The management of a supermarket wants to adopt a new promotional policy of givin

Question

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit to the supermarket. Market analysts have determined that once the promotion goes into effect the expenditures by all customers will be normally distributed with a mean of $95 and a standard deviation of $21.

If management wants to limit giveaways to 5% of all customers, what amount should management set as the level for the giveaway?

In a randomly selected group of 25 customers, what is the probability the mean amount spent during their visits was more than $105?

Explanation / Answer

Mean is 95 and standard deviation s is 21

z is found as (x-mean)/s

a) 5% limit from the bottom will have z value corresponding to 1-0.05 or 95% but with a negative sign. Looking at the normal distribution table we get it as -1.65

Thus answer is 95-1.65*21 =60.35

b) for 25 customers, the standard error , se is 21/sqrt(25)=4.2

P(x>105) =P(z>(105-95)/4.2)=P(z>2.38) or 1-P(z<2.38)

from normal distribution table we get 1-0.9913=0.0087

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