An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 147 lb and a standard deviation of 28.7 lb.

a. If a pilot is randomly selected, find the probability that his weight is between 140 lb and 181 lb. (Round to four decimal places as needed.)

b. If 38 different pilots are randomly selected, find the probability that their mean weight is between 140 lb and 191 lb. (Round to four decimal places as needed.)

c. When redesigning the ejection seat, which probability is more relevant?
A. Part (b) because the seat performance for a sample of pilots is more important.
B. Part (a) because the seat performance for a single pilot is more important.
C. Part (a) because the seat performance for a sample of pilots is more important.
D. Part (b) because the seat performance for a single pilot is more important.

The probability that a randomly selected pilot's weight is between 140 lb and 181 lb is approximately 0.4782. The probability that the mean weight of 38 randomly selected pilots is between 140 lb and 191 lb is approximately 0.9332.

Part (b) is more relevant because it considers the seat performance for a sample of pilots, which is more important in accommodating the weight range of the overall pilot population.

a. To find the probability that a randomly selected pilot's weight is between 140 lb and 181 lb, we need to calculate the area under the normal distribution curve between these two values. We can standardize the values using the formula z = (x - μ) / σ, where x is the weight, μ is the mean, and σ is the standard deviation.

For 140 lb: z = (140 - 147) / 28.7 ≈ -0.2447
For 181 lb: z = (181 - 147) / 28.7 ≈ 1.1833

Using the standard normal distribution table, we can find the corresponding probabilities:
For -0.2447, the probability is 0.4032
For 1.1833, the probability is 0.8814

To find the probability between these two values, we subtract the smaller probability from the larger one: 0.8814 - 0.4032 ≈ 0.4782

Therefore, the probability that a randomly selected pilot's weight is between 140 lb and 181 lb is approximately 0.4782.

b. To find the probability that the mean weight of 38 randomly selected pilots is between 140 lb and 191 lb, we use the Central Limit Theorem. The mean of the sample means will be the same as the population mean, and the standard deviation of the sample means will be the population standard deviation divided by the square root of the sample size.

For a sample size of 38, the standard deviation of the sample means is 28.7 / √38 ≈ 4.6539.

We can then standardize the values using the formula z = (x - μ) / (σ / √n), where x is the weight, μ is the mean, σ is the standard deviation, and n is the sample size.

For 140 lb: z = (140 - 147) / (4.6539) ≈ -1.5011
For 191 lb: z = (191 - 147) / (4.6539) ≈ 9.4624

Using the standard normal distribution table, we find that the probability of a z-value less than -1.5011 is approximately 0.0668, and the probability of a z-value less than 9.4624 is approximately 1.

Therefore, the probability that the mean weight of 38 randomly selected pilots is between 140 lb and 191 lb is approximately 1 - 0.0668 ≈ 0.9332.

c. In the context of redesigning the ejection seat, the more relevant probability is Part (b), where we calculate the probability of the mean weight of a sample of pilots. This is because the performance of the seat needs to account for the average weight of a group of pilots, rather than just a single pilot. By considering the mean weight of a sample, we can ensure that the seat is designed to accommodate the weight range of a typical group of pilots, rather than focusing on the weight of an individual pilot. Therefore, Part (b) is more important for designing an ejection seat that meets the needs of the overall pilot population.

To know more about probability visit:

brainly.com/question/32117953

#SPJ11