Questions: Suppose the heights of 18-year-old men are approximately normally distributed, with mean 66 inches and standard deviation 5 inches. (a) What is the probability that an 18-year-old man selected at random is between 65 and 67 inches tall? (Round your answer to four decimal places.) (b) If a random sample of twenty-five 18-year-old men is selected, what is the probability that the mean height x̄ is between 65 and 67 inches? (Round your answer to four decimal places.) (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? The probability in part (b) is much higher because the standard deviation is smaller for the x̄ distribution.

Solution Steps

(a) To find the probability that an 18-year-old man is between 65 and 67 inches tall, we use the properties of the normal distribution. We calculate the z-scores for 65 and 67 inches and then find the probability between these z-scores using the standard normal distribution.

(b) For a sample of twenty-five 18-year-old men, the distribution of the sample mean is also normal with the same mean but a smaller standard deviation (standard deviation of the population divided by the square root of the sample size). We calculate the z-scores for the sample mean of 65 and 67 inches and find the probability between these z-scores.

(c) Compare the probabilities from parts (a) and (b) to determine which is higher and explain why based on the standard deviation of the sample mean.

To find the probability that an 18-year-old man is between 65 and 67 inches tall, we first calculate the z-scores for these heights using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where \( X \) is the height, \( \mu = 66 \) is the mean, and \( \sigma = 5 \) is the standard deviation.

For \( X = 65 \): \[ z_{65} = \frac{65 - 66}{5} = -0.2 \]

For \( X = 67 \): \[ z_{67} = \frac{67 - 66}{5} = 0.2 \]

The probability that a randomly selected 18-year-old man is between 65 and 67 inches tall is given by the difference in cumulative distribution function (CDF) values:

\[ P(65 < X < 67) = \Phi(0.2) - \Phi(-0.2) = 0.1585 \]

For a sample of 25 men, the standard deviation of the sample mean is:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{25}} = 1 \]

We calculate the z-scores for the sample mean heights of 65 and 67 inches:

For \( \bar{X} = 65 \): \[ z_{65\_sample} = \frac{65 - 66}{1} = -1 \]

For \( \bar{X} = 67 \): \[ z_{67\_sample} = \frac{67 - 66}{1} = 1 \]

The probability that the sample mean height is between 65 and 67 inches is:

\[ P(65 < \bar{X} < 67) = \Phi(1) - \Phi(-1) = 0.6827 \]

The probability in part (b) is much higher than in part (a) because the standard deviation of the sample mean is smaller, leading to a narrower distribution and a higher probability for the same range.

Final Answer