, μ = 90 and σ = 2

`Z=(95-90)/(2)=2.5`

Then 2.5 is the value we are going to search in our table, using the complementary cumulative table for 2.5 we get 0.00621, which means

`P(X>95)=0.00621`

And the total percentage will be

`P(X<85)+P(X>95)=0.01242`

We can write it in percentage

`0.01242=1.242\%`

Therefore, only 1.24% will be rejected.

It's a very low value, but it's expected because it's more than 2 standard deviations (95%).