## Question

The compressive strength of samples of cement is assumed to be normally distributed with a mean of 6000(kg)/(cm^2) and a standard deviation of 100(kg)/(cm^2).

1. What is the probability that a sample's strength is less than 5850(kg)/(cm^2)?
2. What compressive strength is exceed by 95% of the samples?
3. What is the probability that at least one sample, among 4 random samples, will have a compressive strength less than 5850(kg)/(cm^2)?
4. If the quality control section wants to construct a bar(x) chart based on samples of size 6, what will be the 3sigma control limits for the chart? If the process mean suddenly shifts to 6120(kg)/(cm^2), what is the probability that the shift will be detected on the next sample taken? Note that a sample mean falling above the upper control limit or below the lower control limit would signal a shift in the process mean on the bar(x) chart.

## Answer

1. P(X<5850)=P(Z<(5850-6000)/100)=P(Z<-1.5)=Phi(-1.5)=0.0668072.

2. We need to find such x that

P(X>x)=0.95.

P(X>x)=P(Z>(x-6000)/100)=0.95.

(x-6000)/100=Phi^-1(0.95)=1.64485.

x-6000=164.485.

x=6164.485(kg)/(cm^2).

3. Probability that at least one sample will have a compressive strength less than 5850 (kg)/(cm^2) equals to 1 minus probability that all 4 samples will have a compressive strength greater than 5850(kg)/(cm^2).

p=1-(P(X>5850))^4=1-(1-P(X<5850))^4=1-(1-0.0668072)^4=0.241622.

4. Since sample is of size 6, then sample mean will be distributed with mean mu=6000 and standard deviation sigma=100/sqrt(6)=40.8248.

So, 3sigma control limit is 3*40.8248=122.4745.

If sample mean will be outside limits then shift will be detected.

We need to find

P(bar(X)>mu+3sigma|bar(X)<mu-3sigma)=P(Z>(mu+3sigma-6120)/sigma|Z<(mu-3sigma-6120)/sigma)=

=P(Z>(6000+122.4745-6120)/40.8248|Z<(6000-122.4745-6120)/40.8248)=P(Z>0.0606|Z<-5.9394)=

=P(Z>0.0606)+P(Z<-5.9394)=1-P(Z<0.0606)+P(Z<-5.9394)=

=1-Phi(0.0606)+Phi(-5.9394)=0.4758.