The president of Doerman Distributors, Inc., believes that 26% of the firm\'s or

Question

The president of Doerman Distributors, Inc., believes that 26% of the firm's orders come from first-time customers. A simple random sample of 80 orders will be used to estimate the proportion of first-time customers.


Assume that the president is correct and p = 0.26. What is the sampling distribution of for this study?
- Select your answer -A 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 .25 and .35 (to 4 decimals)?

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

Explanation / Answer

ANSWER: Between 0.20 and 0.40 the probability 88.72% [0.8872] Why??? POPULATION PROPORTION, NORMAL DISTRIBUTION, STANDARDIZED VARIABLE z STANDARDIZED VARIABLE: z = (x - µ)/(s/v(n)) 0.20: z = (0.2-0.26)/SQRT(0.26*(1-0.26)/80) [-1.2235] 0.40: z = (0.4-0.26)/SQRT(0.26*(1-0.26)/80) [2.8548] The Table for Standard Normal Distribution is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. For STANDARDIZED VARIABLE z = -1.2235 the corresponding area = 0.1106 and for z = 2.8548. For the interval (-1.2235

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