Question 1085208: SUPPOSE THAT DURING ANY HOUR IN A LARGE DEPARTMENT STORE, THE AVERAGE NUMBER OF SHOPPERS IS 448 , WITH A STANDARD DEVIATION OF 21 SHOPPERS. WHAT IS THE PROBABILITY OF RANDOMLY SELECTING 49 DIFFERENT SHOPPING HOURS, COUNTING THE SHOPPERS, AND HAVING THE SAMPLE MEAN FALL BETWEEN 441 AND 446 SHOPPERS, INCLUSIVE ?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! SUPPOSE THAT DURING ANY HOUR IN A LARGE DEPARTMENT STORE, THE AVERAGE NUMBER OF SHOPPERS IS 448 , WITH A STANDARD DEVIATION OF 21 SHOPPERS. WHAT IS THE PROBABILITY OF RANDOMLY SELECTING 49 DIFFERENT SHOPPING HOURS, COUNTING THE SHOPPERS, AND HAVING THE SAMPLE MEAN FALL BETWEEN 441 AND 446 SHOPPERS, INCLUSIVE ?
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z(441) = (441-448)/(21/sqrt(49)) = -7/7 = -1
z(446) = (446-448)/(21/sqrt(49)) = -2/7
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Ans:: P(441<= x <= 446) = P(-1<= z <=-2/7) = normalcdf(-1,-2/7) = 0.2289
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Cheers,
Stan H.
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