PLZ HELPP 10 POINTS PLZZZ

There is a cool easy way to solve this kind of problem. If a range is given, and the entire range increases in incrememnts of 1, then you can find the number intergers inclusive by taking the first number and subtracting it from the last and adding 1. Since in this problem you want it exclusive, we can rewrite the range to 21-99.

99-21 +1 = 78 + 1 = 79

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3 years ago

Comparing observations from different populations: The heights of adult men in America are normally distributed, with a mean of

Schach [20]

Answer:

a. Z = 1.99

b. 97.67%

c. Z = 2.56

d. 0.52%

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.66 inches, and thus, we have $\mu = 69.7, \sigma = 2.66$

6 feet 3 inches = 6*12 + 3 = 75 inches, which means that we have to find z when X = 75. So

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

$Z = \frac{75 - 69.7}{2.66}$

$Z = 1.99$

b. What percentage of men are SHORTER than 6 feet 3 inches?

The proportion is the p-value of Z = 1.99.

Looking at the z-table, Z = 1.99 has a p-value of 0.9767.

0.9767*100% = 97.67%, which is the answer.

c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches, and thus, we have $\mu = 64.5, \sigma = 2.54$ . We have to find Z when X = 5*12 + 11 = 71. So