document.write( "Question 1137717: 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 3 minutes.
\n" ); document.write( "​(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?
\n" ); document.write( "​(b) If the automotive center does not want to give the discount to more than 7​% of its​ customers, how long should it make the guaranteed time​ limit?
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Algebra.Com's Answer #755604 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
the mean is 17 minutes and the standard deviation is 3 minutes.\r
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\n" ); document.write( "\n" ); document.write( "the probability that it will take longer than 20 minutes is .1587.\r
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\n" ); document.write( "\n" ); document.write( "that means that almost 16% of the customers would receive half price.\r
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\n" ); document.write( "\n" ); document.write( "if the automotive service does not want to give the discount to more than 7% of its customers, then the guarenteed time limits has to be greater than 21.428 minutes.\r
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\n" ); document.write( "\n" ); document.write( "these figures were derived using the following normal distribution calculator.\r
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\n" ); document.write( "\n" ); document.write( "http://davidmlane.com/hyperstat/z_table.html\r
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\n" ); document.write( "\n" ); document.write( "here's a display of the results.\r
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\n" ); document.write( "\n" ); document.write( "the first display calculates area from a value.\r
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\n" ); document.write( "\n" ); document.write( "the second display calculates value from an area.\r
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\n" ); document.write( "\n" ); document.write( "if you use the calculator with x-scores, the mean is 0 and the standard deviation is 1.\r
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\n" ); document.write( "\n" ); document.write( "if you use the calculator with raw scores, as i did above, the mean is the mean of your distribution and the standard deviation is the standard deviation of your distribution.\r
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