Question

One measure of general health is your body mass index (BMI). Adults with a BMI between 18.5 and 24.9 are generally considered to have an ideal body weight for their height. The mean BMI for adults in a certain country is 24.5. Suppose the BMI for adults in this country is normally distributed with standard deviation 1.2. (Round your answers to four decimal places.) Suppose one adult from this country is selected at random. What is the probability that the person's BMI is more than 25.6?

Answer 1

Answer: 0.1797

Step-by-step explanation:

Given : $\mu=24.5\ \ ; \sigma=1.2$

Let x be the random variable that represents BMI for adults in this country.

We assume that  BMI for adults in this country is normally distributed.

Then, z-score for x=25.6 ,$\because z=\dfrac{x-\mu}{\sigma}$

$\Rightarrow\ z=\dfrac{25.6-24.5}{1.2}\approx0.9167$

Using z-value table , we have

The probability that the person's BMI is more than 25.6 :

$P(x>25.6)=P(z>0.9167)=1-P(z\leq0.9167)\\\\=1- 0.8203413=0.1796587\approx0.1797$

Hence, the probability that the person's BMI is more than 25.6 = 0.1797

Hence, the probability that the person's BMI is more than 25.6 = 0.1796

Hence, the  probability of the stick's weight being 2.23 oz or greater = 0.0174