A major department store has determined that its customers charge an average of $500 per month, with a standard deviation of $80. Assume the amounts of charges are normally distributed.

a. what percentage of customers charges more than $380 per month?
b. what percentage of customers charges less than $340 per month?
c. What percentage of customers charges between $644 and $700 per month?

Step-by-step explanation:

Since the amounts of charges are assumed to be normally distributed,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the amounts of charges.

µ = mean amount

σ = standard deviation

From the information given,

µ = $500

σ = $80

a) the probability that customers charges more than $380 per month is expressed as

P(x > 380) = 1 - P(x ≤ 380)

For x = 380,

z = (380 - 500)/80 = - 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.067

P(x > 380) = 1 - 0.067 = 0.933

The percentage of customers that charges more than $380 per month is

0.933 × 100 = 93.3%

b) the probability that customers charges less than $340 per month is expressed as

P(x < 340)

For x = 340,

z = (340 - 500)/80 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

The percentage of customers that charges less than $340 per month is

0.023 × 100 = 2.3%

c) the probability that customers charges between $644 and $700 per month is expressed as

P(644 ≤ x ≤ 700)

For x = 644,

z = (644 - 500)/80 = 1.8

Looking at the normal distribution table, the probability corresponding to the z score is 0.96

For x = 700,

z = (700 - 500)/80 = 2.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.994

P(644 ≤ x ≤ 700) = 0.994 - 0.96 = 0.034

The percentage of customers that charges between $644 and $700 per month is

0.034 × 100 = 3.4%