Question

You work for a pharmaceuticals company as a statistical process analyst. Your job is to analyze processes and make sure they are in 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 with a standard deviation of 0.05.) The acceptable range amounts of the compound added is 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 to each sample. A coworker asks why you take 3 samples of size 5 and find the mean instead of randomly choosing and measuring the amounts in 15 vials individually to check the machine's settings. (Note: Both samples are chosen without replacement.) Question: Assume the machine shifts and the distribution of the amount of the compound asses now has a mean of 9.96 milligrams and a standard deviation of 0.05 milligram. You select one vial and determine how much of the compound was added. What is the probability that you select a vial that is within the acceptable range (in other words, you do not detect that the machine has shifter)? (See figure below) a. b. You randomly select 15 vials. What is the probability that you select at least one vial that is within the acceptable range?

I am a bit lost, totally forgot how to do this.

Answer 1

Let x be the amount of compound asses .

x follows normal distribution with mean ( µ ) = 9.96 milligams and standard deviation (σ) = 0.05 milligrams

Acceptable range is between 9.65 milligrams and 9.95 milligrams.

a) P( 9.65 ≤ x ≤ 9.95 )

= P( x ≤ 9.95 ) - P( x ≤ 9.65 )

= $P\left ( \frac{x-\mu }{\sigma } \leq\frac{9.95-9.96}{0.05}\right ) -P\left ( \frac{x-\mu }{\sigma } \leq\frac{9.65-9.96}{0.05}\right )$

= P( z ≤ -0.2 ) - P ( z ≤ -6.2 )

= 0.4207 - 0 ---- ( from z score table, table value less than -3.50 would be 0 )

= 0.4207

So probability that you select a vital is within acceptable range is 0.4207

b) y is number of vital within acceptable range

y follows binomial distribution with n = 15 and p = 0.4207 , q = 1- 0.4207 = 0.5793

P(y) =$\binom{n}{y}*p^{y}*q^{(n-y)}$ ; y = 0,1,2,…n

We are asked to find P( y ≥ 1 )

P( y ≥ 1 ) = 1 - P( y = 0 )

P( 0 ) = $\binom{15}{0}*0.4207^{0}*0.5793^{(15-0)}$

= 1*1*0.0003

Probability = 0.0003