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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(customer charged between $70 and $83)

$P(70 \leq x \leq 83) = P(\displaystyle\frac{70 - 70}{10} \leq z \leq \displaystyle\frac{83-70}{10}) = P(0 \leq z \leq 1.3 )\\\\= P(z \leq 1.3) - P(z < 0)\\= 0.9032 - 0.500 = 0.4032= 40.32\%$

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

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