Mathematics, 18.06.2021 08:50 usabmx34

12.) The FAA has to periodically revise their rules regarding weight estimates that vary by season. Airlines currently use an estimate for the weight of an adult male passenger of 190 lbs.
in the summer that assumes
he carries sixteen pounds of baggage and has five pounds of
clothing. According to your textbook, assume that men (not clothed nor carrying baggage) have
weights that are normally distributed with a mean of 188.6 lbs. and a standard deviation of 38.9
lbs.

a.) If one adult male is randomly selected and is assumed to be also carrying 21 lbs. due to
clothing and baggage, find the probability that the total weight is greater than 190 lbs.

b.) What total weight would separate the lower 20% of men carrying 21 lbs. of clothing and
baggage from the top 80%?

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Answer from: natalie2sheffield

a) 0.4856 = 48.56% probability that the total weight is greater than 190 lbs.

b) A total weight of 155.9 lbs.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 188.6 lbs. and a standard deviation of 38.9 lbs.

This means that $\mu = 188.6, \sigma = 38.9[/etx]21 lbs of baggage are taken into consideration in this estimate.a.) If one adult male is randomly selected and is assumed to be also carrying 21 lbs. due to clothing and baggage, find the probability that the total weight is greater than 190 lbs.This is 1 subtracted by the the p-value of Z = 190. So[tex]Z = \frac{X - \mu}{\sigma}$

$Z = \frac{190 - 188.6}{38.9}$

Z = 0.036

Z = 0.036 has a p-value of 0.5144.

1 - 0.5144 = 0.4856.

0.4856 = 48.56% probability that the total weight is greater than 190 lbs.

b.) What total weight would separate the lower 20% of men carrying 21 lbs. of clothing and baggage from the top 80%?

This is the 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

$-0.84 = \frac{X - 188.6}{38.9}$

X - 188.6 = -0.84*38.9

X = 155.9

A total weight of 155.9 lbs.

Answer from: Quest

use an equation of ratios to determine the amount of gas needed to fill the tank.

0.2 0.8

288.75 x

231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal $3

* = $3465 (answer)

1 1 gal

hope this you

Answer from: Quest

step-by-step explanation:

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