document.write( "Question 878682: The length of time that it takes Ken to drive to work represents a normal distribution with a mean of 25 minutes and a standard deviation of 4.5 minutes.If Ken allows 35 minutes to get to work , what percent of the time can he except to be late? \n" ); document.write( "
Algebra.Com's Answer #530758 by ewatrrr(24785)\"\" \"About 
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document.write( "Population: mean is 25 and the deviation is 4.5.
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document.write( "P(x ≥  35) = 1 -  P(z ≤ 2.222) = 1 - .9869 = .0131 0r 1.31 % chance of being late\r
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document.write( "For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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document.write( "Area under the standard normal curve to the left of the particular z is P(z)
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document.write( "Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50%  to the right
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document.write( "one  standard deviation from the mean accounts for about 68% of the set 
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document.write( "two standard deviations from the mean account for about 95%
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document.write( "and three standard deviations from the mean account for about 99.7%. \n" );
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