Question

There is concern about the thickness of bearing plates being produced. If the thickness exceeds
150 mm then the production line must be shut down and the machine recalibrated. Suppose the
thickness produced varies considerably throughout a shift, and has a normal distribution with mean
µ = 160 mm and standard deviation σ = 10 mm

a. What is the probability that the thickness measurement of a single random sample will fail
to detect that the machine is out of calibration?

b. If 5 samples are randomly drawn throughout a shift, what is the probability that the average
of the 5 measurements will be less than 150 mm and hence fail to indicate that the machine
is out of calibration?

c. How many measurements are required in a single shift so that there is a 1 percent
probability of failing to detect that the machine is out of calibration?

Answer 1

Let X =  the thickness of bearing plates .

So from the given information X follows normal distribution with mean µ = 160 mm and standard deviation σ = 10 mm

a ) If the thickness exceeds 150 mm then the production line must be shut down and the machine recalibrated .

Therefore here we want to find P( X < 150)

Let's use excel:

P( X < 150) = "=NORMDIST(150,160,10,1)" = 0.1587

b. If 5 samples are randomly drawn throughout a shift, what is the probability that the average
of the 5 measurements will be less than 150 mm and hence fail to indicate that the machine
is out of calibration?

For sample mean we have mean =  µ = 160 mm and standard deviation σ/ = 10/ mm

Therefore,  P( < 150) = "=NORMDIST(150,160,10/SQRT(5),1)" = 0.0127

c) How many measurements are required in a single shift so that there is a 1 percent
probability of failing to detect that the machine is out of calibration?

Here we want to find sample size (n), such that  

Z_0.01 = "=NORMSINV(0.01)" = -2.326

therefore n = 2.326*2.326 = 5.41 = 6 ( because n is a whole number).

So answer is 6