Homework #3

You can also get the PDF format of Wenqiang’s Homework.

Using the exact binomial distribution

• \(When P=0.1, P(y\leq1) = pbinom(1, 20, 0.1) =\) 0.391747
• \(When P=0.3, P(y\leq1) = pbinom(1, 20, 0.3) =\) 0.0076373

Using normal approximation without continuity correction

• When P = 0.1, y = 1

  • \(\mu = n\pi = 20(0.1) =\) 2

  • \(\sigma = \sqrt{n\pi (1-\pi)} = \sqrt{20(0.1)(1-0.1)}\) = 1.3416408

  • \(z = \frac{y-\mu}{\sigma} = \frac{1-2}{1.341641}\) = -0.7453559

  • \(P(y\leq1) \approx P(z< -0.7453559) = pnorm(-0.7453559) =\) 0.2280283

• When P = 0.3, y = 1

  • \(\mu = n\pi = 20(0.3) =\) 6

  • \(\sigma = \sqrt{n\pi (1-\pi)} = \sqrt{20(0.3)(1-0.3)}\) = 2.0493902

  • \(z = \frac{y-\mu}{\sigma} = \frac{1-6}{2.0493902}\) = -2.4397504

  • \(P(y\leq1) \approx P(z< -2.43975) = pnorm(-2.43975) =\) 0.0073487

Using normal approximation with continuity correction

• When P = 0.1, y = 1

  • \(\mu = n\pi = 20(0.1) =\) 2

  • \(\sigma = \sqrt{n\pi (1-\pi)} = \sqrt{20(0.1)(1-0.1)}\) = 1.3416408

  • \(z = \frac{y-\mu}{\sigma} = \frac{1.5-2}{1.341641}\) = -0.3726779

  • \(P(y\leq1) \approx P(z< -0.3726779) = pnorm(-0.3726779) =\) 0.3546941

• When P = 0.3, y = 1

  • \(\mu = n\pi = 20(0.3) =\) 6

  • \(\sigma = \sqrt{n\pi (1-\pi)} = \sqrt{20(0.3)(1-0.3)}\) = 2.0493902

  • \(z = \frac{y-\mu}{\sigma} = \frac{1.5-6}{2.0493902}\) = -2.1957753

  • \(P(y\leq1) \approx P(z< -2.195775) = pnorm(-2.195775) =\) 0.014054

Does the continuity correction always yield a better approximation?

• No. From our case we can see that the continuity correction yields a better approximation for one (P=0.1) but not for the other (P=0.3).

Exercise 4.54

• 1. find the area under the normal curve between z = 0.7 and z = 1.7
  • P = pnorm(1.7) - pnorm(0.7) = 0.1973982
• 2. find the area under the normal curve between z = -1.2 and z = 0
  • P = pnorm(0) - pnorm(-1.2) = 0.3849303

Exercise 4.64

• 1. \(P(y > 100) = 1 - P(y\leq100)\) = 1 - pnorm(100, 100, 8) = 0.5
• 2. \(P(y>105) = 1 - P(y\leq105)\) = 1 - pnorm(105, 100, 8) = 0.2659855
• 3. \(P(y<110) = 1 - P(y\leq110)\) = pnorm(110, 100, 8) = 0.8943502
• 4. \(P(88<y<120)=\) pnorm(120, 100, 8) - pnorm(88, 100, 8) = 0.9269831
• 6. \(P(100<y<108)=\) pnorm(108, 100, 8) - pnorm(100, 100, 8) = 0.3413447