Mathematics, 03.07.2020 15:01 brandonkelly104

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 181 lb. The new population of pilots has normally distributed weights with a mean of 140 lb and a standard deviation of 27.3 lb. Required:
a. If a pilot is randomly selected, find the probability that his weight is between 150 lb and 201 lb. The probability is approximately.
b. If 39 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 201 lb. The probability is approximately.
c. When redesigning the ejection seat which probability is more relevant

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Answer from: ambercable3

(a) The probability that his weight is between 150 lb and 201 lb is 0.3428.

(b) The probability that the sample mean weight is between 150 lb and 201 lb is 0.011.

Step-by-step explanation:

We are given that the seat was designed for pilots weighing between 130 lb and 181 lb.

The new population of pilots has normally distributed weights with a mean of 140 lb and a standard deviation of 27.3 lb.

Let $\bar X$ = sample mean price for a movie in the United States

SO, X ~ Normal( $\mu=140,\sigma^{2} =27.3^{2}$ )

(a) The z-score probability distribution for the normal distribution is given by;

Z = $\frac{ X-\mu}{\sigma}} }$ ~ N(0,1)

where, $\mu$ = population mean weights = 140 lb

$\sigma$ = standard deviation = 27.3 lb

Now, the probability that his weight is between 150 lb and 201 lb is given by = P(150 lb < X < 201 lb) = P(X < 201 lb) - P(X $\leq$ 150 lb)

P(X < 201 lb) = P( $\frac{ X-\mu}{\sigma}} }$ < $\frac{ 201-140}{27.3}} }$ ) = P(Z < 2.23) = 0.9871

P(X $\leq$ 150 lb) = P( $\frac{ X-\mu}{\sigma}} }$ $\leq$ $\frac{ 150-140}{27.3}} }$ ) = P(Z $\leq$ 0.37) = 0.6443

The above probability is calculated by looking at the value of x = 2.23 and x = 0.37 in the z table which has an area of 0.9871 and 0.6443.

Therefore, P(150 lb < X < 201 lb) = 0.9871 - 0.6443 = 0.3428.

(b) Let $\bar X$ = sample mean weight

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean weight = 140 lb

$\sigma$ = standard deviation = 27.3 lb

n = sample of pilots = 39

Now, the probability that the sample mean weight is between 150 lb and 201 lb is given by = P(150 lb < $\bar X$ < 201 lb) = P( $\bar X$ < 201 lb) - P( $\bar X$ $\leq$ 150 lb)

P( $\bar X$ < 201 lb) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{201-140}{\frac{2.73}\sqrt{39} } }$ ) = P(Z < 13.95) = 0.9999

P( $\bar X$ $\leq$ 150 lb) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{150-140}{\frac{2.73}\sqrt{39} } }$ ) = P(Z $\leq$ 2.29) = 0.9889

Therefore, P(150 lb < $\bar X$ < 201 lb) = 0.9999 - 0.9889 = 0.011.

(c) When redesigning the ejection seat, the probability of a single pilot is more relevant as discussed in part (a) because it is important to look after the safety of each and every pilot not of a particular sample.

Answer from: Quest

idk

step-by-step explanation:

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Answer from: Quest

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step-by-step explanation:

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