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Question Help Library An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 171 lb. The new population of pilots has normally distributed weights with a mean of 139 lb and a standard deviation of 29.7 ib. a. If a pilot is randomly selected, find the probability that his weight is between 130 lb and 171 lb. The probability is approximately (Round to four decimal places as needed.) b. 1937 different pilots are randomly selected, find the probability that their mean weight is between 130 lib and 171 lb. The probability is approximately (Round to four decimal places as needed.) c. When redesigning the election sont, which probability is more relevant? O A Port (b) because the seat performance for a single pilot is more important B. Part(a) because the seat performance for a sample of plots is more important OC. Part (b) because the seat performance for a sample of pilots is more important OD. Part(a) because the seat performance for a single pilot is more important dhid ntents Resources Success Success Options

Answer 1

Answer: An engineer is going to redesign an ejection seat for an airplane. The seat for an airplane. The seat was designed for pilots weighing between 130 lb and 171 lb. The new population of pilots has normally distributed weights with a mean of 139 lb and a standard deviation of 29.7 lb.

Solution:

Mean, μ = 139 lb

Standard deviation, σ = 29.7 lb

a) If a pilot is randomly selected, find the probibility that his weight is between 130 lb and 171 lb.

P(130 < X < 171) = P(130-139/29.7 < Z < 171-139/29.7)

P(130 < X < 171) = P(- 0.3030 < Z < 1.0774)

P(130 < X < 171) = P(Z < 1.0774) - P(Z < -0.3030)

P(130 < X < 171) = 0.8593 - 0.3809 (from z table)

P(130 < X < 171) = 0.4784

Therefore, the probability that their mean weight is between 130 lb and 171 lb is 0.4784

b) If 37 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 171 lb.

P(130 < X < 171) = P(130-139/29.7/√37 < Z < 171-139/29.7/√37)

P(130 < X < 171) = P(- 1.8433 < Z < 6.5538)

P(130 < X < 171) = P(Z < 6.5538) - P(Z < -1.8433)

P(130 < X < 171) = 1 - 0.032643 (from z table)

P(130 < X < 171) = 0.9674

Therefore, the probability that their mean weight is between 130 lb and 171 lb is 0.9674.

c) When redesigning the ejection seat, which probability is more relevant?

The option C is correct answer.

Part (b) because the seat performance for a sample of pilots is more important.

Answer: A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 600 babies were born and 330 of them were girls.

Solution:

p = x/n = 330/600 = 0.55

At 99% confidence interval, α = 0.01

Z critical =Zα/2 = 2.5758

the 99% confidence interval :

CI = p ± Z critical * √p(1-p)/n

CI = 0.55 ± 2.5758 * √0.55(1-0.55)/600

CI = 0.55 ± 0.0523

CI = (0.4977, 0.6023)

Therefore, 99% confidence interval is

0.4977 < p < 0.6023.

No, the proportion of girls is not significantly different from 0.5.

As per HOMEWORKLIB RULES only first question to be answered but I solve one more. Please repost the other question.

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