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The cross section of a plastic tube for use in pulmonary resuscitators is normally distributed with a mean of 12.5mm and a st

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Answer #1

Solution :- Given : Mean (u) = 12.5 Standard deviation (0) = 0.2 P(12 < X < 13 ) = P(12 - 12.5 / 0.2 < X - u/o < 13 - 12.5 /

Probability is 0.9876

Ans -- Option ' E ' is Correct.

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