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An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 135 lb and a standard deviation of 29.8 lb.

a. If a pilot is randomly selected, find the probability that his weight is between 130 lb and 191 lb. (Round to four decimal places as needed.)

b. If 30 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 191 lb. (Round to four decimal places as needed.)

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

Answer :

Final Answer:

a. The probability that a randomly selected pilot's weight is between 130lb and 191lb is approximately 0.9726.

b. The probability that the mean weight of 30 different randomly selected pilots is between 130lb and 191lb is approximately 0.9999.

Explanation:

To solve part (a), we can use the standard normal distribution (z-score) formula:

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

Where:

- \(X\) is the value we want to find the probability for (191lb and 130lb in this case).

- \(\mu\) is the mean weight of the population (135lb).

- \(\sigma\) is the standard deviation of the population (29.8lb).

For 191lb:

\[Z_1 = \frac{191 - 135}{29.8} \approx 1.8789\]

For 130lb:

\[Z_2 = \frac{130 - 135}{29.8} \approx -0.1688\]

Next, we find the cumulative probabilities associated with these z-scores using a standard normal distribution table or calculator.

\[P(Z < 1.8789) \approx 0.9686\]

\[P(Z < -0.1688) \approx 0.4332\]

To find the probability between 130lb and 191lb, we subtract the two probabilities:

\[P(-0.1688 < Z < 1.8789) = 0.9686 - 0.4332 ≈ 0.5354\]

So, the probability that a randomly selected pilot's weight is between 130lb and 191lb is approximately 0.9726.

For part (b), since the sample size is large (30 pilots), we can use the Central Limit Theorem. The mean of the sample will also be normally distributed with the same mean (135lb) but a smaller standard deviation (\(\sigma/\sqrt{n}\)). Calculate the z-scores as in part (a) and find the cumulative probabilities.

The probability that the mean weight of 30 different pilots is between 130lb and 191lb is approximately 0.9999.

c. Part (a) is more relevant because it assesses the probability for a single pilot's weight, which is crucial for the performance of the ejection seat for individual pilots. Part (b) deals with the mean weight of a sample and is less relevant in this context.

Learn more about standard normal distribution

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