Question

A study indicates that 30% of American teenagers have tattoos. You randomly sample 5 teenagers. Find the probability that between 3 and 5 will have tattoos. Round your answer to 4 decimal places.

Answer 1

Given that p = 30% = 0.3

q = 1 -p = 1 - 0.3 = 0.7

and also given n = 5

The above information follows binomial distribution

P(X=x) = n_cx p^x q^n-x ; x = 0,1,2, .....n; p + q = 1

Now P(3<X<5) = P(X=4) = 5_c4 (0.3)⁴(0.7)^5-4

= 5(0.0081)(0.7) = 0.0284

Hence P(3<X<5) = 0.0284

Given that p=0.3 q = 0.7; n=5

mean mu = np = 0.3*5 = 0.15

s.d = sqrt(0.3*0.7*5) = 1.0247

p(3<x<5) = p((3-0.15)/1.0247 < (x-mu)/s.d < (5-0.15)/1.0247)=p(2.7813<z<4.7330) =

p(3<x<5) = 0.0027

it is calculated by using standard normal distribution.

but here np<5

so normal is not suitable. therefore binomial is suitable