Question

6.57 Working on summer vacation. According to an Adweek/ Harris (July 2011) poll of U.S. adults, 35% do not work during their

(a) What is the mean and standard deviation of p? (b) What is the probability that 70% or more of sampled adults do not work

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Answer #1

a) Mean is calculated as

n*p

=500*0.35

=175

p(1-p) n

0.35(1 0.35)- 0.0213 500

b) Since the sample is large enough to consider it as normal distribution hence probability is calculated using Z calculation as:

0.70 0.35 Р—р 16.432 p(1-р) 0.0213 n

P(\hat{p}>0.70) =P(Z>16.432)=0.000 computed using Z table shown below

c) Now P(\hat{p} >Po )=0.01 we need to find Z score at Po    using Z table shown below:

Z=2.33 so using Z formula as:

2.33 - Ро - 0.35 0.0213 Р—р — ро — 0.40 Ро р(1-р) п

Table of Standard Normal Probabilities for Negative Z-scores Table of Standard Normal Probabilities for Positive Z-scores 0.0

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