The mean amount of gasoline and services charged by Key Refining Company credit customers is $70 per month. The distribution of amounts spent 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?
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A. B. C. D. E.A
z = (x-u)/sigma. z1 = 70 - 70/10 = 0 and z2 = 83 - 70/10 = 1.3. For z = 1.3, the area under the curve is 0.4032.
To solve this problem, we need to use the concept of standard normal distribution.
Given: Mean (μ) = $70 per month Standard deviation (σ) = $10
We want to find the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83.
Step 1: Standardize the values To work with the standard normal distribution, we need to convert the values to z-scores. The z-score formula is:
z = (x - μ) / σ
For the lower value of $70: z1 = ($70 - $70) / $10 = 0
For the upper value of $83: z2 = ($83 - $70) / $10 = 1.3
Step 2: Find the probability We need to find the area under the standard normal curve between the z-scores of 0 and 1.3.
To find this area, we can look up the values in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF) of the standard normal distribution.
Using a calculator, we can find that the cumulative probability corresponding to z = 0 is approximately 0.5 (or 50%). The cumulative probability corresponding to z = 1.3 is approximately 0.9032 (or 90.32%).
The probability of selecting a customer charged between $70 and $83 is the difference between these two probabilities:
P(0 ≤ z ≤ 1.3) = P(z ≤ 1.3) - P(z ≤ 0)
P(0 ≤ z ≤ 1.3) ≈ 0.9032 - 0.5 ≈ 0.4032
Therefore, the correct answer is A. 0.4032.