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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Review Problems, Exam 2}
\centerline{November 7, 2005}
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\noindent
3.157 a) Let $Y$ be the number of imperfections.
Using the formula for Poisson probabilities, we get with $\lambda = 4$
$$P(Y \ge 1) = \sum_{y=1}^\infty {4^y\over y!}e^{-4} = 1 - e^{-4} \doteq .982$$
For b), the idea is the same but we take $\lambda = 12$
$$P(Y \ge 1) = \sum_{y=1}^\infty {12^y\over y!}e^{-12} = 1 - e^{-12} \doteq .999994$$
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\noindent
3.158/Here in an 8 square yard bolt, the expected number of 
imperfections is $\lambda = 32$.  So $E(C) = E(10Y) = 10E(Y) = 10\cdot 32 = 320$.
$V(C) = V(10Y) = 10^2V(Y) = 100\cdot 32 = 3200$ so $\sigma = \sqrt{3200} = 56.57$.
(We are using the formulas $E(Y) = V(Y) = \lambda$ for a Poisson random
variable here.)
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\noindent
3.160/We have 
$$m(t) = (pe^t + q)^n = \left(p\left(1 + t + {t^2\over 2} + \cdots\ \right) + q\right)^n
= \left(1 + pt + {pt^2\over 2} + \cdots\ \right)^n$$
Expanding and keeping only terms with $t$ and $t^2$ (all we need for the 
problem), we get:
$$= 1 + npt + \left({n\choose 2} p^2 + {n\over 2} p\right) t^2 + \cdots $$
Hence 
$\mu_1' = m'(0) = np$ and $\mu_2' = m''(0) = n(n-1)p^2 + np$.
(This agrees with calculations we did in class for the binomial random 
variables!)
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\noindent
3.173 a)  Hypergeometric with $N = 100, r = 40$ (the defectives), and $n = 20$,
so 
$$p(10) = {{40\choose 10}{60 \choose 10}\over {100\choose 20}} \doteq .119$$
(Note: The approximate value would be obtained by expressing the 
binomial coefficients as factorials, cancelling as many common factors
as you can, then carefully multiplying everything out -- straightforward
but very tedious.  On an exam, of course, the binomial coefficient form
would be acceptable!!)
b)  Using the binomial distribution (treating the selection of the sample
of 20 as independent events each with a probability $p = 40/100 = .4$ of getting a defective),
$$p(10) \doteq {20\choose 10} (.4)^{10}(.6)^{10} \doteq .117$$
Note that here the binomial probability is a good approximation to
the hypergeometric probability (the exact value).  The idea is that $N = 40$
is large enough so that the probabilities of getting a defective do not
change that much in successive trials.
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\noindent
4.67/See class notes.
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\noindent
4.88/By inspection, this has a gamma distribution with $\alpha = 3$, $\beta = 1/2$.
So $E(Y) = \alpha\beta = 3/2$ and $V(Y) = \alpha\beta^2 = 3/4$.
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\noindent
4.101/a)  This is the form for a beta density function with 
$\alpha = 3$ and $\beta = 5$.  Thus $c = {1\over B(3,5)} = 105$
b)  $E(Y) = {\alpha\over \alpha+\beta} = {3\over 8}$.
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\noindent
4.110/a)  Using the general formula for a uniformly distributed random variable,
with $\theta_1 = 0$ and $\theta_2 = 1$, we have 
$$m_Y(t) = {e^t - 1\over t}$$
For the next parts we use the formula $m_{aY+b}(t) = e^{bt} m_Y(at)$ proved
on Problem Set 8.  b) We get:
$$m_{aY}(t) = {e^{at} - 1\over at}$$
This is the mgf for a uniform r.v. on $[0,a]$.
c)  We get:
$$m_{-aY}(t) = {e^{-at} - 1\over -at} = {1 - e^{-at}\over at}$$
This is the mgf for a uniform r.v. on $[-a,0]$.
d) We get
$$m_{aY+b}(t) = e^{bt}{e^{at} - 1\over at} = {e^{(a+b)t} - e^{bt}\over at}$$
This is the mgf for a uniform r.v. on $[b,a+b]$.
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\noindent
4.128/a)  The distribution function (i.e. cdf) is 
$$\eqalign{
F(y) = \int_{-\infty}^y f(y)\ dy &= \cases{ 0 & if $y < -1$\cr
{2\over \pi}\int_{-1}^y {dy \over y^2 + 1} & if $-1 \le y < 1$\cr
1 & if $y \ge 1$\cr}\cr
&= \cases{ 0 & if $y < -1$\cr
{2\over\pi}\arctan(y) + {1\over 2} & if $-1 \le y < 1$\cr
1 & if $y \ge 1$\cr}\cr}$$
b)  
$$E(Y) = \int_{-1}^1 y\cdot {2\over \pi} {dy\over y^2 + 1} = 0$$
since the integrand is an odd function and the interval is symmetric about
zero.
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\noindent
4.129/Let $Y$ be the time to finish the exam.
We want to find the $z$ so that the standard normal $Z = {Y - 70\over 12}$
satisfies $P(Z > z) = .1$.  From the normal probability table,
we see $z \doteq 1.28$, so $Y = 70 + (1.28)(12) \doteq 85.4$ minutes.
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\noindent
4.52/Let $Y$ be the diameter.  Then we have $Z = {Y - 3.0005\over .001}$
is a standard normal.  The tolerance range is $3.000 \pm .002$, so 
we want 
$$1 - P\left({2.998 - 3.0005\over .001} < Z < {3.002 - 3.0005\over .001}\right)
=1 - P(-2.5 < Z < 1.5)$$
Using the symmetry of the normal density, this is the same as $1 - .927 = .073$.
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\noindent
4.131/Using the binomial probability formula, the answer
is $\sum_{y=1}^5 {5 \choose y} (.927)^y (.073)^{5-y} = 1 - (.927)^5 \doteq .3155$.
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\noindent
4.132/``Standardize'' and use the normal probability table.
a)  $$P(Y < 60) = P\left({Y-75\over 12} < {60 - 75\over 12}\right)= .1056$$
b)  $P(Y \ge 60) = 1 - .1056 = .8944$.
c)  $P(Y \ge 90) = P\left({Y-75\over 12} \ge {90 - 75\over 12}\right) = .1056$.
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\noindent
4.133/By inspection, this is the density for a gamma-distributed r.v.
with $\alpha = 2$ and $\beta = 1/2$.  Hence
a) $c = {1\over \Gamma(2) (1/2)^2} = 4$.  b)  $\mu = \alpha\beta = 1$ and
$\sigma^2 = \alpha\beta^2 = 1/2$.  c) $m(t) = (1 - t/2)^{-2}$.
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\noindent
4.135/For a beta-distributed r.v.
$$\eqalign{
E(Y^k) &= \int_0^1 y^k \cdot {y^{\alpha-1}(1-y)^{\beta-1} \over B(\alpha,\beta)}\ dy\cr
&= \int_0^1 {y^{\alpha+k-1}(1-y)^{\beta-1} \over B(\alpha,\beta)}\ dy\cr 
&={B(\alpha+k,\beta)\over B(\alpha,\beta)}\cr
&={\Gamma(\alpha+k)\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\alpha+\beta+k)}\cr}$$
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\noindent
4.136/The probability that the time $T$ to the first arrival is greater than $t_0$
$P(T > t_0) = P(N = 0)$ at $t = t_0$ which is the Poisson probability
$${(\lambda t_0)^0 \over 0!} e^{-\lambda t_0} = e^{-\lambda t_0}$$
Hence the cdf for $T$ is 
$$F(t) = P(T \le t) = 1 - P(T > t) = 1 - e^{-\lambda t}$$
The pdf $f(t) = F'(t) = \lambda e^{-\lambda t}$, which is exponential with 
$\beta = {1\over \lambda}$.  
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\noindent
4.137/With $\lambda = 10$, $\int_{.25}^\infty 10e^{-10t}\ dt \doteq .082$
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\noindent
4.142/$T$ is exponential witth $\beta = 1/2$, so 
$$\int_{.25}^\infty 2e^{-2y}\ dy = e^{-.5} \doteq .607.$$
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\noindent
4.150/We must split the integral for the mgf at $0$ because
the formula for $|y|$ changes there.  Also in order to get convergence
of these integrals, we must have $|t| < 1$, so $1 + t > 0$ and $1 - t > 0$:
$$\eqalign{
m(t) &= \int_{-\infty}^\infty e^{ty} f(y)\ dy\cr
     &= {1\over 2}\int_{-\infty}^0 e^{ty} e^y\ dy + {1\over 2}\int_0^\infty e^{ty} e^{-y}\ dy\cr
     &= {1\over 2}\cdot {1\over 1+t} + {1\over 2}\cdot {1\over 1 - t}\cr
     &= {1\over 1-t^2}\cr}$$
Hence $E(Y) = m'(0) = {2t\over 1 - t^2}|_{t=0} = 0$.
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\noindent
4.151/Just use our general formulas for all this.  For part a, you must
show that $f(y) \ge 0$ for all $y$ and $\int_{-\infty}^\infty f(y)\ dy = 1$.
This follows clearly since $f_1$ and $f_2$ are densities, so the 
total integral of $f$ is $a\cdot 1 + (1 - a)\cdot 1 = 1$.  The rest follows
from the usual rules
$$E(Y) = \int_{-\infty}^\infty yf(y)\ dy$$
and 
$$V(Y) = E(Y^2) - (E(Y))^2$$    
\bye