I don\'t understand how to work the followingproblem - Suppose the average check

Question

   I don't understand how to work the followingproblem - Suppose the average checkout tab at a large supermarket is$65.12, with a standard deviation of $21.45. 23% of the timewhen a random sample of 45 customer tabs is examined, the sampleaverage should exceed what value? Please help    I don't understand how to work the followingproblem - Suppose the average checkout tab at a large supermarket is$65.12, with a standard deviation of $21.45. 23% of the timewhen a random sample of 45 customer tabs is examined, the sampleaverage should exceed what value? Please help

Explanation / Answer

Given =65.12, =21.45, n=45 P(Xbar > c)=0.23 -->P([Xbar - ]/[/n] >[c-65.12]/[21.45/45])=0.23 --> [c-65.12]/[21.45/45]=-0.7388 (check normal table) --> c= 65.12 -0.7388* (21.45/sqrt(45)) = 62.75763

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