Question

1. Consider all women in the U.S. aged 18-74. Suppose their heights are measured and found to be approximately normally distributed with a mean µ = 63.5” and standard deviation σ = 2.5”. Using the empirical rule (68-95-99.7) to answer the following questions.

a. What percent of women are between 61” and 66” tall?
b. What percent of women are between 56” and 71” tall?

c. What percent of women are between 63.5” and 66” tall?

d. What percent of women are less than 63.5” tall?

e. What percent of women are more than 58.5” tall?

2. Using the standard normal curve answer the following questions

a. What percent lie between 1.24 and 2.38?

b. What percent lie between -1.18 and 2.63?

c. What percent lie above -2.19?

d. What percent lie below -0.46?

e. What percent lie below 0.23?

3. Suppose that people’s weights are normally distributed, with mean 175 pounds and a standard deviation of 6 pounds.

a. What percent of the population would weigh between 165 and 170 pounds?

b. What percent of the population would you expect to weigh more than 182 pounds?

c. Out of a group of 9000 people, how many would you expect to be heavier than 182 pounds?

d. What percent of the population would you expect to weigh between 172 and 180 pounds?

e. What percent of the population would you expect to be heavier than 163 pounds?

4. Suppose that for a certain exam given nationwide is normally distributed with a mean of 80 and a standard deviation of 3?

a. What percent of the population would score higher than an 82?

b. What percent of the population would score between 78 and 85?

c. What percent of the population would score between 74 and 78?

d. What percent of the population would score above a 76?

e. If 5000 students took this exam, how many would score above a 76?

Answer 1

Solution(1)

Given in the question

U.S aged 18-74 heights are approximately normally distributed with

Mean ($\mu$) = 63.5

Standard deviation ($\sigma$) = 2.5

From the empirical rule, we can say that

68% of the data values are between +/-1 standard deviation from the mean

95% of the data values are between +/-2 standard deviation from the mean

99.7% of the data values are between +/-3 standard deviation from the mean

Solution(a)

We need to calculate the percent of women are between 61 and 66 fall

P(61<X<66) = P(X<66) - P(X<61)

First we need to calculate Z score

Z = (X-$\mu$)/$\sigma$ = (61-63.5)/2.5 = -2.5/2.5 = -1

Z = (66-63.5)/2.5 = 1

From Empirical rule we can say that 68% of women are between 61 and 66 tall.

Solution(b)

We need to calculate percent of women are between 56 and 71

P(56<X<71) = P(X<71) - P(X<56)

Z = (56-63.5)/2.5 = -3

Z = (71-63.5)/2.5 = 3

From Empirical rule we can say that 99.7% of women are between 56 and 71 tall.

Solution(c)

We need to calculate percent of women are between 63.5 and 66 tall

P(63.5<X<66) = P(X<66) - P(X<63.5)

Z = (63.5-66)/2.5 = -1

Z = (66-66)/2.5 = 0

From Z table we found p-value

P(63.5<X<66) = 0.5 - 0.1587 = 0.3413

So there is 34.13% of women are between 63.5 and 66 tall.

Solution(d)

Here we need to calculate percent of women are less than 63.5 tall

Z = (63.5-66)/2.5 = -1

From Z table we found p-value

P(X<63.5) = 0.1587

So there is 15.87% of women are less than 63.5 tall.

Solution(e)

Here we need to calculate percent of women ar more than 58.5 tall.

Z = (58.5-66)/2.5 = -3

From Z table we found p-value

P(X>58.5) = 0.9987

So there is 99.87% of women are more than 58.5 tall.

No. Data ovatie =0.68 &value=0.997 Z =-1 Z =0 Z =1 Z =-3 zas z =o Meen=67.5 •c- Mon=63.5 pvalue=0.54L value=0.1587 2=-1 Z= o Meon=63.5 Z=-1 - Z=0 Mon=65.5 2 Valle - 0.9987 ZE 92 Meen 635 Seega