P1. The heights of adult men in the United States are approximately normally distributed with a mean of 70 inches and a standard deviation of 3 inches.

Question

P1. The heights of adult men in the United States are approximately normally distributed with a mean of 70 inches and a standard deviation of 3 inches.

a. Actor Michael B. Jordan is 6’0” (72”) tall. What percentage of adult men are shorter than Michael?

b. Comedian Jack Black is 5’7’ (67”) tall. What percentage of adult men are taller than Jack?

c. What percentage of adult men are of height between Jack and Michael?

P2. Continue with the heights of adult men. What is the 90th percentile of their heights?

P3. Continue with the heights of adult men.

a. For means from samples of size 25, what is the standard deviation of the distribution?

b. What is the probability that the mean weight of 25 randomly selected adult men is over 71 inches?

Answer 1

- First we should declare our known variables: (population mean) µ = 70in, (population s.d.) σ = 3in.

- Let's remember the equation for the Z statistic: Z = (x-µ)/σ

- Make sure you have a Z distribution chart to reference

a. Actor Michael B. Jordan is 6’0” (72”) tall. What percentage of adult men are shorter than Michael?

1. Z = (72-70)/3 = .67 (round to two decimal places because most Z dist. charts only round to two)
2. The corresponding value on the table is .74857 which can be read in this context as the probability of an adult man having a height of 0-72 inches, and no greater.
3. Answer is 74.857%

b. Comedian Jack Black is 5’7’ (67”) tall. What percentage of adult men are taller than Jack?

1. Z = (67-70)/3 = -1
2. The corresponding value is .15866 which can be read in this context as the probability of an adult man having a height of 0-67 inches, and no greater. But since we are concerned with the percentage of those taller than jack, we take the compliment of .15866, which is (1 – .15866 = .84134).
3. Answer is 84.134%

c. What percentage of adult men are of height between Jack and Michael?

1. Another way to word this is what is the probability any given adult man is between the heights of Jack and Michael. This can be written as:
2. P(X < 72) – P(x < 67) = .74857 – .15866 = .58991
3. Answer is 58.991%

P2. Continue with the heights of adult men. What is the 90th percentile of their heights?

1. Instead of using a Z score to find its corresponding probability on the chart, we will use a probability to find the corresponding Z score. Look for .90000 on the chart. Notice it is between 1.28 and 1.29. We can use 1.285 as an approximation.
2. Now we rewrite the Z score formula in terms of x. So we get x = µ + Z*σ
3. Using our numbers we have x = 70 + 1.285*3 = 73.855
4. Answer is 73.855in

P3. Continue with the heights of adult men.

a. For means from samples of size 25, what is the standard deviation of the distribution?

1. The distribution of means of samples of size n from the height data used above is represented as:
2. X ~ N(µ_x, σ_x / √n)
3. µ_x and σ_x are the population mean and population standard deviation respectively.
4. The population s.d. is divided by the square root of the size of the samples taken to get the s.d. of the distribution of the sample mean because as n increases to the population size we'd expect the sample mean to get very close to the true population mean. For example if the population is the size of the U.S. which is 340,000,000 and we graph the distribution of 100 samples of size 339,999,990 then we'd expect them to fall extremely close to each other (like s.d. = .00000001) because those 10 people not included in the sample, even if 100inches tall, will not weigh the sample mean far from the population mean at all.
5. So σ_x / √n = 3 / √25 = 0.6
6. Answer is 0.6

b. What is the probability that the mean weight of 25 randomly selected adult men is over 71 inches?

1. Follow the procedure of P1 just with a s.d. of 0.6
2. Z = (71-70)/0.6 = 1.67
3. Corresponding probability on chart is .95254
4. Complement is 1 - .95254 = .04746
5. Answer is 4.746%