document.write( "Question 885338: The mean number of flaws per square yard in a type of carpet material is claimed to be 2.2 flaws per square yard along with a standard deviation of 1.2 flaws per square yard. The population can be assumed to be normal.
\n" ); document.write( "Find P(X ≤ 4), the percentage of time there are fewer than 4 flaws per square yard.
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Algebra.Com's Answer #535094 by Theo(13342)\"\" \"About 
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m = mean
\n" ); document.write( "sd = standard deviation
\n" ); document.write( "x = raw score
\n" ); document.write( "z = z score
\n" ); document.write( "you get:
\n" ); document.write( "m = 2.2
\n" ); document.write( "sd = 1.2
\n" ); document.write( "x = 4
\n" ); document.write( "population is assumed to be normal.
\n" ); document.write( "z score is equal to (x-m)/sd
\n" ); document.write( "replace x with 4 and m with 2.2 and sd with 1.2 and you get:
\n" ); document.write( "z score is equal to (4 - 2.2) / 1.2 = 1.5
\n" ); document.write( "you are looking for the area under the normal distribution curve that is to the left of a z score of 1.5
\n" ); document.write( "look up the z score in the z score table and you will see that the area to the left of that z score is equal to .9332
\n" ); document.write( "this means that the percentage of time that there are fewer than 4 flaws per square yard of carpet material is 93.32%.\r
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