The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 inches.

Question

The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 inches.

a. 0.9999****
b. 0.0001
c. 0.8577
d. 0.1432

Answer 1

To find the probability that the mean height of the 100 men is less than 72 inches, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means tends to follow a normal distribution, regardless of the shape of the population distribution, as the sample size increases.

In this case, we have a sample size of 100, which is reasonably large, so we can assume that the distribution of sample means will be approximately normal.

The mean height of men is 69.0 inches, and the standard deviation is 2.8 inches. Since we are interested in the mean height of 100 men, the population mean and standard deviation remain the same, but the sample mean and standard deviation can be calculated using the formulas:

Sample Mean = Population Mean = 69.0 inches
Sample Standard Deviation = Population Standard Deviation / √Sample Size = 2.8 inches / √100 = 2.8 inches / 10 = 0.28 inches

Now, we can calculate the z-score, which measures the number of standard deviations a sample mean is away from the population mean. The formula for calculating the z-score is:

z = (x - μ) / (σ / √n)

where:
x = 72 inches (The mean height we are interested in)
μ = 69.0 inches (The population mean)
σ = 0.28 inches (The sample standard deviation)
n = 100 (The sample size)

z = (72 - 69.0) / (0.28 / √100) = 3 / 0.028 = 107.14

Now, we need to find the probability that a z-score is less than 107.14. However, such a large z-score corresponds to an extremely small (almost negligible) probability. Therefore, the correct answer is b. 0.0001.

Note: The answer a. 0.9999**** is incorrect because it suggests a probability of almost 1, which is not possible in this case.