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`.