If you select a random sample of 16 ping-pong balls?

The diameter of ping pong balls manufactured at a large factory is approximately normally distributed with a mean de 1.30 inches and standard deviation of 0.04 inch. If you select a random sample of 16 ping pong balls





a. What is the sampling distribution of the mean?


b. What is the probability that the sample mean is less than 1.28 inches?


c. What is the probability that the sample mean is between 1.31 and 1.33 inches?


d. The probability is 60% that the sample mean will be between what two values symmetrically distributed and the population mean?

If you select a random sample of 16 ping-pong balls?
Let X denote the diameter of a ping pong ball. We are informed that the population mean and population standard deviation of X are μ = 1.3 and σ = 0.04. In other words





X ~ N(1.3, 0.04²)





(a) Let X-bar denote the sample mean of the diameter. Although the sample size is small (n = 16 %26lt; 25), X-bar can be described using a normal distribution because the population variance is known.





X-bar ~ N(1.3, 0.04²/16) or N(1.3, 0.0001)





(b) P(X-bar %26lt; 1.28)


= P(Z %26lt; (1.28 - 1.3)/0.01)


= P(Z %26lt; -2)


= 0.02275





(c) P(1.31 %26lt; X-bar %26lt; 1.33)


= P((1.31 - 1.3)/0.01 %26lt; Z %26lt; (1.33 - 1.3)/0.01)


= P(1 %26lt; Z %26lt; 3)


= P(Z %26lt; 3) - P(Z %26lt; 1)


= 0.99865 - 0.84135


= 0.15730





(d) P(-z %26lt; Z %26lt; z) = 0.6


P(Z %26lt; z) - P(Z %26lt; 0) = 0.3


P(Z %26lt; z) = 0.3 + 0.5 = 0.8


z = 0.84162





x = 1.3 ± z*0.01 = 1.2916 and 1.3084