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An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120 lb and 161 lb. The new population of pilots has normally distributed weights with a mean of 128 lb and a standard deviation of 33.9 lb.

a. If a pilot is randomly selected, find the probability that his weight is between 120 lb and 161 lb. (Round to four decimal places as needed.)

Answer :

Answer:

The probability is approximately nothing is 0.4230 = 42.30%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 128, \sigma = 33.9[/tex]

If a pilot is randomly​ selected, find the probability that his weight is between 120 lb and 161 lb.

This is the pvalue of Z when X = 161 subtracted by the pvalue of Z when X = 120. So

X = 161

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

[tex]Z = \frac{161 - 128}{33.9}[/tex]

[tex]Z = 0.97[/tex]

[tex]Z = 0.97[/tex] has a pvalue of 0.8340

X = 120

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

[tex]Z = \frac{120 - 128}{33.9}[/tex]

[tex]Z = -0.235[/tex]

[tex]Z = -0.235[/tex] has a pvalue of 0.4110

0.8340 - 0.4110 = 0.4230

The probability is approximately nothing is 0.4230 = 42.30%

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