Question

The president of Doerman Distributors, Inc., believes that 30 percent of the firm's orders come from first-time customers. A simple 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 bar for this study?

A non normal distributionA normal distribution because np and n(1-p) are both greater than 5A normal distribution because np and n(1-p) are both less than 5

b.What is the probability that the sample proportion will be between .20 and .40 . and (to 4 decimals)?

c. What is the probability that the sample proportion will be between .25 . and .35 (to 4 decimals)?

Answer 1

Given that h=l00 P 0.30, a) = 30 >5 100 x0.30 np- 100 x O30xo.70 = 70>5 distaf bution There fale, ê Ps a nalmal Me bo th ncl-p) becaure and np goealco tha w Q) meaw P 0.30 0.30X0:70 P(I-P) Standond deviation= loo Y 0.046 b) p (0.20 <p co.40) P-P, 0120 -0.3 0 0.40-0.30 10.30x0.70 /PC(-P) 0.30X0.70 100 100 r(-2.18 <42.18) -0.9.70 % foomm ztable c) plonas<p co.35) P-P, 0,35-0.30 0.25-0.30 0.30 X0.70 PlI-P) 0-30x0.70 100 -P(-1:09<2<1.09) = 07742| from ztable