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Suppose the diameters of lids for aluminum cans produced by a certain manufacturer are normally distributed...

Suppose the diameters of lids for aluminum cans produced by a certain manufacturer are normally distributed with a mean of 4 inches and a standard deviation of 0.012 inch. What proportion of the lids produced are between 3.97 inches and 4.03 inches?

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Answer #1


P ( 3.97 < X < 4.03 )
Standardizing the value

Z = ( 3.97 - 4 ) / 0.012
Z = -2.5
Z = ( 4.03 - 4 ) / 0.012
Z = 2.5
P ( -2.5 < Z < 2.5 )
P ( 3.97 < X < 4.03 ) = P ( Z < 2.5 ) - P ( Z < -2.5 )
P ( 3.97 < X < 4.03 ) = 0.9938 - 0.0062
P ( 3.97 < X < 4.03 ) = 0.9876

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