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Nataly [62]
3 years ago
5

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches and standard deviation 4 inch

es.(a) what is the probability that an 18-year-old man selected at random is between 71 and 73 inches tall? (round your answer to four decimal places.) .2013 incorrect: your answer is incorrect. (b) if a random sample of twenty-three 18-year-old men is selected, what is the probability that the mean height x is between 71 and 73 inches? (round your answer to four decimal places.)

Mathematics
2 answers:
k0ka [10]3 years ago
5 0
(a) 0.1974
(b) 0.7695

Naily [24]3 years ago
5 0

Answer:

a) There is a 19.74% probability that an 18-year-old man selected at random is between 71 and 73 inches tall.

b) There is a 76.98% probability that the mean height x is between 71 and 73 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples 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.

In this problem, we have that:

Mean of 72 inches and standard deviation 4 inches. This means that \mu = 72, \sigma = 4.

(a) what is the probability that an 18-year-old man selected at random is between 71 and 73 inches tall?

This probability is the pvalue of Z when X = 73 and the pvalue of Z when X = 71.

X = 73

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

Z = \frac{73 - 72}{4}

Z = 0.25

Z = 0.25 has a pvalue of 0.5987

X = 71

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

Z = \frac{71 - 72}{4}

Z = -0.25

Z = -0.25 has a pvalue of 0.4013

This mean that there is a 0.5987-0.4013 = 0.1974 = 19.74% probability that an 18-year-old man selected at random is between 71 and 73 inches tall.

b) if a random sample of twenty-three 18-year-old men is selected, what is the probability that the mean height x is between 71 and 73 inches?

Now we use the Central Limit Theorem, so:

s = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{23}} = 0.8341.

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

Z = \frac{73 - 72}{0.8341}

Z = 1.20

Z = 1.20 has a pvalue of 0.8849

X = 71

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

Z = \frac{71 - 72}{0.8341}

Z = -1.20

Z = -1.20 has a pvalue of 0.1151

This means that there is a 0.8849 - 0.1151 = 0.7698 = 76.98% probability that the mean height x is between 71 and 73 inches.

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