Question

A recent study reported that the prevalence of hyperlipidemia (defined as total cholesterol over 200) is 30% in children 2-6 years of age. If 12 children are analyzed.

Required:
a. What is the probability that at least 3 are hyperlipidemic?
b. What is the probability that exactly 3 are hyperlipidemic?
c. How many would be expected to meet the criteria for hyperlipidemia?

Answer 1

(a) 0.7472

(b) 0.2397

(c) 3.6

Explanation:

Let X denote the number of children who are hyperlipidemic.

The proportion of children who are hyperlipidemic is, p = 0.30.

A random sample of n = 12 children are analyzed.

Every child is independent of the others to be expected to meet the criteria for hyperlipidemia.

The random variable X follows a binomial distribution with parameters n = 12 and p = 0.30.

(a)

Compute the probability that at least 3 are hyperlipidemic as follows:

`P(X\geq 3)=1-P(X<3)\\\\=1-\sum\limits^(2)_(0){{12\choose x}(0.30)^(x)(0.70)^(12-x)}\\\\=1-0.25282\\\\=0.74718\\\\\approx 0.7472`

Thus, the probability that at least 3 are hyperlipidemic is 0.7472.

(b)

Compute the probability that exactly 3 are hyperlipidemic as follows:

`P(X=3)={12\choose 3}(0.30)^(3)(0.70)^(12-3)\\\\=220* 0.027* 0.040353607\\\\=0.23970042558\\\\\approx 0.2397`

Thus, the probability that exactly 3 are hyperlipidemic is 0.2397.

(c)

Compute the expected number of children who would meet the criteria for hyperlipidemia as follows:

`E(X)=np=12* 0.30=3.6`

Thus, 3.6 children would be expected to meet the criteria for hyperlipidemia.