Question

The president of Doerman Distributors, Inc., believes that 30% of the firm's orders come from first-time customers. A simpl random sample of 100 orders will be used to estimate the proportion of first-time customers. Use z-table. a. Assume that the president is correct and p = 0.3. What is the sampling distribution of p for this study? Select your answer - b.What is the probability that the sample proportion will be between 20 and .40 (to 4 decimals)? c. What is the probability that the sample proportion will be between 25 and .35 (to 4 decimals)?

Answer 1

Answer)

P = 0.3

N = 100.

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 30

N*(1-p) = 70

Both the conditions are met so we can use standard normal z table to estimate the probability.

So, the sampling distribution of p is normally distributed with mean = 0.3

And s.d = √{p*(1-p)}/√n = 0.14491376746

B)

z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

P(0.2<x<0.4) = P(x<0.4) - P(x<0.2).

P(x<0.4) = P(z<0.69) = 0.7549

P(x<0.2) = P(z<-0.69) = 0.2451

Required probability is 0.7549 - 0.2451 = 0.5098

C)

P(0.25<x<0.35) = P(x<0.35) - P(x<0.25) = P(z<0.35) - P(z<-0.35) = 0.6368 - 0.3632 = 0.2736