Mathematics, 31.03.2020 05:02 22nathanieltimms

The capacity of an elevator is 10 people or 1730 pounds. The capacity will be exceeded if 10 people have weights with a mean greater than 1730 divided by 10 equals 173 pounds. Suppose the people have weights that are normally distributed with a mean of 175 lb and a standard deviation of 31 lb.

a. find the probability that if a person is randomly selected, his weight will be greater than 176 pounds.

The probability is approximately _

b. Find the probability that 10 randomly selected people will have a neam that is greater than 176 pounds.

Statistics and Probability

The probability is approximately _

c. Does the elevator appear to have the correct weight limit?

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Answer from: chessacs2950

a. 48.80%

b. 46.02%

c. 57.93% probability of the sample mean weight being above the weight limit, which is a high probability, meaning that the elevator does not appear to have the correct weight limit

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 175, \sigma = 31$

a. find the probability that if a person is randomly selected, his weight will be greater than 176 pounds.

This is 1 subtracted by the pvalue of Z when X = 176. So

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

$Z = \frac{176 - 175}{31}$

Z = 0.03

Z = 0.03 has a pvalue of 0.5120

1 - 0.5120 = 0.4880

48.80% probability that if a person is randomly selected, his weight will be greater than 176 pounds.

b. Find the probability that 10 randomly selected people will have a neam that is greater than 176 pounds.

Now $n = 10, s = \frac{31}{\sqrt{10}} = 9.8$

This is 1 subtracted by the pvalue of Z when X = 176. So

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

By the Central Limit Thorem

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

$Z = \frac{176 - 175}{9.8}$

Z = 0.1

Z = 0.1 has a pvalue of 0.5398

1 - 0.5398 = 0.4602

46.02% probability that 10 randomly selected people will have a neam that is greater than 176 pounds.

c. Does the elevator appear to have the correct weight limit?

The weight limit is 173 pounds in a sample of 10.

The probability that the mean weight is larger than this is 1 subtracted by the pvalue of Z when X = 173. So

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

$Z = \frac{173 - 175}{9.8}$

Z = -0.2

Z = -0.2 has a pvalue of 0.4207

1 - 0.4207 = 0.5793

57.93% probability of the sample mean weight being above the weight limit, which is a high probability, meaning that the elevator does not appear to have the correct weight limit

Answer from: Quest

Dm me pls mike and

Answer from: Quest

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6

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