Question

Problem 9-22 (modified). A quality control inspector accepts shipments of 500 precision .5" steel rods if the mean diameter of a sample of 81 falls between .4995" and .5005". Previous evaluations have established that the standard deviation for individual rod diameters is .003".

What is the probability the inspector will accept an out-of-tolerance shipment having mu=.5003? (Note: we aren't told the tolerance, but for simplicity assume that it is .0002 so that mu=.5003 is out-of-tolerance. The acceptance standard of between .4995" and .5005" relates to the sample mean, not to the population mean mu.)

Probability=

What is the probability the inspector will reject a near-perfect shipment having mu=.4999?

Probability=

Answer 1

Answer 2

For the first question, we need to calculate the probability of getting a sample mean outside the acceptance range of .4995" to .5005" given that the true population mean is .5003" and the sample size is 81.

Let X be the random variable representing the sample mean diameter. Then, X follows a normal distribution with mean mu = 0.5003 and standard deviation sigma = 0.003/sqrt(81) = 0.00033.

To calculate the probability of getting a sample mean outside the acceptance range, we can standardize X as follows:

Z = (X - mu) / (sigma/sqrt(n)) = (X - 0.5003) / 0.00033

Now, we can calculate the probability of Z being less than (0.4995 - 0.5003) / 0.00033 = -2.42 or greater than (0.5005 - 0.5003) / 0.00033 = 0.61:

P(Z < -2.42 or Z > 0.61) = P(Z < -2.42) + P(Z > 0.61)

Using a standard normal table or calculator, we can find that P(Z < -2.42) = 0.0075 and P(Z > 0.61) = 0.2716. Therefore,

P(Z < -2.42 or Z > 0.61) = 0.0075 + 0.2716 = 0.2791

So the probability of the inspector accepting an out-of-tolerance shipment with mu = 0.5003 is approximately 0.2791.

For the second question, we need to calculate the probability of getting a sample mean outside the acceptance range of .4995" to .5005" given that the true population mean is .4999" and the sample size is 81.

Following the same steps as before, we get:

Z = (X - mu) / (sigma/sqrt(n)) = (X - 0.4999) / 0.00033

The probability of Z being less than (0.4995 - 0.4999) / 0.00033 = -1.21 or greater than (0.5005 - 0.4999) / 0.00033 = 1.82 is:

P(Z < -1.21 or Z > 1.82) = P(Z < -1.21) + P(Z > 1.82)

Using a standard normal table or calculator, we can find that P(Z < -1.21) = 0.1128 and P(Z > 1.82) = 0.0344. Therefore,

P(Z < -1.21 or Z > 1.82) = 0.1128 + 0.0344 = 0.1472

So the probability of the inspector rejecting a near-perfect shipment with mu = 0.4999 is approximately 0.1472.