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The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is%. (Round to two decimal places as needed.) minutes. (b) The guaranteed time limit is (Round up to the nearest minute.)

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is%. (Round to two decimal places as needed.) minutes. (b) The guaranteed time limit is (Round up to the nearest minute.)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Standard Normal Distribution Table (page 2) 0.02 Standard Norma Print istribution 0.9997 0.9997 Done 0.999 0.5319 0.5714 0.6180 0.6844 0.7190 0.751 0.8365 0.8599 0.8810 0.8991 0.0206 09420 0.9535 0.9625 0.0761 0.9854 0.9887 0.9986 0.09 0.3753 OCE 0.6870 0.7224 0.700 08280 n I 4 X Standard Normal Distribution Table (page 1) 0.01 0:000- 0.01 0.02 0.4129 0.02 Standard Normal Distribution 0.03 0.04 0.05 0.3330 04483 0.03 Print 0.06 0.0003 0.0003 0.0004 0.0004 0.0009 0.0000 0.0000 0:000 0.0011 0.0011 0.0016 0.0015 0.0021 0.0020 0.0022 0.0000 0.0054 0.0071 0.0094 0220 0.261 0.2940 0.3300 0.1711 0.2266 0.2578 0.2912 0.3264 04052 04012 0.4443 0.4404 0.4840 0.4801 0.05 0.04 Done 0.2236 03074 0.4364 0.06 0.0375 0.040 0.0838 0.1003 0.1190 01635 0 1804 0.2177 0.2483 0.2810 0.3156 04286 0.468 0.08 0.09 0.0000 www 0.0010 0.0014 0:0019 0.0000 0.0048 0.0064 0.0084 0.0110 0.0197 0.0221 0.0294 0.0367 ww 0.0932 0.0985 0.11A 01067 02148 0.2770 0.3121 0.3850 0.09 I X
Transcribed Image Text:Standard Normal Distribution Table (page 2) 0.02 Standard Norma Print istribution 0.9997 0.9997 Done 0.999 0.5319 0.5714 0.6180 0.6844 0.7190 0.751 0.8365 0.8599 0.8810 0.8991 0.0206 09420 0.9535 0.9625 0.0761 0.9854 0.9887 0.9986 0.09 0.3753 OCE 0.6870 0.7224 0.700 08280 n I 4 X Standard Normal Distribution Table (page 1) 0.01 0:000- 0.01 0.02 0.4129 0.02 Standard Normal Distribution 0.03 0.04 0.05 0.3330 04483 0.03 Print 0.06 0.0003 0.0003 0.0004 0.0004 0.0009 0.0000 0.0000 0:000 0.0011 0.0011 0.0016 0.0015 0.0021 0.0020 0.0022 0.0000 0.0054 0.0071 0.0094 0220 0.261 0.2940 0.3300 0.1711 0.2266 0.2578 0.2912 0.3264 04052 04012 0.4443 0.4404 0.4840 0.4801 0.05 0.04 Done 0.2236 03074 0.4364 0.06 0.0375 0.040 0.0838 0.1003 0.1190 01635 0 1804 0.2177 0.2483 0.2810 0.3156 04286 0.468 0.08 0.09 0.0000 www 0.0010 0.0014 0:0019 0.0000 0.0048 0.0064 0.0084 0.0110 0.0197 0.0221 0.0294 0.0367 ww 0.0932 0.0985 0.11A 01067 02148 0.2770 0.3121 0.3850 0.09 I X
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is %. (Round to two decimal places as needed.) (b) The guaranteed time limit is (Round up to the nearest minute.) minutes. C
Transcribed Image Text:The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The percent of customers that receive the service for half-price is %. (Round to two decimal places as needed.) (b) The guaranteed time limit is (Round up to the nearest minute.) minutes. C
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