Answer:
0.27
Explanation:
The variance of the distribution of pregnancy lengths is 16x16 = 256.
The sum of pregnancy lengths of 6 randomly chosen women follows a normal distribution with mean = 6x266 = 1596 and variance = 6x256 = 1536.
Compute the probability that the sum exceeds 6x270 = 1620.
P(X>1620) = 1-Φ(
) = 1-Φ(
) = 0.27
where Φ is the c.d.f of the standard normal distribution,
μ is mean and σ is the standard deviation of the sum of 6 pregnancy lengths.