document.write( "Question 946898: The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.5 miles per gallon.
\n" ); document.write( "A. What is the probability that a randomly selected hybrid gets more that 61 miles per gallon? 0.127
\n" ); document.write( "B. What is the probability that a randomly selected hybrid gets 52 miles per gallon or less?
\n" ); document.write( "C. What is the probability that a randomly selected hybrid gets between 58 and 61 miles per gallon? 0.261
\n" ); document.write( "D. What is the probability that a randomly selected hybrid gets less than 45 miles per gallon?
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Algebra.Com's Answer #577677 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Find the z score for each scenario, then use the normal distribution function.
\n" ); document.write( "\"z=%28x-mu%29%2Fsigma=%28x-57%29%2F3.5\"
\n" ); document.write( "a) \"z=%2861-57%29%2F3.5=1.143\"
\n" ); document.write( "\"0.8735\", so then it's \"1-0.8735=0.127\" for greater than 61.
\n" ); document.write( "b) \"z=%2852-57%29%2F3.5=-1.429\" so then \"P=0.0766\"
\n" ); document.write( "c) \"z%5B1%5D=%2858-57%29%2F3.5=0.286\" and \"z%5B2%5D=%2861-57%29%2F3.5=1.143\"
\n" ); document.write( "So then \"P=.873451-.612452=0.261\"
\n" ); document.write( "d) \"z=%2845-57%29%2F3.5=-3.429\" so then \"P=0.0003\"
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