Chapter PIV, Problem 15RE

(a)

To determine

To Explain: the indication of the larger standard deviation for the men’s heights.

Explanation of Solution

The following table shows the mean and standard deviation of males and females who are in the Bean club.

	Mean	SD
Male	69.1	2.8
Female	64.0	2.5

The variation in mean heights is more than the difference in the heights of females.

(b)

To determine

To Explain: the men or women are more likely to apply for membership of Beanstalk.

Explanation of Solution

Formula used:

  $Z=\frac{X-{\mu }_M}{{\sigma }_M}$

Calculation:

Suppose that X is that man is qualified for the membership.

The random variable X which follows the normal distribution with mean ${\mu }_M=69.1$ land standard deviation ${\sigma }_M=2.8$ .

Converting feet into inches.

Thus 6 feet = 72 inches

6 feet 2 inches = 74 inches.

For X = 74 the z-score is,

  $\begin{array}{c}Z=\frac{X-{\mu }_M}{{\sigma }_M}\\ =\frac{74-69.1}{2.8}\\ =1.75\end{array}$

Let Y be that women are qualified for the membership.

The random variable Y which is following the normal distribution with mean ${\mu }_W=64$ and standard deviation ${\sigma }_W=2.5$ .

Converting the feet into inches.

Thus, 5 feet = 60 inches

5 feet 10 inches = 70 inches

For X = 70 the z-score is,

  $\begin{array}{c}Z=\frac{X-{\mu }_W}{{\sigma }_W}\\ =\frac{70-64}{2.5}\\ =2.4\end{array}$

Here, the mean is larger likely to qualify since this qualifying height is just 1.75 standard deviation just above mean whereas the qualifying height of the female is 2.4 standard deviation just above mean.

(c)

To determine

To Explain: two random variables and use them to show how many inches taller a man is than a woman.

Explanation of Solution

Suppose

M is a random married man 's height.

W is the random height of married women.

The word of how taller the man is M-W

(d)

To determine

To find: the mean of the difference.

Explanation of Solution

The Hoe Mean is several inches taller than the man,

. $\begin{array}{c}E\left(M-W\right)=E\left(M\right)-E\left(W\right)\\ {\mu }_{M-W}={\mu }_M-{\mu }_W\\ {\mu }_{M-W}=69.1-64\\ {\mu }_{M-W}=5.1\end{array}$

Therefore, the mean of 5.1 inches is how many inches taller than the man.

(e)

To determine

To find: the standard deviation of the difference.

Answer to Problem 15RE

3.75 inches

Explanation of Solution

Finding the Standard deviation of difference.

variance of the difference.

  $\begin{array}{c}\mathrm{var}\left(M-W\right)=\mathrm{var}\left(M-W\right)\\ {\sigma }_{M-W}^2=\mathrm{var}\left(M\right)+\mathrm{var}\left(W\right)\left(M\text{ and W are independent}\right)\\ {\sigma }_{M-W}^2={\left({\sigma }_M\right)}^2+{\left({\sigma }_W\right)}^2\\ {\sigma }_{M-W}^2={\left(2.8\right)}^2+{\left(2.5\right)}^2\\ {\sigma }_{M-W}^2=14.09\end{array}$

Difference of standard deviation is

  $\begin{array}{c}SD\left(M-W\right)={\mu }_{M-W}\\ =\sqrt{{\sigma }_{M-W}^2}\\ =\sqrt{14.09}\\ =3.75\end{array}$

Thus, the standard deviation of difference = 3.75 inches.

(f)

To determine

To Find: the possibility that a man is taller than a woman.

Answer to Problem 15RE

0.9131

Explanation of Solution

Given:

  ${\mu }_{M-W}=5.1$ and ${\sigma }_{M-W}=3.75$

Calculation:

Suppose that the difference is represented by d = M − W.

The Required probability that the man is,

  $\begin{array}{c}P\left(M-W>0\right)=P\left[d>0\right]\\ =P\left[\frac{d-{\mu }_{M-W}}{{\sigma }_{M-W}}>\frac{0-{\mu }_{M-W}}{{\sigma }_{M-W}}\right]\\ =P\left[Z>-1.36\right]\\ =1-0.0879\left(\text{From normal tables}\right)\\ =0.9131\end{array}$

The probability, therefore, that a man is taller than a woman is 0.9131

(g)

To determine

To explain: based on the response to part (f), it is assumed that the choice of spouses by individuals is independent of height.

Explanation of Solution

Yes, the choice of spouses by people is independent of height.

Since it can be seen from part (f) that 91.3 percent of men are taller than the women they marry. So, 92 percent of the survey findings are roughly justified.