Linear combinations of independent normal variables Problem set A Worked solutions

X and Y are independent, \(X\sim N(18,\,5)\) and \(Y\sim N(6,\,3)\). State the distribution of each combination.

1. (a) \(W=X+4Y\)

   • Mean: \(E(W)\) = 18 + 4 × 6 = 42
   • Variance: \(Var(W)\) = 1² × 5 + 4² × 3 = 53

2. (b) \(W=3X-2Y+7\)

   • Mean: \(E(W)\) = 3 × 18 − 2 × 6 + 7 = 49
   • Variance: \(Var(W)\) = 3² × 5 + (−2)² × 3 = 57

X and Y are independent, \(X\sim N(22,\,6)\) and \(Y\sim N(9,\,4)\). Let \(W=3X-2Y+5\). Find \(P(W>50)\).

• E(W) = 3 × 22 − 2 × 9 + 5 = 53
• Var(W) = 3² × 6 + (−2)² × 4 = 70
• From the calculator: \(P(W>50)\) = 0.640 (normal cdf with μ = 53, σ = √70 = 8.37…)

X has mean 50 and standard deviation 4. Y has mean 30 and standard deviation 3. X and Y are independent and both are normal. State the distribution of each.

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

   • Variances: 4² = 16 and 3² = 9
   • Mean = 50 + 30 = 80
   • Variance = 16 + 9 = 25

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

   • Mean = 50 − 30 = 20
   • Variance = 16 + 9 = 25 (variances add)

A delivery driver loads the van and then drives to the customer. The loading time, X minutes, has distribution \(N(18,\,9)\) and the journey time, Y minutes, has distribution \(N(41,\,36)\). The two times are independent. Let T = X + Y. State the distribution of T and find the probability that the whole delivery takes less than 65 minutes.

• E(T) = 18 + 41 = 59
• Var(T) = 1² × 9 + 1² × 36 = 45
• \(T\sim N(59,\,45)\)
• From the calculator: \(P(T<65)\) = 0.814 (normal cdf with μ = 59, σ = √45 = 6.71…)

A plumber charges \$18 for each hour of labour and \$40 for each hour of equipment hire. The labour time on a job, X hours, has distribution \(N(5.5,\,1.2)\). The equipment time, Y hours, has distribution \(N(2.5,\,0.5)\). The two times are independent. The cost of a job is C = 18X + 40Y dollars. State the distribution of C and find the probability that a job costs more than \$250.

• E(C) = 18 × 5.5 + 40 × 2.5 = 199
• Var(C) = 18² × 1.2 + 40² × 0.5 = 1188.8
• From the calculator: \(P(C>250)\) = 0.0695 (normal cdf with μ = 199, σ = √1188.8 = 34.5…)

A cyclist climbs a hill and then rides back down. The time to climb, X minutes, has distribution \(N(14.5,\,2.25)\) and the time to descend, Y minutes, has distribution \(N(6.2,\,0.64)\). The two times are independent. Let D = X − Y. State the distribution of D and find the probability that the climb takes more than 8 minutes longer than the descent.

• E(D) = 14.5 − 6.2 = 8.3
• Var(D) = 1² × 2.25 + (−1)² × 0.64 = 2.89
• From the calculator: \(P(D>8)\) = 0.570 (normal cdf with μ = 8.3, σ = √2.89 = 1.7)

X, Y and Z are independent, \(X\sim N(10,\,4)\), \(Y\sim N(6,\,3)\) and \(Z\sim N(8,\,2)\). Let \(W=X+2Y-Z+1\). Find \(P(W<18)\).

• E(W) = 10 + 2 × 6 − 8 + 1 = 15
• Var(W) = 1² × 4 + 2² × 3 + (−1)² × 2 = 18
• From the calculator: \(P(W<18)\) = 0.760 (normal cdf with μ = 15, σ = √18 = 4.24…)

X and Y are independent, \(X\sim N(40,\,9)\) and \(Y\sim N(25,\,16)\).

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

   • Mean of X+Y = 40 + 25 = 65, variance = 9 + 16 = 25
   • Mean of X−Y = 40 − 25 = 15, variance = 9 + 16 = 25

2. (b) Find \(P(X+Y>70)\).

   • From the calculator: \(P(X+Y>70)\) = 0.159 (normal cdf with μ = 65, σ = √25 = 5)

3. (c) Find \(P(X-Y>20)\). Explain why your answer is the same as in part (b).

   • From the calculator: \(P(X−Y>20)\) = 0.159 (normal cdf with μ = 15, σ = √25 = 5)
   • Both are 5 above the mean and the standard deviation is the same, so the probabilities match.