Question

Based on historical data, your manager believes that 44% of the company's orders come from first-time customers. A random sample of 141 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.48?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Solution:

We are given

n = 141

p = 0.44

q = 1 – p = 1 – 0.44 = 0.56

We have to find P(0.26<P<0.48)

Here, we have to use normal approximation to binomial distribution.

np = 141*0.44 = 62.04

nq = 141*0.56 = 78.96

np and nq > 5, so we can use normal approximation.

Mean = np = 62.04

SD = sqrt(npq) = sqrt(141*0.44*0.56) = 5.894268

P(0.26<P<0.48) = P(141*0.26 < X < 141*0.48) = P(36.66 < X < 67.68)

P(36.66 < X < 67.68) = P(X<67.68) – P(X<36.66)

Find P(X<67.68)

Z = (X – mean) / SD

Z = (67.68 - 62.04) / 5.894268

Z = 0.956862

P(Z<0.956862) = P(X<67.68) = 0.830681

(by using z-table)

Now find P(X<36.66)

Z = (X – mean) / SD

Z = (36.66 - 62.04) / 5.894268

Z = -4.30588

P(Z<-4.30588) = P(X<36.66) = 0.000008316

(by using z-table or excel)

P(36.66 < X < 67.68) = P(X<67.68) – P(X<36.66)

P(36.66 < X < 67.68) = 0.830681 - 0.000008316

P(36.66 < X < 67.68) = 0.830672684

Required probability = 0.8307