Is Alice taller than Bob by two inches?

Question

Alice's and Bob's heights are independent and each normally distributed with mean $70$ inches and standard deviation $3$ inches.

What is the probability that Alice is taller than Bob by more than $2$ inches?

Accepted Answer

Look at the difference D = (	ext{Alice's height}) - (	ext{Bob's height}). A difference of independent normals is again normal. Its mean and variance: E[D] = 70 - 70 = 0, \qquad \operatorname{Var}(D) = 3^2 + 3^2 = 18, so D \sim 	ext{Normal}(0, 18) with standard deviation \sqrt{18} \approx 4.243 inches. (Key point: variances add when you subtract independent variables, you do not subtract them.) We want P(D 2). Convert to a standard z-score: z = \frac{2 - 0}{\sqrt{18}} = \frac{2}{4.243} \approx 0.471. From the standard normal table, P(Z 0.471) \approx 0.319. P(	ext{Alice taller by more than…

Is Alice taller than Bob by two inches?

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