Question

3. Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. Many online BMI calculators allow you to enter weight in pounds and height in inches. High BMI is a common but controversial indicator of overweight or obesity. A study by the National Center for Health Statistics found that the BMI of American young men (ages 20 to 29) is approximately Normal with mean 26.3 and standard deviation 5.4. a. People with BMI less than 18.1 are often classified as ""underweight."" What percent of men aged 20 to 29 are underweight by this criterion? b. People with BMI over 31 are often classified as ""obese."" What percent of men aged 20 to 29 are obese by this criterion?

Answer 1

Answer:

a) $P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

That correspond to approximate 6.44%

b) $P(X>31)=P(\frac{X-\mu}{\sigma}>\frac{31-\mu}{\sigma})=P(Z>\frac{31-26.3}{5.4})=P(Z>0.87)$

And we can find this probability using the normal standard table or excel and we got:

$P(Z>0.87)=1-P(Z$

That correspond to approximate 19.2%

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the BMI of a population, and for this case we know the distribution for X is given by:

$X \sim N(26.3,5.4)$  

Where $\mu=26.3$ and $\sigma=5.4$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

That correspond to approximate 6.44%

Part b

We are interested on this probability

$P(X>31)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>31)=P(\frac{X-\mu}{\sigma}>\frac{31-\mu}{\sigma})=P(Z>\frac{31-26.3}{5.4})=P(Z>0.87)$

And we can find this probability using the normal standard table or excel and we got:

$P(Z>0.87)=1-P(Z$

That correspond to approximate 19.2%