\magnification=\magstep1
\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 2 Solutions}
\centerline{February 6, 2004}
\bigskip
\noindent
7.23/Since $\sigma = 2.5$ and $n = 100$, by the Central Limit Theorem 
we have
$$P(-.5 \le \overline{Y} - \mu \le .5)=
P\left(10\cdot {-.5 \over 2.5} \le 
10\cdot {\overline{Y} - \mu \over 2.5}\le 10\cdot{.5\over 2.5}\right)
\doteq P(-2 \le Z \le 2)$$
(where $Z$ is standard normal -- note that $n = 100$ is in 
the range where we expect 
$$\sqrt{n} \cdot {\overline{Y} - \mu \over \sigma}$$
to be approximately normal).  Hence from the standard normal 
table, the desired probability is $.9544$.

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\noindent
7.24/Continuing from the setup of the previous problem, to get
$$P\left({(-.4)\sqrt{n} \over 2.5} \le 
\sqrt{n}\cdot {\overline{Y} - \mu \over 2.5}\le {(.4)\sqrt{n}\over 2.5}\right) = .95$$
we treat $\sqrt{n}\cdot {\overline{Y} - \mu \over 2.5}$
as approximately standard normal.  Then from the normal table we 
want ${(.4)\sqrt{n}\over 2.5} = 1.96$, so $n = 151$ (rounding up to 
the next largest whole number.

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\noindent
7.25/Since the size of the ethnic group, $n = 64$, is relatively
large, we can answer the given question by finding the probability
that a randomly selected sample of size $n = 64$ has mean wage
less than or equal to \$6.90, applying the Central Limit Theorem:
$$P(\overline{Y} \le 6.90) \doteq 
P\left(Z \le \sqrt{64}\cdot {6.90 - 7.00\over .5}\right) = P(Z \le -1.6) = .0548$$
In other words, there is about a $5.5\%$ chance that a random sample
of size $n = 64$ will have mean wage that small.  In other words,
it's {\it possible}, but
quite unlikely, that the workers from this ethnic group are being treated 
fairly.

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\noindent
7.33/ a)  No, it does not seem that the concentrations are
normally distributed.  With a normal distribution 
and $\mu =12$ and $\sigma = 9$, there
would be a significant chance that the value would be negative
($P(Y < 0) = P(Z < -1.33) = .0918$).  But on physical grounds,
negative values are not possible for concentrations.
b) As in the previous problems, $n = 100$ is large enough
that we expect the Central Limit Theorem to have ``kicked in'':
$$P(\overline{Y} > 14) \doteq P\left(Z \ge 10\cdot {14 - 12\over 9}\right)
= .0132.$$

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\noindent
7.37/Each lamp has mean life $\mu = 50$ and standard deviation
$\sigma = 4$.  There are 25 lamps in all, so writing $Y_i$ for
the life of the $i$th lamp,
we are looking for
$$P\left(\sum_{i=1}^2 Y_i > 1300\right) = P(\overline{Y} > 1300/25) = P(\overline{Y} > 52)$$
As in the previous problems, we will assume that $n$ is large enough
so that the Central Limit Theorem applies (with somewhat less ``ironclad''
justification this time!):
$$P(\overline{Y} > 52) = P\left(Z > \sqrt{25}\cdot{52 - 50\over 4}\right) = P(Z > 2.5) = .0062,$$
by the standard normal table.

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\noindent
7.38/Let $W_i = X_i - Y_i$.  Then the $W_i$ are independent and 
identically distributed with 
$$E(W_i) = \mu_1 - \mu_2\quad V(W_i) = \sigma_1^2 + \sigma_2^2$$
Thus the Central Limit Theorem applies and we get
$$U_n = {{1\over n}\sum_{i=1}^n (X_i - Y_i) - (\mu_1 - \mu_2)
\over \sqrt{ {\sigma_1^2 + \sigma_2^2\over n} }} = 
\sqrt{n}\left( {\overline{X} - \overline{Y} - (\mu_1 - \mu_2) 
\over \sqrt{\sigma_1^2 + \sigma_2^2} }\right) $$
converges to $N(0,1)$ in distribution as $n \to \infty$.

\bigskip
\noindent
7.74/From last semester, the moment generating function of a
Poisson random variable $Y$ with parameter $\lambda$ has the form
$$m_Y(t) = \exp(\lambda(e^t - 1))$$
Hence the moment generating function of 
$Z = (Y - \lambda)/\sqrt{\lambda} = Y/\sqrt{\lambda} - \sqrt{\lambda}$
is (using the formula $m_{aY + b}(t) = e^{bt}m_Y(at)$):
$$\eqalign{
m_Z(t) &= \exp(-t \sqrt{\lambda} ) m_Y(t/\sqrt{\lambda})\cr
&=\exp(-t \sqrt{\lambda}) \exp(\lambda(e^{t/\sqrt{\lambda}} - 1))\cr
&=\exp(\lambda e^{t/\sqrt{\lambda}} - \lambda - t\sqrt{\lambda})\cr}$$
Now we use the Taylor series for the exponential to expand the 
``inside'' term 
$$e^{t\sqrt{\lambda}} = 1 + t/\sqrt{\lambda} + t^2/(2\lambda) 
+ t^3/(6\lambda^{3/2}) + \cdots $$
Substituting this into the above, the first two terms cancel out, 
and we are left with:
$$m_Z(t) = \exp(t^2/2 + t^3/(6\lambda^{1/2}) + \cdots )$$
Hence (since all terms after the first have powers of $\lambda$
in the denominator):
$$\lim_{\lambda\to\infty} m_Z(t) = \exp(t^2/2)$$
which is the mgf for a standard normal.  This shows that
$Z \to N(0,1)$ in distribution as $\lambda \to \infty$.
\bye
 
