A-Level Normal distribution: worked solution

Question

Take sigma = 1.4 so T ~ N(17, 1.4^2), with C ~ N(8, 0.2^2), B ~ N(15, 2.1^2), all independent. Find the probability her total journey takes between 35 and 50 minutes on a random day.

The whole journey is the three legs added together, and they are independent, so add the means and add the variances. Let J = C + T + B. E(J) = 8 + 17 + 15 = 40. Var(J) = 0.2² + 1.4² + 2.1² = 0.04 + 1.96 + 4.41 = 6.41. (Add the VARIANCES, never the standard deviations: 0.2 + 1.4 + 2.1 would be wrong.) A sum of independent normal variables is normal, so J ~ N(40, 6.41), with standard deviation sqrt(6.41) = 2.5318. Standardise both limits: (35 - 40)/2.5318 = -1.9749 and (50 - 40)/2.5318 = 3.9498. P(35 < J < 50) = P(-1.9749 < Z < 3.9498) = P(Z < 3.9498) - P(Z < -1.9749) = 0.99996 - 0.02414 = 0.97582. So the probability is 0.976 (3 s.f.).