Question

BMI (2.2, 5.2, 5.3) Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. Online BMI calculators allow you to enter weight in pounds and height in inches. High BMI is a common but controversial indicator of being overweight or obese. A study by the National Center for Health Statistics found that the BMI of American young women (ages 20 to 29) is approximately Normally distributed with mean 26.8 and standard deviation 7.4.27

a. People with BMI less than 18.5 are often classed as “underweight.” What percent of young women are underweight by this criterion?

b. Suppose we select two American young women in this age group at random. Find the probability that at least one of them is classified as underweight.

Short answer

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

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

Step-by-step solution

Part b) Step 1: Given information

Mean $\mu =26.8$

Standard deviation, $\sigma =7.4$

BMI$x=18.5$

Part b) Step 2: Calculation

Calculate the z - score,

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

To find the corresponding probability, use the normal probability table in the appendix.

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

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

$A:$A young woman from the United States is underweight.

$B:$At least one of the two women from the United States is underweight.

$A^c:$One American young woman is not underweight.

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

One American young woman is now at risk of being underweight.

$P(\mathrm{A})=0.1314$

Apply the complement rule to determine whether or not one American is underweight:

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

Because the young women from the United States were chosen at random, it would be more convenient to assume that they are all independent of one another.

Therefore,

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

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

Apply the complement rule:

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

Therefore,

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

Part a) Step 1: Given information

Mean, $\mu =26.8$

Standard deviation,$\sigma =7.4$

Part b) Step 2: Calculation

Calculate the z-score,

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

To find the corresponding probability, use the normal probability table in the appendix

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

$$\begin{gathered} P(x<18.5)=P(z<-1.12) \\ =0.1314 \\ =13.14\% \end{gathered}$$

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