Question

The mean diameter of a ball bearing produced by a certain manufacturer is 0.80 cm with a standard deviation of 0.03 cm. A sample of 36 ball bearings is randomly selected from a production run. Suppose the diameter is normally distributed.

c. What is the probability that a randomly selected ball bearing will have a diameter of less than 0.798 cm?

c. What is the probability that a randomly selected ball bearing will have a diameter of less than 0.798 cm?
d. What is the probability that the sample of ball bearings will have a sample mean less than 0.798 cm?
							0.3446
e. Explain why your answers in parts c and d are not the same.

Answer 1

X: Ball bearing diameter have a normal distribution with a mean of 0.80 and standard deviation = 0.03

And 36 bearings are randomly selected.

c. P(that a randomly selected ball bearing will have a diameter of less than 0.798) that is P(X < 0.798)

First, find the z score:

That is P(X < 0,798) becomes P(Z < -0.07)

By using z table the probability for z = -0.07 is 0.4721

That is P(X < 0.798) = 0.4721

d) P(Sample of ball bearing will have a sample mean less than 0.798) that is

Here the formula of z score slightly changes since have to find the probability for the sample mean,

That is P(Xbar < 0.798) becomes P(Z < -0.4)

By using z table the probability for z = -0.4 is 0.3446

Therefore,

e. In part c the probability is for single observation of diameter and in part d we found the probability for sample means of a diameter that's why both are different. Since the denominator would change in the z score because the sample mean have a standard deviation while X has a standard deviation only.