HW5 Solutions, Math 320, Spring 98

1. Do Exercise 5.26 on page 191.

Answer: Let X be the distance between holes on a randomly selected sheet of metal. It is given that X ~ N(.68,.01). The plug will not fit if X < .67 or X > .71.
P(X < .67 or X > .71) = 1 - P(.67 <= X <= .71) = 1 - .8400 = .1600 (by calculator).
Hence, 16% of the metal sheets will not allow the plug to fit.
To use Table A2*, use
P(.67 <= X <= .71) = P((.67 - .68)/.01 <= Z <= (.71 - .68)/.01) = P(Z <= 3) - P(Z <= -1) = .9987 - .1587, by Table A2*.

2. (Like Ex. 5.30 page 191). A judge is scheduled to hear 20 appeals for traffic tickets. Each appeal has a probability of .25 of being approved, independent of the other appeals. Find the probability that the number of appeals approved is no less than 3 and no more than 7, i.e., between 3 and 7 inclusively. First do this exactly, using Table A1. Next use the normal approximation to the binomial. Compare the two answers.

Answer: Let X be the number of appeals approved by the judge of the 20 he heard. From the information given, X ~ B(20,.25). Then
P(3 <= X <= 7) = P(X <= 7) - P(X <= 2) (discrete R.V.)
= .8982 - .0913 = .8069 from Table A1 (n = 20, p = .25).
Using the normal approximation to the binomial:
P(X <= 7) - P(X <= 2) approximately equals P(Z <= (7+.5-20x.25)/sqrt(20x.25x.75)) - P(Z <= (2+.5-5)/sqrt(3.75))
= P(Z <= 1.291) - P(Z <= -1.291) = .9015 - .0985 = .8030 by Table A2*.

3. Replace the probability function of Exercise 5.40 on page 219 with f(1) = .25 = f(5), f(2) = .2 = f(4), f(3) = .1. With this probability function do exercise 5.40 on page 219.

Answer: Graph of probability function: 

  |
.3|
  |  *       *
.2|    *   *
  |
.1|      *
  |_____________ x
     1 2 3 4 5
mu = sum x f(x) = 3, sigma = sqrt(2.4) = 1.549

4. Using the probability function in problem 3 above, do exercise 5.41 on page 219.

Answer: For typographical reasons I'll write Y for X-bar = (X1 + X2)/2, where X1 and X2 are independent samples of the random variable X in problem 3. The values of Y are given in the following table: 

      1   2   3   4   5
     -------------------
    1|1   1.5 2   2.5 3
    2|1.5 2   2.5 3   3.5
    3|2   2.5 3   3.5 4
    4|2.5 3   3.5 4   4.5
    5|3   3.5 4   4.5 5
Using the independence of X1 and X2, we find the probability function g(y) of Y from the probability function f(x) of X:
g(1) = f(1)f(1) = .0625
g(1.5) = 2f(1)f(2) = .1
g(2) = 2f(1)f(3) + f(2)f(2) = .09
g(2.5) = 2f(1)f(4) + 2f(2)f(3) = .14
g(3) = 2f(1)f(5) + 2f(2)f(4) + f(3)f(3) = .215
g(3.5) = 2f(2)f(5) + 2f(3)f(4) = g(2.5) = .14
g(4) = 2f(3)f(5) + f(4)f(4) = g(2) = .09
g(4.5) = 2f(4)f(5) = g(1.5) = .1
g(5) = f(5)f(5) = g(1) = .0625

5. Using the probability function in problem 3 above, do exercise 5.42 on page 219.

Answer: The graph of g(y) is similar to the graph in Fig. 5.37, except for the "dip" at y = 2. 

.25|
   |
   |     *
   |
.20|
   |
   |
   |
.15|
   |    * *
   |
   |
.10|  *     *
   |   *   *
   |
   | *       *
.05|
   |
   |
   |___________
     1 2 3 4 5

6. Using the probability function in problem 3 above, do exercise 5.43 on page 2.19.

Answer: From equations 4.12 and 4.14 (i.e., the definitions of population mean and standard deviation):
muY = 3, sigmaY = 1.095 (by calculator).
By equations 5.14 and 5.16,
muY = muX = 3, sigmaY = sigmaX/sqrt(2) = 1.549/sqrt(2) = 1.095
The answers agree.