Question

Suppose that the store chain of Electronics Plus sells extended warranties to 20% of customers who purchase electronic devices. A local Seattle store samples 300 customers from their data base. What is the probability that those sampled purchased between 50 and 65 warranties?

Answer 1

Solution:

We are given n = 300, p = 0.20, q = 1 – p = 0.80

We have to use normal approximation to binomial distribution.

Mean = n*p = 300*0.20 = 60

SD = sqrt(n*p*q) = sqrt(300*0.20*0.80) = 6.92820323

We have to find P(50<X<65)

By using continuity correction we have to find P(49.5<X<65.5)

P(49.5<X<65.5) = P(X<65.5) – P(X<49.5)

Z = (X – mean) / SD

For P(X<65.5)

Z = (65.5 - 60) / 6.92820323

Z = 0.79385662

P(Z<0.79385662) = 0.786360547 (by using z-table/excel)

P(X<65.5) = 0.786360547

Find P(X<49.5)

Z = (49.5 - 60)/ 6.92820323

Z = -1.515544457

P(Z<-1.515544457) = 0.06481729 (by using z-table/excel)

P(X<49.5) = 0.06481729

P(49.5<X<65.5) = P(X<65.5) – P(X<49.5)

P(49.5<X<65.5) = 0.786360547 - 0.06481729

P(49.5<X<65.5) = 0.721543257

Required probability = 0.721543257