Chapter 5.3, Problem 107E

(a)

To determine

Percent of young women are underweight according to the criterion.

Answer to Problem 107E

Around 13.14% of the young women are underweight.

Explanation of Solution

Given information:

Mean, $\mu =26.8$

Standard deviation, $\sigma =7.4$

BMI, $x=18.5$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{18.5-26.8}{7.4}\approx -1.12$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.1 and the column that starts with .02 of the standard normal probability table for $P\left(z<-1.12\right)$ .

  $\begin{array}{l}P\left(x<18.5\right)=P\left(z<-1.12\right)\\ =0.1314\\ =13.14\%\end{array}$

Thus,

Around 13.14% of the young women have a BMI (Body Mass Index) less than 18.5 and are underweight.

(b)

To determine

Probability for at least one of the two American young women is classified as underweight.

Answer to Problem 107E

Probability that at least one of the two American young women is classified as underweight is 0.2455.

Explanation of Solution

Given information:

Mean, $\mu =26.8$

Standard deviation, $\sigma =7.4$

BMI, $x=18.5$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{18.5-26.8}{7.4}\approx -1.12$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.1 and the column that starts with .02 of the standard normal probability table for $P\left(z<-1.12\right)$ .

  $P\left(x<18.5\right)=P\left(z<-1.12\right)=0.1314$

A: One American young woman is underweight

B: At least one of the two American women is underweight

A^c: One American young women is not underweight

B^c: Neither of the two American young women is underweight

Now,

The probability for one American young woman is underweight,

  $P\left(\text{A}\right)=0.1314$

The probability for one American is not underweight,

Apply the complement rule:

  $P\left({\text{A}}^{\text{c}}\right)=1-P\left(\text{A}\right)=1-0.1314=0.8686$

Since the American young women are selected at random, it would be more convenient to assume that the women are independent of each other.

Thus,

For probability that neither of the two American young women is underweight, apply multiplication rule for independent events:

  $P\left({\text{B}}^{\text{c}}\right)=P\left({\text{A}}^{\text{c}}\right)\times P\left({\text{A}}^{\text{c}}\right)={\left(P\left({\text{A}}^{\text{c}}\right)\right)}^2={\left(0.8686\right)}^2\approx 0.7545$

Apply the complement rule:

  $P\left(\text{B}\right)=P\left({\left({\text{B}}^{\text{c}}\right)}^{\text{c}}\right)=1-P\left({\text{B}}^{\text{c}}\right)=1-0.7545=0.2455$

Thus,

Probability for at least one of the two American young women is classified as underweight is 0.2455.