Linear combinations of independent normal variables Problem set B Worked solutions

X and Y are independent, \(X\sim N(25,\,7)\) and \(Y\sim N(11,\,2)\). State the distribution of each combination.

1. (a) \(W=2X+3Y\)

   • Mean: \(E(W)\) = 2 × 25 + 3 × 11 = 83
   • Variance: \(Var(W)\) = 2² × 7 + 3² × 2 = 46

2. (b) \(W=4X-3Y-2\)

   • Mean: \(E(W)\) = 4 × 25 − 3 × 11 − 2 = 65
   • Variance: \(Var(W)\) = 4² × 7 + (−3)² × 2 = 130

X and Y are independent, \(X\sim N(35,\,10)\) and \(Y\sim N(14,\,3)\). Let \(W=2X-3Y+4\). Find \(P(W>40)\).

• E(W) = 2 × 35 − 3 × 14 + 4 = 32
• Var(W) = 2² × 10 + (−3)² × 3 = 67
• From the calculator: \(P(W>40)\) = 0.164 (normal cdf with μ = 32, σ = √67 = 8.19…)

X has mean 60 and standard deviation 5. Y has mean 45 and standard deviation 2. X and Y are independent and both are normal. State the distribution of each.

1. (a) \(X+Y\)

   • Variances: 5² = 25 and 2² = 4
   • Mean = 60 + 45 = 105
   • Variance = 25 + 4 = 29

2. (b) \(X-Y\)

   • Mean = 60 − 45 = 15
   • Variance = 25 + 4 = 29 (variances add)

A student gets ready for school and then walks there. The getting-ready time, X minutes, has distribution \(N(25,\,16)\) and the walking time, Y minutes, has distribution \(N(18,\,9)\). The two times are independent. Let T = X + Y. State the distribution of T and find the probability that the student needs less than 40 minutes in total.

• E(T) = 25 + 18 = 43
• Var(T) = 1² × 16 + 1² × 9 = 25
• \(T\sim N(43,\,25)\)
• From the calculator: \(P(T<40)\) = 0.274 (normal cdf with μ = 43, σ = √25 = 5)

A photographer charges \$60 for each hour of shooting and \$35 for each hour of editing. The shooting time for a job, X hours, has distribution \(N(3.5,\,0.8)\). The editing time, Y hours, has distribution \(N(5,\,2.25)\). The two times are independent. The cost of a job is C = 60X + 35Y dollars. State the distribution of C and find the probability that a job costs more than \$420.

• E(C) = 60 × 3.5 + 35 × 5 = 385
• Var(C) = 60² × 0.8 + 35² × 2.25 = 5636.25
• From the calculator: \(P(C>420)\) = 0.321 (normal cdf with μ = 385, σ = √5636.25 = 75.1…)

Water flows into a tank through one pipe and out through another over one hour. The inflow, X litres, has distribution \(N(210,\,64)\) and the outflow, Y litres, has distribution \(N(185,\,49)\). The two flows are independent. Let D = X − Y be the net gain in the tank. State the distribution of D and find the probability that the tank gains more than 35 litres.

• E(D) = 210 − 185 = 25
• Var(D) = 1² × 64 + (−1)² × 49 = 113
• From the calculator: \(P(D>35)\) = 0.173 (normal cdf with μ = 25, σ = √113 = 10.6…)

X, Y and Z are independent, \(X\sim N(15,\,5)\), \(Y\sim N(9,\,2)\) and \(Z\sim N(12,\,3)\). Let \(W=X+3Y-2Z-4\). Find \(P(W<22)\).

• E(W) = 15 + 3 × 9 − 2 × 12 − 4 = 14
• Var(W) = 1² × 5 + 3² × 2 + (−2)² × 3 = 35
• From the calculator: \(P(W<22)\) = 0.912 (normal cdf with μ = 14, σ = √35 = 5.92…)

X and Y are independent, \(X\sim N(50,\,16)\) and \(Y\sim N(30,\,4)\).

1. (a) State the distributions of \(X+Y\) and \(X-Y\).

   • Mean of X+Y = 50 + 30 = 80, variance = 16 + 4 = 20
   • Mean of X−Y = 50 − 30 = 20, variance = 16 + 4 = 20

2. (b) Find \(P(X+Y>85)\).

   • From the calculator: \(P(X+Y>85)\) = 0.132 (normal cdf with μ = 80, σ = √20 = 4.47…)

3. (c) Find \(P(X-Y>25)\). Explain why your answer is the same as in part (b).

   • From the calculator: \(P(X−Y>25)\) = 0.132 (normal cdf with μ = 20, σ = √20 = 4.47…)
   • Both are 5 above the mean and the standard deviation is the same, so the probabilities match.