Question

The president of Doerman Distributors, Inc., believes that 32% 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.

Assume that the president is correct and p = 0.32. What is the sampling distribution of  for this study?
SelectA 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 5A non normal distributionItem 1  

What is the probability that the sample proportion will be between .24 and .40 (to 4 decimals)?
  

What is the probability that the sample proportion will be between .29 and .35 (to 4 decimals)?

Answer 1

What is the probability that the sample proportion will be between .24 and .40

The approximation to Normal with   and standard deviation  

P ( 0.24 < X < 0.40 )

Standardizing the value

Z = -1.71

Z = ( 0.4 - 0.32 ) / 0.04665

Z = 1.71

P ( -1.71 < Z < 1.71 )

P ( 0.24 < X < 0.4 ) = P ( Z < 1.71 ) - P ( Z < -1.71 )

P ( 0.24 < X < 0.4 ) = 0.9568 - 0.0432

P ( 0.24 < X < 0.4 ) = 0.9136

What is the probability that the sample proportion will be between .29 and .35

P ( 0.29 < X < 0.35 )

Standardizing the value

Z = ( 0.29 - 0.32 ) / 0.04665

Z = -0.64

Z = ( 0.35 - 0.32 ) / 0.04665

Z = 0.64

P ( -0.64 < Z < 0.64 )

P ( 0.29 < X < 0.35 ) = P ( Z < 0.64 ) - P ( Z < -0.64 )

P ( 0.29 < X < 0.35 ) = 0.7399 - 0.2601

P ( 0.29 < X < 0.35 ) = 0.4798