Let X be a normally distributed random variable with mean 20 and standard deviation 6. Find: a) P(X < 27); and b) the value of x such that P(X < x) = 0.3015.

a) 27 is higher than the mean. So we can simply calculate the z value. z = (27 - 20)/6 ≈ 1.17. Using the table in the formula booklet, we find that P(Z < 1.17) = 0.8790, so P(X < 27) = 0.8790. b) Let's give an expression for our z value: z = (x - 20)/6. So P(Z < z) = P(X < x) = 0.3015. But this is lower than 0.5, so to find the value of z we first need to find -z. We find that P(Z < -z) = 1 - 0.3015 = 0.6985. This, from the table, corresponds to a z value of 0.52, so z = -0.52. Hence from our first expression for z, we can deduce that x = (-0.52 × 6) + 20 = 16.88.