Statistics for laboratory scientists

Solutions for the homework problems for lecture 9

1. X ~ normal(mean=5, SD=3). Let Z = (X - 5)/3.

   1. Pr(X < 6) = Pr[Z < (6 - 5)/3] = Pr(Z < 1/3) = (approx) 63%.

   2. Pr(X > 0) = Pr(Z > -5/3) = Pr(Z < 5/3) = (approx) 95%.

   3. Pr(0 < X < 5) = Pr(X > 0) - 1/2 = (approx) 45%.

   4. Pr(2 < X < 8) = Pr[(2-5)/3 < Z < (8-5)/3] = Pr(-1 < Z < 1) = (approx) 68%.

   5. Pr(|X - 5| > 2) = Pr(|Z| > 2/3) = 2 × Pr(Z < -2/3) = (approx) 50%.

2. Y ~ normal(mean=200, SD=18). Let Z = (Y - 200)/18.

   1. Pr(Y > 250) = Pr[Z > (250-200)/18] = 1 - Pr[Z < (250-200)/18] = (approx) 3/1000.

   2. Pr(180 < Y < 220) = Pr(-20/18 < Z < 20/18) = Pr(Z < 20/18) - Pr(Z < -20/18) = (approx) 73%.

   3. Pr(|Y - 180| > 20) = Pr(Y > 200) + Pr(Y < 160) = 50% + Pr[Z < (160 - 200)/18] = (approx) 51%.

3. Let L ~ normal(mean=3.2, SD=0.8). Let Z = (L - 3.2)/0.8.

   1. Pr(L > 4.5) = Pr[Z > (4.5 - 3.2)/0.8] = 1 - Pr(Z < 1.625) = (approx) 5%.

   2. Pr(L > 1.78) = 1 - Pr[Z < (1.78 - 3.2)/0.8] = (approx) 96%.

   3. Pr(2.9 < L < 3.6) = Pr[(2.9-3.2)/0.8 < Z < (3.6-3.2)/0.8] = (approx) 34%.

4. X and Y are independent, X ~ binomial(n=5, p=0.1), and Y ~ binomial(n=5, p=0.4).

   Thus, E(X) = 0.5 and SD(X) = sqrt{5 × 0.1 × 0.9} = (approx) 0.67.

   Similarly, E(Y) = 2.0 and SD(Y) = (approx) 1.10.

   1. E(X + Y) = E(X) + E(Y) = 0.5 + 2.0 = 2.5.

   2. Since X and Y are independent, SD(X + Y) = sqrt{SD(X)² + SD(Y)² = sqrt{0.67² + 1.10²} = (approx) 1.28.

   3. E[(X+Y)/2] = 2.5/2 = 1.25

   4. SD[(X+Y)/2] = (approx) 1.28/2 = 0.64.

   5. Note that X - Y = X + (-Y), and that E(-Y) = -E(Y) = -2.5 and SD(-Y) = E(Y) = (approx) 1.10.

      Thus, E(X-Y) = 0.5 - 2.0 = -1.5.

   6. SD(X - Y) = SD(X + Y) = (approx) 1.28.

5. X₁, X₂, X₃, ..., X₁₀ are iid with mean=3 and SD=3.

   1. E(X₁ + X₂ + ... + X₁₀) = 10 × 3 = 30.

   2. SD(X₁ + X₂ + ... + X₁₀) = sqrt{10} × 3 = (approx) 9.5.

   3. E[(X₁ + X₂ + ... + X₁₀)/10] = 3.

   4. SD[(X₁ + X₂ + ... + X₁₀)/10] = 3 / sqrt{10} = (approx) 0.95.