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Assume that customers expect a transaction to be completed in less than 7.5 minutes. If the process completion time is Normally distributed with a mean of 10 minutes and a standard deviation of 1 minute, What percentage of customers will be satisfied? (Hint: Draw a picture of the Normal distribution described in the question).

A. Greater than 5.0%
B. Less than 2.5%
C. Greater than 7.5%
D. No customers will be satisfied
E. Cannot be determined

User Ben Hunter
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Final answer:

When transaction times are normally distributed with a mean of 10 minutes and a standard deviation of 1 minute, the percentage of customers who will be satisfied with a transaction time of less than 7.5 minutes is less than 2.5%.

Step-by-step explanation:

The question asks to find the percentage of customers who will be satisfied with a transaction time of less than 7.5 minutes when the transaction times are normally distributed with a mean of 10 minutes and a standard deviation of 1 minute. To find this probability, we need to calculate the z-score of 7.5 minutes using the formula z = (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation. The z-score represents how many standard deviations an element is from the mean. In this case, z = (7.5 - 10) / 1 = -2.5. We then look up the z-score in the standard normal distribution table to find the probability of a customer being satisfied. Since a z-score of -2.5 corresponds to a very small probability (less than 0.5%), we can conclude that the percentage of satisfied customers will be less than 2.5%.

User Zephni
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