Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 7 October 29, 2009

4.41. From Theorem 4.6 (and work in class), we know that E(Y) = \frac{θ_{1}+θ_{2}}{2} . Now compute

E(Y^{2}) = ∫_{θ_{1}}^{θ_{2}} y^{2} · \frac{1}{θ_{2} − θ_{1}} dy

= \frac{1}{θ_{2} − θ_{2}} \frac{y^{3}}{3} |_{θ_{1}}^{θ_{2}}

= \frac{θ_{2}^{3} − θ_{1}^{3}}{3(θ_{2} − θ_{1})}

= \frac{θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}}{3} .

Then

V (Y) = E(Y^{2}) − E(Y)^{2}

= \frac{θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}}{3} − (\frac{θ_{1} + θ_{2}}{2} )^{2}

= \frac{4(θ_{2}^{2} + θ_{2}θ_{1} + θ_{1}^{2}) − 3(θ_{2}^{2} + 2θ_{2}θ_{1} + θ_{1}^{2})}{12}

= \frac{(θ_{2} − θ_{1})^{2}}{12} ,

which is what we wanted to show.

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, or else use the computations from 4.41 above: E(Y^{2}) = \frac{25+5+1}{3} = \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}.

4.58. Using the standard normal table, (a) P (0 ≤ Z ≤ 1.2) = .5 − P (Z > 1.2) = .5 − .1539 − .3461. (b) By symmetry, P (−.9 ≤ Z ≤ 0) = P (0 ≤ Z ≤ .9) = .5 − P (Z > .9) = .5 − .1841 = .3159.

(c) P (.3 ≤ Z ≤ 1.56) = P (Z ≥ .3) − P (Z > 1.56) = .3821 − .0594 = .3227.

(d) By symmetry, P (−.2 ≤ Z ≤ .2) = 2P (0 ≤ Z ≤ .2) = 2(.5 − P (Z > .2)) = 1 − .8414 = .1586.

(e) By symmetry, P (−1.56 ≤ Z ≤ −.2) = P (.2 ≤ Z ≤ 1.56) = P (Z ≥ .2) − P (Z > 1.56) = .4207 − .0594 = .3613.

4.59. (a) P (Z > z_{0}) = .5 for z_{0} = 0.

(b) P (Z < z_{0}) = .8643 ⇔ P (Z ≥ z_{0}) = 1 − .8643 = .1357, so z_{0} = 1.1.

(c) P (−z_{0} < Z < z_{0}) = .9 ⇔ P (Z ≥ z_{0}) = \frac{1}{2}(1 − .9) = .05. Hence, z_{0} ≐ 1.645. (We need to interpolate between two entries in the table here.)

(d) P (−z_{0} < Z < z_{0}) = .99 ⇔ P (Z ≥ z_{0}) = \frac{1}{2}(1 − .99) = .005. Hence, z_{0} ≐ 2.575. (We also need to interpolate between two entries in the table here.)

4.61. The median is the y_{0} such that P (Y ≥ y_{0}) = \frac{1}{2}. By the symmetry of the normal density function, this happens when y_{0} = μ.

4.68. This is given by the normal probability P (Y > 3) = P (Z > \frac{3−2.4}{.8}) = P (Z > .75) = .2266. So about 23% of the students will have GPA’s above 3. (Note: This problem was written either before the current state of 1 “grade inflation” or else at a large university where the student body is not as strong as at a college like HC!)

4.69. P (Y < 1.9) = P (Z < \frac{1.9−2.4}{.8}) = P (Z < −.625) = P (Z > .625) (by symmetry). This is ≐ .2660 from the table. So about 27% of the students would be dropped(!)

4.70. This is a binomial probability question with the “success probability” obtained from the normal distribution. We have P (Y > 3) = .2266 from question 4.68. So if we pick three students at random, the probability that all three have GPA’s greater than 3.0 is \binom{3}{3}(.2266)^{3}(.7731)^{0} ≐ .012.

4.75. Let Y be the amount of soda dispensed, which is normally distributed with mean μ and standard deviation σ = .3. We want to set μ so P (Y > 8) = .01. So after standardizing, P (Z > \frac{8−μ}{.3}) = .01. From the standard normal table, \frac{8−μ}{.3} ≐ 2.33 so μ ≐ 7.3 ounces.

4.76. The question is asking what σ will make P (μ − 1 < Y < μ + 1) ≥ .95. Standardizing, this is equivalent to P (−\frac{1}{σ} < Z < \frac{1}{σ}) ≥ .95. Hence, by symmetry, P (Z ≥ \frac{1}{σ}) ≤ .025, so \frac{1}{σ} ≥ 1.96 from the standard normal table. So σ ≤ \frac{1}{1.96} ≐ .51.

4.89. (a) We have

.0821 = P (Y > 2)

= ∫_{2}^{∞} \frac{1}{β} e^{\frac{−y}{β}} dy

= e^{\frac{−2}{β}} .

So we solve by taking logarithms: β = \frac{−2}{ln(.0821)} ≐ .8.

(b) Then

P (Y ≤ 1.7) = ∫_{0}^{1.7} \frac{1}{.8}e^{\frac{−y}{.8}} dy ≐ .8806.

4.91. The mean is 100 cfs, so β = 100 in the exponential density.

(a) P (Y > 200) = ∫_{200}^{∞} \frac{1}{100}e^{\frac{−y}{100}} dy = e^{−2} ≐ .1353.

(b) Let c be the desired capacity. The question is asking: What c will make P (Y > c) = .01. As in part (a), P (Y > c) = ∫_{c}^{∞} \frac{1}{100}e^{\frac{−y}{100}} dy = e^{\frac{−c}{100}} . So we want .01 = e^{\frac{−c}{100}} . Solve by taking logarithms: c = −100 ln(.01) ≐ 460.5.

4.95. (a) We have by the definition of the discrete random variable X:

P (X = k) = ∫_{k−1}^{k} \frac{1}{β} e^{\frac{−y}{β}} dy
= −e^{\frac{−y}{β}} |_{k−1}^{k}
= −e^{\frac{−k}{β}} + e^{\frac{−(k−1)}{β}}
= (e^{\frac{−(k−1)}{β}} ) (1 − e^{\frac{1}{β}} )
= (e^{\frac{−1}{β}} )^{k−1} (1 − e^{\frac{−1}{β}} ) .

(b) This is the form (q^{k−1}p) for the probability function for a geometric random variable with p = 1 − e^{\frac{−1}{β}}.

4.109. First, let’s derive the general formula for E(Y^{k}) when Y has a Gamma-distribution with parameters α, β. We apply the substitution u = \frac{y}{β}, so dy = β du, then apply the functional equation Γ(x + 1) = xΓ(x) repeatedly:

E(Y^{k}) = \frac{1}{Γ(α)β^{α}} ∫_{0}^{∞} y^{k} · y^{α−1}e^{\frac{−y}{β}} dy

= \frac{1}{Γ(α)β^{α}} ∫_{0}^{∞} y^{α+k−1}e^{\frac{−y}{β}} dy

= \frac{1}{Γ(α)β^{α}} · β^{α+k} ∫_{0}^{∞} u^{α+k−1}e^{−u} du

= \frac{1}{Γ(α)β^{α}} · β^{α+k} · Γ(α + k)

= β^{k}(α + k − 1)(α + k − 2) ⋯ (α + 1)(α).

In our problem, with α = 3, β = 2, we get

E(Y) = αβ = 6
E(Y^{2}) = (α + 1)(α)β^{2} = 48
E(Y^{3}) = (α + 2)(α + 1)αβ^{3} = 480
E(Y^{4}) = (α + 3)(α + 2)(α + 1)αβ^{4} = 5760.

Then

E(L) = 30E(Y) + 2E(Y^{2}) = 30 · 6 + 2 · 48 = 276

and

E(L^{2}) = 900E(Y^{2}) + 120E(Y^{3}) + 4E(Y^{4}) = 900 · 48 + 120 · 480 + 4 · 5760 = 123840. Hence

V (C) = E(L^{2}) − (E(L))^{2} = 123840 − 276^{2} = 47664.

Additional Problem A) 1) We must show that when α is a positive integer,

\frac{1}{Γ(α)} ∫_{λ}^{∞} y^{α−1}e^{−y} dy = ∑_{n=0}^{α−1} \frac{λ^{n}}{n!} e^{−λ}.  (1)

We will use induction on α. The base case is α = 1. Then the left of (1) is

1 · ∫_{λ}^{∞} e^{−y} dy = e^{−λ}.

With α = 1, the sum on the right of (1) is:

∑_{n=0}^{0} \frac{λ^{n}}{n!} e^{−λ} = e^{−λ}.

So the base case is established.

Now for the induction step, assume that (1) is true for α = k. In other words, our induction hypothesis is:

\frac{1}{Γ(k)} ∫_{λ}^{∞} y^{k−1}e^{−y} dy = ∑_{n=0}^{k−1} \frac{λ^{n}}{n!} e^{−λ}.  (2)

Consider the integral on the left of (1) for α = k + 1:

\frac{1}{Γ(k + 1)} ∫_{λ}^{∞} y^{k}e^{−y} dy.

Apply integration by parts with u = y^{k}, and dv = e^{−y} dy. We get:

\frac{1}{Γ(k + 1)} ∫_{λ}^{∞} y^{k}e^{−y} dy = \frac{1}{Γ(k + 1)} (−y^{k}e^{−y}|_{λ}^{∞} + k ∫_{λ}^{∞} y^{k−1}e^{−y} dy)

= \frac{λ^{k}}{k!} e^{−λ} + \frac{1}{Γ(k)} ∫_{λ}^{∞} y^{k−1}e^{−y} dy.

Now we use the induction hypothesis (2) on the remaining integral. We get that this is equal to

\frac{λ^{k}}{k!} e^{−λ} + ∑_{n=0}^{k−1} \frac{λ^{n}}{n!} e^{−λ} = ∑_{n=0}^{k} \frac{λ^{n}}{n!} e^{−λ}.

Hence our formula also works with α = k + 1, and by induction we conclude that it works for all positive integers α.

2) Using the Poisson table in the text with λ = 5, α = 3

∑_{n=0}^{2} \frac{5^{n}}{n!} e^{−5} = .125