Mathematics 375 – Probability Theory Solutions for Problem Set 6 October 28, 2011

4.21. The mean is

E(Y) = ∫_{−∞}^{∞} yf(y) dy

= ∫_{0}^{1} \frac{3y^{3}}{2} + y^{2} dy

= \frac{3y^{4}}{8} + \frac{y^{3}}{3} |_{0}^{1}

= \frac{3}{8} + \frac{1}{3} = \frac{17}{24}.

Then the variance is V (Y) = E(Y^{2}) − E(Y)^{2}.

E(Y^{2}) = ∫_{−∞}^{∞} y^{2}f(y) dy

= ∫_{0}^{1} \frac{3y^{4}}{2} + y^{3} dy

= \frac{3y^{5}}{10} + \frac{y^{4}}{4} |_{0}^{1}

= \frac{3}{10} + \frac{1}{4}

= \frac{11}{20}.

So V (Y) = \frac{11}{20} − (\frac{17}{24})^{2} = \frac{139}{2880}.

4.22. We have

E(Y) = ∫_{−1}^{0} y(.2) dy + ∫_{0}^{1} y(.2 + 1.2y) dy = .4.

Then

E(Y^{2}) = ∫_{−1}^{0} y^{2}(.2) dy + ∫_{0}^{1} y^{2}(.2 + 1.2y) dy ≐ .433.

So V (Y) ≐ .433 − (.4)^{2} = .273.

4.32. (a) E(Y) = \frac{3}{64} ∫_{0}^{4} y^{3}(4−y) dy = \frac{3}{64} ∫_{0}^{4} 4y^{3}−y^{4} dy = 2.4. Then E(Y^{2}) = \frac{3}{64} ∫_{0}^{4} 4y^{4}−y^{5} dy = 6.4 and V (Y) = E(Y^{2}) − E(Y)^{2} = .64.

(b) Easiest method here is to use general facts about expected values of linear combinations as in 4.26 above: E(C) = 200(2.4) = $480. V (C) = 200^{2}(.64) = 25600 (units of variance are dollars^{2}), so σ = 160.

(c) P (C > 600) = P (Y > 3) = \frac{3}{64} ∫_{3}^{4} y^{2}(4 − y) dy ≐ .26, so this can certainly happen. 4.34. We have by properties of iterated and double integrals,

E(Y) = ∫_{0}^{∞} yf(y) dy
= ∫_{0}^{∞} (∫_{0}^{y} dt) f(y) dy
= ∫_{0}^{∞} ∫_{0}^{y} f(y) dt dy
= ∫_{0}^{∞} ∫_{t}^{∞} f(y) dy dt
= ∫_{0}^{∞} 1 − F (t) dt.

(The inner integral represents the probability P (Y > t) = 1 − F (t).) The integration variable may be renamed to y to match the formula to be proved. 4.39. Closer to A means that Y = location she lands satisfies Y < \frac{A+B}{2} , the midpoint.

P (Y < \frac{A + B}{2} ) = ∫A^{\frac{A+B}{2}} \frac{1}{B − A} dy = \frac{1}{2}.

Similarly, distance to A is more than three times the distance to B means \frac{A+3B}{4} ≤ Y ≤ B. The probability is \frac{1}{4}, since this is one quarter of the interval from A to B (this can be done with an integral too, of course!) 4.40. This is a binomial probability, Y = number that land closer to B is binomial with n = 3, p = 1/2. So P (Y = 1) = \binom{3}{1}(1/2)^{1}(1/2)^{2} = \frac{3}{8} = .375. 4.42. We have

.5 = P (Y ≤ ϕ_{.5}) = ∫_{θ_{1}}^{ϕ_{.5}} \frac{1}{θ_{2} − θ_{1}} dy = \frac{ϕ_{.5} − θ_{1}}{θ_{2} − θ_{1}} .

Solving this for ϕ_{.5}, we get ϕ_{.5} = \frac{θ_{1} + θ_{2}}{2} . This is the median, by definition. 4.43. The mean of the area is \overline{A} = E(A) = ∫_{0}^{1} πr^{2} dr = \frac{π}{3} , and the variance is

V (A) = E(A^{2}) − E(A)^{2}

= ∫_{0}^{1}(πr^{2})^{2} dr − (\frac{π}{3} )^{2}

= \frac{π^{2}}{5} − \frac{π^{2}}{9}

= \frac{4π^{2}}{45} .

4.47. The density function for Y is f(y) = \frac{1}{4} for 1 ≤ y ≤ 5, and zero otherwise. Hence (a) P (Y ≥ 2) = ∫_{2}^{5} \frac{1}{4} dy = \frac{3}{4}.

(b) The expected cost is E(C) = E(c_{0} +c_{1}y^{2}) = c_{0} +c_{1}E(Y^{2}). We can compute E(Y^{2}) directly:

E(Y^{2}) = ∫_{1}^{5} y^{2} · \frac{1}{5 − 1} dy = \frac{y^{3}}{12}| 1^{5} = \frac{31}{3}.

(Alternative method:

E(Y^{2}) = V (Y) + E(Y)^{2} = \frac{(5 − 1)^{2}}{12} + 9 = \frac{31}{3}.)

Therefore,

E(C) = c_{0} + \frac{31c_{1}}{3} .

4.51. This is a conditional probability:

P (Y > 65 | Y > 55) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.

4.59. Using the standard normal table, (a) z_{0} = 0. (b) z_{0} = 1.1. (c) z_{0} ≐ 1.645. (d) z_{0} ≐ 2.575.

4.62. (a) We have P (Z^{2} < 1) = P (−1 < Z < 1) = 1 − 2(.1587) = .6826. (b) Similarly, P (Z^{2} < 3.84146) = P (−1.96 < Z < 1.96) = 1 − 2(0.025) = .95. (Note that these correspond with the 68% and 95% from the “empirical rule” we saw in Chapter 1.)

4.71. (a) Let Y be resistance of a wire produced by company A. We compute the desired probability by “standardizing” and consulting the standard normal table:

P (.12 < Y < .13) = P (\frac{.12 − .13}{.005} < Z < \frac{.14 − .13}{.005}) = P (−2 < Z < 2) ≐ 1 − 2(.0228) = .9544.

(b) This can be thought of as a binomial probability: \binom{4}{4}(.9544)^{4}(.0456)^{0} = .8297.

4.74. (a) We have σ = 6, so P (Y > 72) = P (Z > −1) = .8413. (b) We want b so P (Z > \frac{b−78}{6}) ≐ .1. From the normal table, \frac{b−78}{6} ≐ 1.28 so b ≐ 85.68. If the scores are to be whole numbers, then b ≥ 86. (c) P (Z > z_{0}) = .281 when z_{0} = .58. So we want \frac{b−78}{6} = .58 and the cutoff is b = 81.48 (or 82 if scores must be whole numbers). (d) The cutoff for the lowest 25% occurs at the where P (Z > z_{0}) = .75, so z_{0} ≐ −0.675. This corresponds to a score of 74. The proportion of students with scores > 74 + 5 = 79 is about 100% × P (Z > \frac{79−78}{6}) = 100% × P (Z > .167) ≐ 43%. (f) P (Y > 84 | Y > 72) = P (Z > 1 | Z > −1) = .1587/.8413 ≐ .1886.

4.80. We have E(A) = E(LW) = E(3|Y |^{2}) = 3E(Y^{2}) = 3(σ^{2} + μ^{2}).