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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Problem Set 5 Solutions}
\centerline{October 22, 2003}
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\noindent
Problems from Wackerly, Mendenhall, Scheaffer:
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3.85/  The number is 
$${{4\choose 0}{6\choose 5}\over {10\choose 5}} = {1\over 42}$$
(hypergeometric pdf with $N = 10, n = 5, r = 4, y = 0$).
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3.86/  The repair cost is $C = 50 Y$, where $Y$ is the number
of defectives.  The mean cost is 
$$E(C) = 50E(Y) = 50 {nr \over N} = 50 {5\cdot 4\over 10} = 100$$
The variance of the cost is 
$$V(C) = (50)^2 V(Y) = 2500 \cdot n\cdot {r \over N}\cdot {N - r\over N}\cdot {N-n \over N-1} = 
2500\cdot 5\cdot {4\over 10}\cdot{6\over 10}\cdot { 5\over 9} = 1666.67$$
(Note: $V(aY) = a^2 V(Y)$.)
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3.87/  $Y$ is hypergeometric with $N = 6, n = 2, r = 4$, so
$$\eqalign{
P(Y = 0) &= {{4\choose 0}{2\choose 2} \over {6\choose 2}} = {1\over 15}\cr
P(Y = 1) &= {{4\choose 1}{2\choose 1} \over {6\choose 2}} = {8\over 15}\cr
P(Y = 2) &= {{4\choose 2}{2\choose 0} \over {6\choose 2}} = {6\over 15}\cr}$$
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3.89/ For a) hypergeometric with $N = 10, n = 3, r = 2$ so table
of values of the pdf is
$$\matrix{ y &   0   &  1    & 2 \cr
       P(Y=y)& 14/30 & 14/30 & 2/30\cr}$$
For b), $N = 10, n = 3, r = 4$, so
 $$\matrix{ y &   0   &  1    & 2   & 3 \cr
       P(Y=y)& 5/30 & 15/30 & 9/30\ & 1/30 \cr}$$
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3.95/ The number of {\it improperly drilled gear boxes} $Y$ is a 
hypergeometric random variable with $N = 20, n = 5, r = 2$.
For a), we want 
$$P(Y = 0) = {{2\choose 0}{18\choose 5}\over {20\choose 5}} =42/76 \doteq .553$$ 
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 
$$E(T) = 9 E(Y) + 5  = 9\cdot {5\cdot 2\over 20} + 5 =  9.5$$
The standard deviation is obtained from 
$$V(T) = 81 V(Y) = 81 \cdot 5\cdot {2\over 20}\cdot {18\over 20}\cdot 
{15\over 19} \doteq 28.78$$
Then $\sigma \doteq \sqrt{28.78} \doteq 5.365$.
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3.97/ $Y$ is Poisson with $\lambda = 2$.
\item{a)} $P(Y = 4) = {2^4\over 4!} e^{-2} \doteq .090$.
\item{b)} Using the table of Poisson probabilities in text:
$P(Y \ge 4) = 1 - P(Y \le 3) = 1 - .857 = .143$.
\item{c)}  $P(Y < 4) = P(Y \le 3) = .857$.
\item{d)}  $P(Y\ge 4 | Y \ge 2) = .143/.594 = .241$ (using definition of
conditional probability).
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3.98/  $Y$ is Poisson with $\lambda = 7$, so
\item{a)} $P(Y \le 3) = .082$
\item{b)} $P(Y \ge 2) = 1 - P(Y \le 1) = 1 - .007 = .993$
\item{c)} $P(Y = 5) = P(Y \le 5) - P(Y \le 4) = .301-. 173 = .127$.
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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 $P(T > 150) = P(Y > 15) = 1 - P(Y \le 15) = 1 - .998 = .002$,
so no, this is quite unlikely.
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3.100/ Both probabilities equal ${14^2 \over 2!}e^{-14} \doteq .000082$.
Note:
$${7^2 \over 2!}e^{-7}\cdot {7^0 \over 0!}e^{-7} + 
{7^1 \over 1!}e^{-7}\cdot {7^1 \over 1!}e^{-7} + 
{7^0 \over 0!}e^{-7}\cdot {7^2 \over 2!}e^{-7} = {14^2 \over 2!}e^{-14}$$
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3.111/ $P = 50 - 2Y - Y^2$, where $Y$ is Poisson with mean $\lambda = 2$.
Hence $E(P) = 50 - 2E(Y) - E(Y^2)$.  We know $E(Y) = \lambda$ and
$E(Y^2) = \lambda^2 + \lambda$ from the derivation of the
formula for the variance of Poisson variables.  Hence
$E(P) = 50 - 4 - 6 = 40$.
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3.114/  
\item{a)}  We have 
$$p(y)/p(y-1) = {{\lambda^y\over y!} e^{-\lambda} \over 
{\lambda^{y-1}\over (y-1)!} e^{-\lambda}} = {\lambda \over y}$$
\item{b)}  
Hence $p$ is increasing for $y < \lambda$, and decreasing for $y > \lambda$.
\item{c)}  $p(y)$ is maximized with $y = \lfloor \lambda\rfloor$
(the greatest integer less than or equal to $\lambda$).
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3.117/  For a geometric random variable, we have 
$$\eqalign{m_Y(t) = E(e^{tY}) &= \sum_{y=1}^\infty e^{ty} q^{y-1} p \cr
&=pe^t \sum_{y=1}^\infty \left(q e^t\right)^{y-1}\cr
&={pe^t\over 1 - qe^t}\cr}$$
(by the geometric series summation formula).
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3.118/  Continuing from the formula in 3.117,
$$\eqalign{E(Y) = \mu_1' &= {d\over dt} \left({pe^t\over 1 - qe^t}\right)\vert_{t=0}\cr
&= {pe^t\over (1 - qe^t)^2}\vert_{t=0}\cr
&= {p\over (1-q)^2}\cr
&= {1\over p}\cr}$$
(the derivative is computed with the quotient rule and simplified).
Then 
$$\eqalign{E(Y^2) = \mu_2' &= {d^2\over dt^2} \left({pe^t\over 1 - qe^t}\right)\vert_{t=0}\cr
&={d\over dt} \left({pe^t\over (1 - qe^t)^2}\right)\vert_{t=0}\cr
&={(1-q)^2p+2pq(1-q)\over (1-q)^4}\cr
&={p+2q\over p^2}\cr
&={1+q\over p^2}\cr}$$
So 
$$V(Y) = E(Y^2) - (E(Y))^2 = {1+q\over p^2} - {1\over p^2} = {1-p\over p^2}.$$
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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,
\item{a)} Binomial, $n = 5$, $p = 1/3$.
\item{b)} Geometric, $p = 1/2$.
\item{c)} Poisson, $\lambda = 2$.
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3.126/  Letting $r(t) = \ln m(t)$, we have
$$r'(0) = {m'(0)\over m(0)} = {\mu_1' \over 1} - \mu_1' = E(Y)$$
Then by the quotient rule for derivatives,
$$r''(0) = {m(0) m''(0) - (m'(0))^2 \over m(0)^2} = \mu_2' - \mu_1'^2
= E(Y^2) - (E(Y))^2 = V(Y)$$ 
\bye