Mathematics 375 – Probability and Statistics 1 Problem Set 5 Solutions October 22, 2003

Problems from Wackerly, Mendenhall, Scheaffer: 3.85/ The number is

\begin{matrix}\binom{4}{0} & \binom{6}{5} \\ ( & {10 \atop 5} & ) \end{matrix} = \frac{1}{42}

(hypergeometric pdf with N = 10, n = 5, r = 4, y = 0). 3.86/ The repair cost is C = 50Y, where Y is the number of defectives. The mean cost is

E(C) = 50E(Y) = 50\frac{nr}{N} = 50\frac{5 · 4}{10} = 100

The variance of the cost is

V (C) = (50)^{2}V (Y) = 2500 · n · \frac{r}{N} · \frac{N − r}{N} · \frac{N − n}{N − 1} = 2500 · 5 · \frac{4}{10} · \frac{6}{10} · \frac{5}{9} = 1666.67

(Note: V (aY) = a^{2}V (Y).) 3.87/ Y is hypergeometric with N = 6, n = 2, r = 4, so

\begin{aligned}P (Y = 0) & = \begin{matrix}\binom{4}{0} & \binom{2}{2} \\ ( & {6 \atop 2} & ) \end{matrix} = \frac{1}{15} \\ P (Y = 1) & = \begin{matrix}\binom{4}{1} & \binom{2}{1} \\ ( & {6 \atop 2} & ) \end{matrix} = \frac{8}{15} \\ P (Y = 2) & = \begin{matrix}\binom{4}{2} & \binom{2}{0} \\ ( & {6 \atop 2} & ) \end{matrix} = \frac{6}{15} \end{aligned}

3.89/ For a) hypergeometric with N = 10, n = 3, r = 2 so table of values of the pdf is

\begin{matrix}y & 0 & 1 & 2 \\ P (Y = y) & 14/30 & 14/30 & 2/30 \end{matrix}

For b), N = 10, n = 3, r = 4, so

\begin{matrix}y & 0 & 1 & 2 & 3 \\ P (Y = y) & 5/30 & 15/30 & 9/30 & 1/30 \end{matrix}

3.95/ The number of improperly drilled gear boxes Y is a hypergeometric random variable with N = 20, n = 5, r = 2. For a), we want

P (Y = 0) = \begin{matrix}\binom{2}{0} & \binom{18}{5} \\ ( & {20 \atop 5} & ) \end{matrix} = 42/76 ≐ .553

For b, the time T to install the 5 is

T = 10Y + (5 − Y) = 9Y + 5

(the installations are done in sequence, not “in parallel”, so the times add). The mean time is

E(T) = 9E(Y) + 5 = 9 · \frac{5 · 2}{20} + 5 = 9.5

The standard deviation is obtained from

V (T) = 81V (Y) = 81 · 5 · \frac{2}{20} · \frac{18}{20} · \frac{15}{19} ≐ 28.78

Then σ ≐ \sqrt{28.78} ≐ 5.365.

3.97/ Y is Poisson with λ = 2.

a) P (Y = 4) = \frac{2^{4}}{4!} e^{−2} ≐ .090.

b) Using the table of Poisson probabilities in text: P (Y ≥ 4) = 1−P (Y ≤ 3) = 1−.857 = .143.

c) P (Y < 4) = P (Y ≤ 3) = .857.

d) P (Y ≥ 4|Y ≥ 2) = .143/.594 = .241 (using definition of conditional probability).

3.98/ Y is Poisson with λ = 7, so

a) P (Y ≤ 3) = .082

b) P (Y ≥ 2) = 1 − P (Y ≤ 1) = 1 − .007 = .993

c) P (Y = 5) = P (Y ≤ 5) − P (Y ≤ 4) = .301 − .173 = .127.

3.99/ The service time is T = 10Y with Y from 3.98. So E(T) = 10E(Y) = 70 minutes. The variance is V (T) = 100V (Y) = 700 minutes. The probability P (T > 150) = P (Y > 15) = 1 − P (Y ≤ 15) = 1 − .998 = .002, so no, this is quite unlikely.

3.100/ Both probabilities equal \frac{14^{2}}{2!} e^{−14} ≐ .000082. Note:

\frac{7^{2}}{2!} e^{−7} · \frac{7^{0}}{0!} e^{−7} + \frac{7^{1}}{1!} e^{−7} · \frac{7^{1}}{1!} e^{−7} + \frac{7^{0}}{0!} e^{−7} · \frac{7^{2}}{2!} e^{−7} = \frac{14^{2}}{2!} e^{−14}

3.111/ P = 50 − 2Y − Y^{2}, where Y is Poisson with mean λ = 2. Hence E(P) = 50 − 2E(Y) − E(Y^{2}). We know E(Y) = λ and E(Y^{2}) = λ^{2} + λ from the derivation of the formula for the variance of Poisson variables. Hence E(P) = 50 − 4 − 6 = 40.

3.114/

a) We have

p(y)/p(y − 1) = \frac{\frac{λ^{y}}{y!} e^{−λ}}{\frac{λ^{y−1}}{(y−1)!}e^{−λ}} = \frac{λ}{y}

b) Hence p is increasing for y < λ, and decreasing for y > λ.

c) p(y) is maximized with y = ⌊λ⌋ (the greatest integer less than or equal to λ).

3.117/ For a geometric random variable, we have

m_{Y} (t) = E(e^{tY}) = ∑_{y=1}^{∞} e^{ty}q^{y−1}p

= pe^{t} ∑_{y=1}^{∞} (qe^{t})^{y−1}

= \frac{pe^{t}}{1 − qe^{t}}

(by the geometric series summation formula).

3.118/ Continuing from the formula in 3.117,

E(Y) = μ′_{1} = \frac{d}{dt} (\frac{pe^{t}}{1 − qe^{t}} ) |_{t=0}

= \frac{pe^{t}}{(1 − qe^{t})^{2}} |_{t=0}

= \frac{p}{(1 − q)^{2}}

= \frac{1}{p}

(the derivative is computed with the quotient rule and simplified). Then

E(Y^{2}) = μ′_{2} = \frac{d^{2}}{dt^{2}} (\frac{pe^{t}}{1 − qe^{t}} ) |_{t=0}

= \frac{d}{dt} (\frac{pe^{t}}{(1 − qe^{t})^{2}} ) |_{t=0}

= \frac{(1 − q)^{2}p + 2pq(1 − q)}{(1 − q)^{4}}

= \frac{p + 2q}{p^{2}}

= \frac{1 + q}{p^{2}}

So

V (Y) = E(Y^{2}) − (E(Y))^{2} = \frac{1 + q}{p^{2}} − \frac{1}{p^{2}} = \frac{1 − p}{p^{2}}.

3.121/ By the Uniqueness Theorem for the moment generating function, and the formulas from class and the text for moment generating functions for standard types of random variables,

a) Binomial, n = 5, p = 1/3.

b) Geometric, p = 1/2.

c) Poisson, λ = 2.

3.126/ Letting r(t) = ln m(t), we have

r′(0) = \frac{m′(0)}{m(0)} = \frac{μ′_{1}}{1} − μ′_{1} = E(Y)

Then by the quotient rule for derivatives,

r′′(0) = \frac{m(0)m′′(0) − (m′(0))^{2}}{m(0)^{2}} = μ′_{2} − μ_{1}^{′2} = E(Y^{2}) − (E(Y))^{2} = V (Y)