Math 341 Exam 2 Review Answers
  1. Moose and squirrel get on busses.
    1. Probability of any point is zero.
    2. 5 / 18.
    3. The number of waiting days is geometric, so 18 / 5.
    4. This is binomial, n*p = 30 * 5 / 18 = 25 / 3.
    5. I get 7 / 36.
  2. p(y) = (y - 3)^2 / 5 for y = 3, 4 or 5.
    1. 0/5 + 1/5 +4/5 =1, all terms are non-negative.
    2. E(Y) = 3*0 + 4*(1 / 5) + 5*(4 / 5) = 24 / 5.

      3. E(Y^2) = 9*0 + 16*(1 / 5) + 25*(4 / 5) = 116 / 5.
      V(Y) = 116 / 5 - 24^2 / 25 = 4 / 25.
    3. mY(t) = (1/5)*(e^4t + 4e^5t).
    4. mY'(t) = 0.8e^4t + 4e^5t and mY''(t) = 3.2e^4t + 20e^5t. Evaluate these at t = 0 to get E(Y) and E(Y^2).
  3. The lifetime has density f(y) = 10 / y^2 for y > 10.
    1. An antiderivative G(Y) is -10 / y, let I = infinity. G(I) - G(10) = 1.
    2. This can be done by integrating f(y) from 20 to I, or by 1 - G(20) = 1 - 1/2 = 1/2.
    3. F(y) = 1 -10 / y for y > 10. F(15) = probability that the device lasts less than 15 years = 1 / 3.
    4. Yes, since 2/3 of them would still be alive after 15 years. If t is in hours, then NO, or 1/3 of your ovens would require repairs after only 15 hours of use.
    5. This is binomial, n = 6, p = 2 / 3 and y = 3, so the probability is about 0.2195.
    6. The expected value is the integral from 10 to I of y*f(y) = 10 / y. This improper integral diverges, so the expected lifetime is infinite!
  4. Homer goes to the cities.
    1. 2/3
    2. No, the first 10 minutes of the 8 o'clock hour are identical to the first 10  minutes of the 7 o'clock hour.
    3. Yes, now NY is 35/50 and Boston is 15/50.
  5. 15 straight letters and 11 curved.
    1. Geometric, p(y) = (11/26)^(y-1)*15/26.
    2. About 0.1033.
    3. Adding from 1 to 6 yields about 0.9943.
    4. Expected value of a geometric with p = 15 / 104 is 104 / 15, or about 6.93 key presses.
    5. X is binomial (if the 9 key presses were forced to be different, X would be hypergeometric), n = 9, p = 11/26. p(x) = (9 choose x)(11/26)^x*(15 / 26)^(9-x). p(3) is about 0.2346, and P(x <= 6) is about 0.965.
  6. Corner market sales:
    1. 42
    2. 6
    3. 6
    4. 0.10326
    5. 0.0614
    6. No, about 20% of the time weekly sales are 48 cases or more, so this is NOT very surprising.
  7. Usual glaucoma levels cover about 0.8176 of the population.
  8. Otits Media.
    1. 2.02
    2. 2.0216
    3. Yes, very close in fact.
    4. All probabilities seem small.
    5. Poisson with lambda = 2.02 or so.
    6. Using the probability density of a Poisson with lambda = 2.02, I get:

      7.  
      y 0 1 2 3 4 5 6
      p(y) 0.132655 0.267964 0.270644 0.182233 0.0920279 0.0371793 0.012517
      Seems pretty reasonable to me.
  9. Finding F(y) and solving F(y) = 0.99 gives about 602 gallons.
  10. Rain in Iowa.
    1. The interval has a radius of 2.5 standard deviations, so no less that 0.84 of all years has rainfall in the interval.
    2. 0.9876
    3. 0.00621
    4. 37.5 inches in Alsonormal has the same standardized value (Z = 2.5) as 50 inches in Randomcity.