You work for a pharmaceuticals company as a statistical process analyst. Your jo

Question

You work for a pharmaceuticals company as a statistical process analyst. Your job is to analyze processes and make sure statistical control. In one process, a machine is supposed to add 9.8 milligrams of a compound to a mixture in a vial. (Assume this process can be approximated by a normal distribution.) The acceptable range of amounts of the compound added milligrams 9.65 milligrams to 9.95 milligrams, inclusive. Because of an error with the release valve, the setting on the machine "shifts" from 9.8 milligrams. To check that the machine is adding the correct amount of the compound into the vials, you select at random three samples of five vials and find the mean amount of the compound added for each sample. A coworker asks why you take the 3 samples of size 5 and find the mean instead randomly choosing and measuring the amounts in 15 vials individually the machine's settings. You select five vials and find the mean amount of compound added. Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligram. (a) What is the probability that you select a sample of five vials that has a mean that is not outside the acceptable range? (See figure.) (b) You randomly select three samples of five vials. What is the probability that you select at least one sample of five vials that has a mean that is mu outside the acceptable range? (c) What is more sensitive to change-an individual measure or the mean?

Explanation / Answer

Let X be the amount of compound added.

Given that X has mean (mu) = 9.96 milligrams and standard deviation (sigma) = 0.05 milligrams

Here from the figure we can say that the acceptable range is 9.7 to 10.

a) Here we have to find P(9.7 < Xbar < 10)

n = 5

By using central limit theorem XBar follows normal distribution with mean is mu standard deviation is sigma/sqrt(n).

Now convert xbar = 9.7 and xbar=10 into z-score.

z-score is defined as,

z = (xbar - mu) / (sigma/sqrt(n))

z = (9.7 - 9.96) / (0.05/sqrt(5)) = -11.63

z = (10 - 9.96) / (0.05/sqrt(5)) = 1.79

That is we have to find P(-11.63 < X < 1.79).

P(-11.63 < X < 1.79) = P(Z < 1.79) - P(Z < -11.63)

This probability we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where z is z-score.

P(-11.63 < X < 1.79)= 0.9632 - 0.0000 = 0.9632

The probability that all three samples we select has a mean that is outside the acceptable range equals

probability = (1 - 0.9632)^3 = 4.984E-05

Therefore the probability that at least one sample has a mean that is not outside the acceptable range equals

1 - 4.984E-05 = 1.000E+00

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