Solution :
Given
Mean=270
Standard deviation =15
a) P(309 or longer)
= P(z > (309 - 270)/15)
= P(z > 2.6)
= 0.0046612 ( from z table)
The probability is 0.0047
b)
we have to find length of pregnancy ( X) corresponding to lowest 2%
X = + * Z0.02
Z0.02 = -2.05
X = 270 + 15 *(-2.05)
X = 239. 25
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