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The president of Doerman Distributors, Inc., believes that 26% 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.

1. Assume that the president is correct and p = 0.26. 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
   
2. What is the probability that the sample proportion will be between .12 and .40 (to 4 decimals)?
   
   
3. What is the probability that the sample proportion will be between .17 and .35 (to 4 decimals)?

Answer 1

Given that, population proportion (p) = 0.26 and

sample size (n) = 100

a) np = 100 * 0.26 = 26 and n(1-p) = 100 * (1 - 0.26) = 74

Therefore, sampling distribution of sample proportion is approximately normal because np and n(1-p) are both greater than 5.

b) We want to find,

Therefore, required probability is 0.9986

c) We want to find,

Therefore, required probability is 0.9596