Need Help in Statistics!!

Question

Suppose body mass index (BMI) varies approximately to the normal distribution in a population of boys aged 2-20.

A national survey analyzed the BMI for American adolescents in this age range and found the µ=17.8 and the σ=1.9.

 

1. At most, what percentage of BMI is less than 10.2? (Using Chebyshev's theorem.)
2. What is the probability of randomly selecting an obese child (whose BMI is greater than 21.6) based on a sample of 15 children?

 

Answer 1

1. We need to first define the probability that we are looking for:

X - (random variable X) is the BMI for boys aged 2-20

µ=17.8  

σ=1.9

P(X ≤ µ-a)≤ σ²/(σ² + a²)  This is one sided Chebyshev’s theorem

First we need to find the value of a:

µ-a = 10.2

17.8-a = 10.2

a = 7.6

P(X ≤ 17.8-7.6)≤1.9²/(1.9² + 7.6²) = .0588

2. This problem does not say to use Chebyshev’s theorem, so we will use the normal distribution with the appropriate z score.

P(X ≥ 21.6)

P(x ≥ 21.6) = 1 - P(z ≤ (21.6 - 17.8)/(1.9/√15))

= 1 - P(z ≤ 7.746) = 0.000

If we use Chebyshev’s theorem:

a = 3.8

P(X ≥ 17.8+3.8)≤1.9²/(1.9² + 3.8²) = .20