Question

Problem 9.6

The diameter of ping-pong balls manufactured at a large factory is expected to be approximately normally distributed with a mean of 2.30 inches and a standard deviation of .04 inch. What is the probability that a randomly selected ping-pong ball will have a diameter.....

1. Between 2.28 and 2.30 inches?

2. Between 2.31 and 2.33 inches?

3. Between what two values (symmetrically distributed around the mean) will 60% of the balls fall (in terms of diameter)?

4. If many random samples of 16 balls are selected.....

4.1. What would be the mean and standard error of the mean be expected to be?

4.2 What distribution would the sample means follow?

4.3 What proportion of the sample mean would be between 2.28 and 2.30 inches?

4.4 What proportion of the sample mean would be between 2.31 and 2.33 inches

4.5 60% of the sample means would be between what two values?

5. Compare the answers of (1) and (4.3) and (2) with (4.4). Explain.

6. Explain the difference in results of (3) and (4.5).

7. Which is more likely to occur - an individual value about 2.34 inches, a sample mean above 2.32 inches in a sample size of 4, or a sample mean above 2.31 inches in a sample of size 16? Explain

8. Referring to Problem 9.6 (above), if the population consisted of 200 ping-pong balls, what would be your answer to part (4.4) of that problem?

-----I got the answers 1) 0.1915, 2) -0.0867, 4.1) The mean would be the same as the sample which is 2.30. Standard error of the mean is 0.01. 4.2) Normally distributed

Answer 1

μ = 2.3, σ = 0.04

(1) x1 = 2.28, x2 = 2.3

z1 = (x1 - μ)/σ = (2.28 - 2.3)/0.04 = -0.5 and z2 = (2.3 - 2.3)/0.04 = 0

P(2.28 < x < 2.3) = P(-0.5 < z < 0) = 0.1915

(2) x1 = 2.31, x2 = 2.33

z1 = (2.31 - 2.3)/0.04 = 0.25 and z2 = (2.33 - 2.3)/0.04 = 0.75

P(2.31 < x < 2.33) = P(0.25 < z < 0.75) = 0.1747

(3) z- scores containing the middle 60% of the values are ± 0.8416

x1 = μ - 0.8416 * σ = 2.3 - 0.8416 * 0.04 = 2.266 and x2 = μ + 0.8416 * σ = 2.3 + 0.8416 * 0.04 = 2.334

(4.1) x-bar = μ = 2.3, SE = σ/√n = 0.04/√16 = 0.01

(4.2) Normal distribution

(4.3) z1 = (2.28 - 2.3)/0.01 = -2 and z2 = (2.3 - 2.3)/0.01 = 0

P(2.28 < x-bar < 2.3) = P(-2 < z < 0) = 0.4772

(4.4) z1 = (2.31 - 2.3)/0.01 = 1 and z2 = (2.33 - 2.3)/0.01 = 3

P(2.31 < x-bar < 2.33) = P(1 < z < 3) = 0.1573

(4.5) z- scores containing the middle 60% of the values are ± 0.8416

x1-bar = 2.3 - 0.8416 * 0.01 = 2.292 and x2-bar = 2.3 + 0.8416 * 0.01 = 2.308