A study indicates that 4 of American teenagers have tattoos

A study indicates that 4% of American teenagers have tattoos. You randomly sample 30 teenagers. What is the likelihood that exactly 3 will have a tattoo? What is the likelihood that at least 3 will have a tattoo?

Solution

Let X be a random variable American teenagers have tattoos.

X follows Binomial distribution with parameters n and p.

where n are number of trials = 30

p = probability of success = 4% = 0.04

P(X = x) = (n C x) * p^x * q^(n-x) (C is used for combination)

q = 1 - p = 1 - 0.04 = 0.96

X ~ Bin( 30, 0.04 )

Here likelihood means we have to calculate probability.

a) likelihood that exactly 3 will have a tattoo.

P(X = 3) =  (30 C 3) * 0.04^3 * 0.96^(30-3)

= 4060 * 0.000064 * 0.3321

= 0.0863

likelihood that at least 3 will have a tattoo.

That is we have to find the P(X 3) = 1 - P(X < 3)

= 1 - [ P(X = 0) + P(X = 1) + P(X = 2) ]

P(X = 0) = (30 C 0) * 0.04^0 * 0.96^(30-0) = 0.2939

P(X = 1) = (30 C 1) * 0.04^1 * 0.96^(30-1) = 0.3673

P(X = 2) = (30 C 2) * 0.04^2 * 0.96^(30-2) = 0.2219

= 1 - [ 0.2939 + 0.3673 + 0.2219 ]

= 1 - 0.8831

P(X 3) = 0.1169