The mean and standard deviation of the amount customers spen

The mean and standard deviation of the amount customers spend in a high end clothing store is $600 and $150 respectively. Assume the amount of money customers spend is normally distributed.

(a) What is the probability that 5 randomly selected customers spend over $3500?

(b) If we randomly sample 49 people, what is the probability they spend less than $500 on average?

(c) What is the 90th percentile of total amount spent for 10 randomly selected people?

(d) Did we need to assume the population was normally distributed in parts (a) and (b)? Explain why or why not.

Solution

Mean =600 and sigma = 150

P(spending > 3500) = P(Z>2900/150/rt5) = 0.00

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b) n =49

P(X<500) = P(Z<-1.49) = 0.0694

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c) 90th percentile z = 2.33

x = 600+150*2.33/rt10 = 710.53

Yes, we assumed population was normally distributed as we used std normal variate score for that.