Question

The president of Doerman Distributors, Inc., believes that 30% of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers.

(a) Assume that the president is correct and p = 0.30. What is the sampling distribution of p for n = 100? (Round your answer for σp to four decimal places.)

σp =

E(p) =

Since np = and n(1 − p) = , approximating the sampling distribution with a normal distribution appropriate (IS or IS NOT) in this case.

(b) What is the probability that the sample proportion p will be between 0.20 and 0.40? (Round your answer to four decimal places.)

(c) What is the probability that the sample proportion will be between 0.25 and 0.35? (Round your answer to four decimal places.)

Answer #1

This is a binomial distribution question with

n = 100

p = 0.30

q = 1 - p = 0.7

This binomial distribution can be approximated as Normal distribution since

np > 5 and nq > 5

a)

Since we know that

b)

P(0.20 < p < 0.40) = P(0.20*100 < p*n < 0.40*100) = P(20.0 < x < 40.0) = ?

This implies that

P(20.0 < x < 40.0) = P(-2.1822 < z < 2.1822) =
**0.9709**

c) P(25.0 < x < 35.0)=?

This implies that

P(25.0 < x < 35.0) = P(-1.0911 < z < 1.0911) =
**0.7248**

Please hit thumps up if the answer helped you

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