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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions -- Midterm Exam 2}
\centerline{November 23, 2003}
\bigskip
\noindent
I.  If $Y$ denotes the number of underage students who get carded,
then $Y$ is hypergeometric with $N = 10$, $r = 4$, $n = 5$.  So
$$P(Y = 2) = {{4 \choose 2} {6\choose 3}\over {10 \choose 5}}$$
\bigskip
\noindent
II.  By examining the form of the density function, we see
that $Y$ has a {\it beta distribution} with $\alpha = \beta = 3$.
Hence:
\item{A)}  The constant $c$ must be 
$$c = {1\over B(3,3)} = {\Gamma(6)\over \Gamma(3)\Gamma(3)} = {5!\over 2!2!} = 30$$
\item{B)}  The variance is 
$$V(Y) = {\alpha\beta\over (\alpha+\beta)^2(\alpha+\beta+1)} = {3\cdot 3\over 36\cdot 7}
 = {1\over 28}$$
\item{C)}  The cumulative distribution function is the antiderivative
$F(y)$ of the density with $F(-\infty) = 0$ and $F(+\infty) = 1$.  This
is
$$F(y) = \cases{0 & if $y < 0$\cr
                10y^3 - 15y^4 + 6y^5 & if $0 \le y < 1$\cr
                1 & if $y \ge 1$\cr}$$
\item{D)} This is 
$$\int_{.1}^{.25} 30y^2(1-y)^2\ dy = F(.25) - F(.1) \doteq .09496.$$
\bigskip
\noindent
III.  The lifetime $Y$ of a single switch has the exponential density
$$f(y) = \cases{{1\over 2}e^{-y/2} & if $y > 0$\cr
                    0 & otherwise\cr}$$
\item{A)}  The probability that a single switch fails during the 
first year is the same as the probability that its life is less
than 1:
$$P(Y < 1) = \int_0^1 {1\over 2}e^{-y/2}\ dy = -e^{-y/2}|_0^1 = 1 - e^{-1/2}
\doteq .3935$$
\item{B)}  The probability that at most 30 out of 100 of these switches
will fail in the first year is computed by the binomial probability 
formula (since we assume the switches operate independently):
$$\sum_{y=0}^{30} {100\choose y} (.3935)^y (.6065)^{100-y}.$$
\bigskip
\noindent
IV.  
\item{A)}  Let $Y$ be the weight of one poodle
$$P(7.3 < Y < 9.1) = P\left({7.3-8\over .9} < {Y-8\over .9} < {9.1-8\over .9}\right)$$
This is the same as 
$$P\left(-.78 < {Y-8\over .9} < 1.22\right)$$
We know $Z = {Y-8\over .9}$ is a standard normal, so we can use
the {\it standard normal table} (p. 792) to find this probability.  
By the symmetry of the normal density,
$$P(-.78 < Z < 0) = P(0 < Z < .78) = .5 - .2177 = .2823$$
We also have 
$$P(0 < Z < 1.22) = .5 - .1112 = .3888$$
Hence
$$P(-.78 < Z < 1.22) = .2823 + .3888 = .6711$$
\bigskip
\item{B)}  The only reasonable way to do this problem is to think of 
the moment-generating function approach.  $E(Y^3) = m'''(0)$ where
$m(t)$ is the mgf for a normal random variable.  The  general form
of this is
$$m(t) = e^{{\sigma^2 t\over 2} + \mu t}$$
Hence differentiating with the chain and product rules:
$$\eqalign{
m'(t) &= e^{{\sigma^2 t\over 2} + \mu t} ( \sigma^2 t + \mu)\cr
m''(t) &= e^{{\sigma^2 t\over 2} + \mu t} (( \sigma^2 t + \mu)^2 + \sigma^2)\cr
m'''(t) &= e^{{\sigma^2 t\over 2} + \mu t} (( \sigma^2 t + \mu)^3 + 3\sigma^2(\sigma^2t + \mu))\cr
\Rightarrow m'''(0) &= \mu^3 + 3\sigma^2 \mu\cr}$$
With $\mu = 8$ and $\sigma = .9$, this yields $E(Y^3) = 531.44$.
\bigskip
\noindent
{\it Extra Credit.}  This is the density for a Gamma-distributed random 
variable with $\alpha = 2$ and $\beta = 1$.  Hence
$$m(t) = (1 - t)^{-2}$$
(This can also be computed directly as
$$E(e^{tY}) = \int_0^\infty e^{ty}\cdot y e^{-y}\ dy.)$$
\bye