Question

Customers of the Key Refining Company charge an average of $70 per month. The distribution of amounts charged is approximately normal, with a standard deviation of $10. What is the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83

Answer 1

0.4032 is the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83.

Explanation:

We are given the following information in the question:

Mean, μ = $70

Standard Deviation, σ = $10

We are given that the distribution of amounts charged is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

P(customer charged between $70 and $83)

`P(70 \leq x \leq 83) = P(\displaystyle(70 - 70)/(10) \leq z \leq \displaystyle(83-70)/(10)) = P(0 \leq z \leq 1.3 )\\\\= P(z \leq 1.3) - P(z < 0)\\= 0.9032 - 0.500 = 0.4032= 40.32\%`

`P(70 \leq x \leq 83) = 40.32\%`

Thus, 0.4032 is the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83.