Question

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. A sample of four metal sheets is randomly selected from a batch. What is the probability that the average length of a sheet is between 30.25 and 30.35 inches long

Answer 1

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 30.05 inches

           $\sigma$   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } }$ ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214