mean 285 SD = 13
x = days of pregnancy
a) P(x < 259) = P(z < (259 - 285)/13))
b) P(x > 295) = 1 - P(z < (295 - 285)/13))
c) P(268 < x < 288) = P( (268 - 285)/13 < z < (288 - 285)/13)
d) P(x > 300) or (x < 252) = (z > (300 - 295)/13) + {z > (252 - 285)/13)
e) 30th percentile, z = -0.52
-0.52 = (x - 285)/13, solve for x
f) 42 weeks = 294 days. P(x > 294) = P(z > (294 - 285)/13). If that probability is < 0.05, then it is rare.