Mathematics, 28.07.2019 07:30 Rileyb101207

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 places.)

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(a) 0.1974
(b) 0.7695
Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches a

Answer from: junkmailemail42

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.

Answer from: Quest

step-by-step explanation:

10 ptsfind the area of the inside of the frame where the picture would belong. then find the area of

Answer from: Quest

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[ (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

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