Question

A machine at Katz Steel Corporation makes 5-inch-long nails. The probability distribution of the lengths of these nails is approximately normal with a mean of 5 inches and a standard deviation of 0.12 inch. The quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than 4.95 inches or greater than 5.05 inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment?

Round your answer to 4 decimal places.

Answer #1

Given,

The length of the nails follows normal distribution with mean 5 inches and standard deviation 0.12 inch

The interval for acceptance has been given i.e 4.95-5.05 inches

Sample Size is 25.

To calculate the probability that the machine needs an adjustment is equivalent to calculate the probability of rejection(or Significance level), which can be calculated by going reverse in calculating the confidence interval.

Now,

For the sample, expected mean= 5 inches

Sample Standard deviation= standard deviation of the length/sqrt(sample size)=0.12/sqrt(25)=0.024

Now, we have been given the interval values from that we can calculate the z values

and,

Now,

The probability that the machine needs adjustment=Probability that the sample mean is less than 4.95+Probability that the sample mean is k=more than 5.05=P(z<-2.08333)+P(z>2.08333)=0.01862+0.01862=0.03724 3.7% (Ans)

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