Question

I rlly need help with this :(

Answer 1

Using the normal distribution, it is found that:

a) The pilot is at the 72th percentile.

b) 19.13% of pilots are unable to fly.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 72.6, \sigma = 2.7$.

Item a:

The percentile is the p-value of Z when X = 74.2, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{74.2 - 72.6}{2.7}$

Z = 0.59

Z = 0.59 has a p-value of 0.7224.

72th percentile.

Item b:

The proportion that is able to fly is the p-value of Z when X = 78 subtracted by the p-value of Z when X = 70, hence:

X = 78:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{78 - 72.6}{2.7}$

Z = 2

Z = 2 has a p-value of 0.9772.

X = 70:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{70 - 72.6}{2.7}$

Z = -0.96

Z = -0.96 has a p-value of 0.1685.

0.9772 - 0.1685 = 0.8087 = 80.87%.

Hence the percentage that is unable to fly is:

100 - 80.87 = 19.13%.

More can be learned about the normal distribution at brainly.com/question/4079902

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