A-Level Normal distribution

Rays mass X ~ N(100, sigma²). Require P(X <= 100 - sigma²) < 0.2. Standardize: Z = (X-100)/sigma, so P(Z <= (100 - sigma² - 100)/sigma) = P(Z <= -sigma) < 0.2. Since P(Z <= -0.8416) = 0.2 and P(Z <= t) is increasing, need -sigma < -0.8416, i.e. sigma > 0.8416. (sigma>0 automatically.)

Step 1 - build the distribution of the TOTAL duration. A routine call is first categorised (time X) and then resolved (time T), so the duration of the whole call is T + X. The sum of independent normal variables is itself normal. Its mean is the sum of the means, which the question gives as 7 minutes. Its variance is the sum of the VARIANCES - never the sum of the standard deviations. T contributes k and X contributes 0.2 squared = 0.04. T + X ~ N(7, k + 0.04) Step 2 - write the given probability as a standardised statement. Let sigma = sqrt(k + 0.04) be the standard deviation of T + X. P(T + X > 8) = 0.254 Standardise with Z = (T + X - 7)/sigma, which is N(0, 1): P(Z > (8 - 7)/sigma) = 0.254 P(Z > 1/sigma) = 0.254 Step 3 - turn the tail probability into a z-value. Tables give the area to the LEFT, so convert: P(Z < 1/sigma) = 1 - 0.254 = 0.746 Reading the inverse normal for 0.746 gives 1/sigma = 0.6620 (4 d.p.) Step 4 - solve for sigma. sigma = 1 / 0.6620 = 1.5107 Step 5 - square to recover the variance, then subtract the part you already know. k + 0.04 = sigma squared = 1.5107 squared = 2.2821 k = 2.2821 - 0.04 = 2.2421 k = 2.24 (2 d.p.), as required.

Let X be the time to resolve a routine call and Y the time to resolve a complex call before the review, X and Y independent normal variables as defined earlier in the question. From the earlier parts E(Y) = 20, Var(Y) = 5² = 25, E(X) = 5, Var(X) = 2.4² = 5.76. Step 1: model the effect of the review. Cutting a complex-call time by 20% leaves 80% of it, so a reviewed complex call takes 0.8Y. Step 2: turn the wording into an inequality. "Two complex calls take more than twice the time of one routine call" means 0.8Y1 + 0.8Y2 > 2X, where Y1, Y2 are the times of the two complex calls. Rearrange so that one variable is compared with zero, because that is what we can standardise: 0.8(Y1 + Y2) − 2X > 0. So define W = 0.8(Y1 + Y2) − 2X and find P(W > 0). Step 3: find E(W). Expectation is linear, so constants come straight out: E(W) = 0.8[E(Y1) + E(Y2)] − 2E(X) = 0.8(20 + 20) − 2(5) = 32 − 10 = 22. Step 4: find Var(W). This is where marks are lost. A constant multiplier is SQUARED when it passes through a variance, and for independent variables the variances ADD even when the combination is a difference: Var(W) = 0.8²[Var(Y1) + Var(Y2)] + (−2)²Var(X) = 0.64(25 + 25) + 4(5.76) = 0.64(50) + 23.04 = 32 + 23.04 = 55.04. Step 5: W is a linear combination of independent normal variables, so W is normal: W ~ N(22, 55.04), and sd(W) = √55.04 = 7.4189. Step 6: standardise and evaluate. P(W > 0) = P(Z > (0 − 22)/7.4189) = P(Z > −2.9655) = P(Z < 2.9655) = 0.99849 = 0.998 (3 s.f.).