Question

1. BMI or body mass index is a way to classify people according to their height and weight. For all state employees, the state of North Carolina is making everyone who has a BMI greater than 30 pay extra for their health insurance. Assume that BMI is approximately a normal distribution with a mean of 27.73 and a standard deviation of 6.10 .

a. What percent of the state employees would have to pay extra for health insurance?

b. Suppose that the state decides to make the top 5% of all BMI scores to pay extra for health insurance. What BMI score separates the top 5% from the bottom 95%?

Answer 1

Solution :

Given that,

mean = = 27.73

standard deviation = = 15

a ) P (x > 30 )

= 1 - P (x < 30 )

= 1 - P ( x -  / ) < ( 30 - 27.73 / 6.10 )

= 1 - P ( z <- 2.27/ 6.10 )

= 1 - P ( z < -0.37)

Using z table

= 1 - 0.3557

= 0.6443

Probability = 0.6443 =64.43%

P(Z < z) = 95%

= P(Z < z) = 0.95  

= P(Z < 1.645 ) = 0.95

z =1.64

Using z-score formula,

x = z * +

x = 1.64* 6.10 + 27.73

x = 37.734

BMI score = 37.73