A popular retail store knows that the distribution of purchase amounts by its cu

Question

A popular retail store knows that the distribution of purchase amounts by its customers is approximately normal with a mean of $30 and a standard deviation of $9. Below you will find probability and percentile calculations related to the customer purchase amounts.

a.         What is the probability that a randomly selected customer will spend less than $15 at this store?

b.         What is the probability that a randomly selected customer will spend $20 or more at this store?

c.         What is the probability that a randomly selected customer will spend $30 or more at this store?

d.         What is the probability that a randomly selected customer will spend between $20 and $35 at this store?

e.         What two dollar amounts, equidistant from the mean of $30, such that 90% of all customer purchases are between these values?

f.          What two dollar amounts, equidistant from the mean of $30, such that 98% of all customer purchases are between these values?

Explanation / Answer

a) The Distribution is normal.

The formula for the z value of the distribution is

z= (x- mean)/ standard deviation

The area under the normal curve represents the total probability of the event occuring.

In the present case, x= $15, mean = $30 , standard deviation= $9

Therefore z= (15-30)/ 9 = - 1.66

The area on the left of z represents , represents the desired probability.

Looking at the normal distribution tables, we find the probability to be = 0.5- .4505 =.0495

b) x= $20 , mean= $ 30 , standard deviation = $9

z= (20-30)/9 = -1.1

The area on the right of z represent the desired probability.

Looking at the z probability distribution table, we find that the required probability is = .3643 + 0.5 = 0.8643

c) $30 represents the center of the distribution

z= 30-30/9 = 0

Thus the desired probability is = 0+0.5 =0.5

d) x=$ 20

z= (20-30)/9 = -1.1

x= $ 35

z= (35-30)/9 = 0.55

Required probability = 0.3643+ .2088= .5731 ( the values have been found by looking at the z probability distribution table)

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