Question

A study revealed that 75% of students from heavy-smoking families showed signs of nasal allergies on physical exams. Consider a sample of 30 students exposed daily to heavy smoking.

What is the probability that fewer than 15 of the students will have nasal allergies?

What is the probability that more than 20 of the students will have nasal allergies?

On average, how many of the 30 students would you expect to show signs of nasal allergies?

Answer 1

X ~ binomial (n,p)

Where n = 30, p = 0.75

Mean = np = 30 * 0.75 = 22.5

Standard deviation SD = Sqrt(np(1-p))

= Sqrt( 30 * 0.75 * 0.25 )

= 2.3717

a)

Using normal approximation to binomial distribution,

P( X < x) = P( Z < x-0.5 - Mean / SD )

So,

P( X < 15) = P( Z < 14.5 - 22.5 / 2.3717)

= P( Z < -3.3731)

= 1 - P( Z < 3.3731)

= 1 - 0.9996

= 0.0004

b)

P( X > x) = P( Z > x+0.5 - Mean / SD)

So,

P( X > 20) = P( Z > 20.5 - 22.5 / 2.3717)

= P( Z > -0.8433)

= P( Z < 0.8433)

= 0.8005

c)

Expected value = np

= 30 * 0.75

= 22.5