document.write( "Question 1137290: On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with a mean of 45.1 and a standard deviation of 0.5.
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document.write( "a)find the breaking distance that corresponds to z=1.8
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document.write( "b)find the breaking distance that represents the 91st percentile
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document.write( "c)find the z-score for a breaking distance of 46.1m
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document.write( "d)find the probability that the breaking distance is less than or equal to 45m
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document.write( "e)find the probability that the breaking distance is greater than 46.8m \n" );
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Algebra.Com's Answer #755158 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "z=1.8=(x-45.1)/0.5 \n" ); document.write( "0.9=x-45.1 \n" ); document.write( "x=46 m ANSWER a\r \n" ); document.write( "\n" ); document.write( "91st percentile z=1.341 \n" ); document.write( "Here, 1.341=(x-45)/0.5 \n" ); document.write( "0.67=x-45.1 \n" ); document.write( "x=45.8 m (rounded) ANSWER b\r \n" ); document.write( "\n" ); document.write( "less than 45 has a z=(45-45.1)/0.5 or a z<-.1/.5 or z<-0.2 \n" ); document.write( "That probability is 0.4207 ANSWER c\r \n" ); document.write( "\n" ); document.write( "greater than 46.8 is z>1.7/0.5 or z> 3.4 \n" ); document.write( "This probability is 0.0003 \n" ); document.write( " |