Mathematics: GIVING BRAINLEST TO FIRST PERSON TO ANSWER!

GIVING BRAINLEST TO FIRST PERSON TO ANSWER!

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teseladavisp07jif

24.04.2022

Mathematics

a) The average weight is 199 pounds.

b) The probability that the maximum safe weight will be exceeded is 0.4840 = 48.40%.

c) The probability that the maximum safe weight will be exceeded is 0.3539 = 35.39%.

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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

(a) If 56 people are on the elevator, and their total weight is 11,144 pounds, what is their average weight?

This is 11144 divided by 56, that is 11144/56 = 199 pounds.

Their average weight is 199 pounds.

(b) If a random sample of 56 adult men ride the elevator, what is the probability that the maximum safe weight will be exceeded?

The weights of adult U.S. men have mean 196 pounds and standard deviation 71 pounds, which means that $\mu = 196, \sigma = 71$

11144 pounds is an average of 199 pounds, so this probability is 1 subtracted by the pvalue of Z when X = 199.

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

$Z = \frac{199 - 196}{71}$

Z = 0.04

Z = 0.04 has a pvalue of 0.5160

1 - 0.5160 = 0.4840

The probability that the maximum safe weight will be exceeded is 0.4840 = 48.40%.

(c) If a random sample of 56 adult women ride the elevator, what is the probability that the maximum safe weight will be exceeded?

The weights of adult U.S. women have mean 169 pounds and standard deviation 80 pounds, which means that $\mu = 169, \sigma = 80$

11144 pounds is an average of 199 pounds, so this probability is 1 subtracted by the pvalue of Z when X = 199.

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

$Z = \frac{199 - 169}{80}$

Z = 0.375

Z = 0.375 has a pvalue of 0.6461

1 - 0.6461 = 0.3539

The probability that the maximum safe weight will be exceeded is 0.3539 = 35.39%.

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