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The average waiting time for a drive-in window at a local bank is 9.8 minutes, with a standard deviation of 2.8 minutes. What percentage of customers will have to wait the given time? Assume the random variable is normally distributed. Refer to the table of values (Area Under the Standard Normal Distribution) as needed. Part: 0/2 Part 1 of 2 (a) Between 4.2 and 15.4 minutes The percentage of customers who will have to wait between 4.2 and 15.4 minutes is %.

The average waiting time for a drive-in window at a local bank is 9.8 minutes, with a standard deviation of 2.8 minutes. What percentage of customers will have to wait the given time? Assume the random variable is normally distributed. Refer to the table of values (Area Under the Standard Normal Distribution) as needed. Part: 0/2 Part 1 of 2 (a) Between 4.2 and 15.4 minutes The percentage of customers who will have to wait between 4.2 and 15.4 minutes is %.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.CR: Chapter 13 Review
Problem 58CR
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The average waiting time for a drive-in window at a local bank is 9.8 minutes, with a standard deviation of 2.8 minutes. What percentage of customers will have to wait the given time? Assume the random variable is normally distributed. Refer to the table of values Area Under the Standard Normal Distribution) as needed. Part: 0/2 Part 1 of 2 (a) Between 4.2 and 15.4 minutes The percentage of customers who will have to wait between 4.2 and 15.4 minutes is X S
Transcribed Image Text:The average waiting time for a drive-in window at a local bank is 9.8 minutes, with a standard deviation of 2.8 minutes. What percentage of customers will have to wait the given time? Assume the random variable is normally distributed. Refer to the table of values Area Under the Standard Normal Distribution) as needed. Part: 0/2 Part 1 of 2 (a) Between 4.2 and 15.4 minutes The percentage of customers who will have to wait between 4.2 and 15.4 minutes is X S
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