Question

A university reported its annual findings about sexually transmitted HPV. The results proved to be quite disheartening, specifically pertaining the HPV rates observed throughout the year. Suppose 25% of women living on campus that year tested positive for HPV, and 10 female students were chosen at random for a new prevention program promoting safe sex practices.

(a) What is the probability of finding at least 3 females with HPV in the sample?

(b) What is the expected number of female students based on the sample?

(c) What is the variance?

Answer 1

Here we have, 25% of women living on campus that year tested positive for HPV. So p = 0.25

10 female students were chosen at random. So n = 10

This is Binomial experiment, pmf is given by,

a) We need to find,

p ( X 3 )

= 1 - p ( x < 3 )

= 1 - { p ( x =0 ) + p ( x = 1 ) + p ( x = 2 )

= 1 - { ¹⁰C₀ * 0.25⁰ * ( 1 - 0.25 ) ^10-0 + ¹⁰C₁ * 0.25¹ * ( 1 - 0.25 ) ^10-1  + ¹⁰C₂ * 0.25² * ( 1 - 0.25 ) ^10-2 }

= 1 - { 0.0563 + 0.1877 + 0.2816 }

= 1 - 0.5256

= 0.4744

b) Expected number = E ( x) = np = 10 * 0.25 = 2.5

c) The variance is given by,

v ( x ) = n* p* ( 1- p ) = 10 * 0.25 * 0.75 = 1.875