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

bStep-by-step explanation:i did it i got it right...

Suppose the heights of 18-year-old men are approximately

Ответ:

arjava1008

15.03.2022

Mathematics

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 $\mu = 72, \sigma = 4$ .