Question

The first National Health and Nutrition Exam Survey reported that the average body mass index(BMI) in the 25 to 39 age group for women was found to be 24 with a standard deviation of 5. If awoman from this age group is selected at random, what is the probability she has a BMI between 32and 35.Write your probability as a decimal rounded to the nearest ten-thousandth.

Answer 1

the average body mass index (BMI) in the 25 to 39 age group for women was found to be 24 with a standard deviation of 5

From the above, we can obtain this information:

mean: μ = 24

standard deviation: σ = 5

Question

a woman from this age group is selected at random, what is the probability she has a BMI between 32 and 35.

P ( 32

In order to find P ( 32

First, we need to calculate z-value

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

P(x<35)

$Z=\frac{35-24}{5}=2.2$

now, we can use a Z-score table to find the probability having the above z-value of 2.2

P(z<2.2) = 0.9861

Therefore, P(x<35) = 0.9861

P(x<32)

$Z=\frac{32-24}{5}=1.6$

Again, using the Z-score table, for a value of Z=1.6, the probability is

P(z<1.6) = 0.9452

Therefore, P(x<32) = 0.9452

Finally, we must solve P(x<35) - P(x<25) to find P(32this is: 0.9861 - 0.9452 = 0.0409

Which is the same as: P(1.6or P(32

In conclusion, the probability of selecting a woman with an IBM between 32 and 35 is P = 0.0409