A-Level Normal distribution: worked solution

Question

Fasting glucose (mmol/L) of a random man is N(5.4,0.5²) and of a random woman is N(5.0,0.3²). Choosing one of each at random, find the probability the man's reading exceeds the woman's, and state an assumption used.

Let X be the man's fasting glucose reading and Y the woman's: X ~ N(5.4, 0.5²) and Y ~ N(5.0, 0.3²). Step 1: Turn the comparison into a single random variable. "The man's reading exceeds the woman's" is the event X > Y, which is the same event as X - Y > 0. So work with the difference X - Y instead of the two readings separately. Step 2: Find the distribution of X - Y. A linear combination of independent normal variables is itself normal. Mean: E(X - Y) = E(X) - E(Y) = 5.4 - 5.0 = 0.4 Variance: Var(X - Y) = Var(X) + Var(Y) [note the PLUS sign: variances ADD even for a difference] = 0.5² + 0.3² = 0.25 + 0.09 = 0.34 So X - Y ~ N(0.4, 0.34), and its standard deviation is sqrt(0.34) = 0.58310. Step 3: Find the probability by standardising. P(X > Y) = P(X - Y > 0) = P(Z > (0 - 0.4)/sqrt(0.34)) = P(Z > -0.68599) = 0.75364 = 0.754 (3 s.f.) Step 4: State the assumption. Step 2 added the variances, and Var(X - Y) = Var(X) + Var(Y) is only valid when the two variables are independent. So the assumption used is that the man's reading and the woman's reading are independent of each other.