## Problem 7.4 (2 points)

We are given an observable random variable that is normally distributed with standard deviation $$\sigma = 4$$ square inches. Thus \begin{align*} P(-1 \le \overline{Y} - \mu \le 1) &= P\left( -\frac{\sqrt{n}}{4} \le \sqrt{n}\frac{\overline{Y} - \mu}{\sigma} \le \frac{\sqrt{n}}{4} \right) \\ &= P\left( -\frac{\sqrt{n}}{4} \le Z \le \frac{\sqrt{n}}{4} \right) \\ &= 0.90 \end{align*}

is satisfied for $$\sqrt{n}/4 = z_{0.05} \doteq 1.645$$, or $$n \doteq [(4)(1.645)]^2\doteq 43.30$$. Since $$n$$ must be an integer, we may take $$n=44$$ to guarantee that the probability is at least $$0.90$$.

## Problem 7.8 (2 points)

Let $$X = \overline{Y}_A - \overline{Y}_B$$. We are given $$E[Y_A] = E[Y_B]$$, and $$V[Y_A] = 0.4$$, $$V[Y_B] = 0.8$$, and $$n_A = n_B = 10$$. Then $$E[X] = 0$$ and $$V[X] = (0.4/10) + (0.8/10) = 0.12$$. Thus, \begin{align*} P(\overline{Y}_A > \overline{Y}_B + 1) &= P(X > 1) \\ &= P(X/\sqrt{0.12} > 1/\sqrt{0.12}) \\ &\doteq P(Z > 2.89) \\ &\doteq 0.0019. \end{align*}

## Problem 7.10 (4 points)

1. Let $$U$$ have a $$\chi^2$$ distribution with $$\nu$$ degrees of freedom. Then $$U$$ has a Gamma distribution with parameters $$\alpha=\nu/2$$ and $$\beta=2$$. Thus (see Theorem 4.8) \begin{align*} E[U] &= \alpha \beta = \nu, \\ V[U] &= \alpha \beta^2 = 2\nu. \end{align*}
2. $$U = (n-1) S^2 / \sigma^2$$ has a $$\chi^2$$ distribution with $$(n-1)$$ degrees of freedom. Thus, \begin{align*} E[S^2] &= E\left[\frac{\sigma^2}{(n-1)} U \right] = \frac{\sigma^2}{(n-1)} E[U] = \frac{\sigma^2}{(n-1)} (n - 1) = \sigma^2, \\ V[S^2] &= V\left[\frac{\sigma^2}{(n-1)} U \right] = \frac{\sigma^4}{(n-1)^2} V[U] \\ &= \frac{\sigma^4}{(n-1)^2} 2 (n - 1) = \frac{2\sigma^4}{(n-1)}. \end{align*}

## Problem 7.12 (2 points)

We are given $$n=9$$, thus $T = \sqrt{9}\cdot\frac{\overline{Y} - \mu}{S}$ has a $$t$$-distribution with $$\nu=8$$ degrees of freedom. Since $P( -t_{0.05,8} \le T \le t_{0.05,8}) = 0.90$ and $$t_{0.05,8} \doteq 1.86$$ we see that $P\left( -\frac{t_{0.05,8}}{\sqrt{9}}\cdot S \le \overline{Y} - \mu \le \frac{t_{0.05,8}}{\sqrt{9}}\cdot S\right) = 0.90,$ so we may take \begin{align*} g_1 &= -\frac{t_{0.05,8}}{\sqrt{9}}\cdot S \doteq -0.620\cdot S \\ g_2 &= \frac{t_{0.05,8}}{\sqrt{9}}\cdot S \doteq 0.620\cdot S. \end{align*}

## Problem 7.14 (4 points)

1. $$E[Z]=0$$, $$E[Z^2] = V[Z] + E[Z]^2 = 1 + 0^2 = 1$$.
2. $$T = \sqrt{\nu}\cdot Z\cdot Y^{-1/2}$$. By problem 4.90, if $$\nu>1$$ then $$E[Y^{-1/2}]$$ exists. Since $$Z$$ and $$Y$$ are independent, if $$\nu>1$$ then $E[T] = \sqrt{\nu}\cdot E[Z]\cdot E[Y^{-1/2}] = \sqrt{\nu}\cdot 0\cdot E[Y^{-1/2}] = 0.$ By problem 4.90, if $$\nu > 2$$ then $$E[Y^{-1}] = 1/(\nu - 2)$$. Thus, for $$\nu > 2$$ \begin{align*} V[T] &= E[T^2] - E[T]^2 = E[T^2] - 0^2 = E[T^2] \\ &= E[\nu\cdot Z^2\cdot Y^{-1}] \\ &= \nu\cdot E[Z^2]\cdot E[Y^{-1}] \qquad\text{(since $$Z$$ and $$Y$$ are independent)} \\ &= \nu\cdot 1 \cdot \frac{1}{\nu-2} \\ &= \frac{\nu}{\nu -2}. \end{align*} Therefore, if $$\nu>2$$ then a $$t$$-distributed random variable with $$\nu$$ degrees of freedom has mean 0 and variance $$\nu/(\nu-2)$$.

## Problem 7.16 (4 points)

Let $$W_1$$ and $$W_2$$ be independent $$\chi^2$$ distributed random variables with $$\nu_1$$ and $$\nu_2$$ degrees of freedom, respectively. Let $F = \frac{W_1/\nu_1}{W_2/\nu_2} = \nu_2\cdot \nu_1^{-1}\cdot W_1\cdot W_2^{-1}.$
1. By problem 4.90, if $$\nu_2 > 2$$ then $$E[W_2^{-1}] = 1/(\nu_2 - 2)$$. Since expected value is linear and since $$W_1$$ and $$W_2$$ are independent random variables, \begin{align*} E[F] &= E[\nu_2\cdot \nu_1^{-1}\cdot W_1\cdot W_2^{-1}] \\ &= \nu_2\cdot \nu_1^{-1}\cdot E[W_1]\cdot E[W_2^{-1}] \\ &= \nu_2\cdot \nu_1^{-1}\cdot \nu_1\cdot \frac{1}{\nu_2 - 2} \\ &= \frac{\nu_2}{\nu_2 -2}. \end{align*}
2. By problem 4.90, $$E[W_1^2]=\nu_1\cdot(\nu_1+2)$$, and if $$\nu_2 > 4$$ then $$E[W_2^{-2}] = 1/[(\nu_2 - 2)(\nu_2-4)]$$. Since expected value is linear and since $$W_1$$ and $$W_2$$ are independent random variables, \begin{align*} E[F^2] &= E[\nu_2^2\cdot \nu_1^{-2}\cdot W_1^2\cdot W_2^{-2}] \\ &= \nu_2^2\cdot \nu_1^{-2}\cdot E[W_1^2]\cdot E[W_2^{-2}] \\ &= \nu_2^2\cdot \nu_1^{-2}\cdot \nu_1\cdot(\nu_1+2)\cdot \frac{1}{(\nu_2 - 2)\cdot(\nu_2 - 4)} \\ &= \frac{\nu_2^2\cdot(\nu_1 + 2)}{\nu_1\cdot(\nu_2 -2)\cdot(\nu_2 - 4)}. \end{align*} Finally, the variance of $$F$$ is $$V[F] = E[F^2] - E[F]^2$$ which is, after simplification, $V[F] = \frac{2\cdot\nu_2^2\cdot(\nu_1 + \nu_2 -2)}{\nu_1\cdot (\nu_2 - 2)^2\cdot(\nu_2 - 4)}.$

## Problem 7.24 (2 points)

We are given that the population standard deviation is $$\sigma = 2.5$$ inches. We are trying to find a sample size $$n$$ so that $$P( |\overline{Y} - \mu | \le 0.4) \ge 0.95$$. This is equivalent to the condition $P\left(-\frac{0.4\cdot \sqrt{n}}{2.5} \le \sqrt{n}\cdot\frac{\overline{Y} - \mu}{\sigma} \le \frac{0.4\cdot \sqrt{n}}{2.5}\right) \ge 0.95.$ For a large enough sample size, $$\sqrt{n}\cdot (\overline{Y}-\mu)/\sigma$$ is well approximated by the standard normal random variable $$Z$$. Recall that $$z_{\alpha}$$ is the number defined by the property $$P(Z \ge z_{\alpha}) = \alpha$$, where $$0<\alpha<1$$. Since $$P(-z_{0.025} \le Z \le z_{0.025}) = 0.95$$, and since $$z_{0.025}\doteq 1.96$$, our estimated $$n$$ should satisfy $$0.4\cdot\sqrt{n}/2.5 \ge 1.96$$, so it suffices to take $n \ge (1.96\cdot 2.5 / 0.4)^2 \doteq (12.25)^2 \doteq 150.06.$ We will take $$n=151$$.

## Problem 7.26 (2 points)

We are given a sample size of $$n=40$$ and we are given that the range for pH is $$8-5=3$$, therefore we can estimate the standard deviation by $$\sigma \doteq 3/4 = 0.75$$. Using the estimate provided by the central limit theorem, we calculate \begin{align*} P(| \overline{Y} - \mu | \le 0.2) &= P\left( -\frac{0.2\cdot \sqrt{40}}{\sigma} \le \sqrt{40}\cdot \frac{\overline{Y} - \mu}{\sigma} \le \frac{0.2\cdot \sqrt{40}}{\sigma} \right) \\ &\doteq 1 - 2\cdot P( Z \ge 0.2\cdot \sqrt{40}/\sigma) \\ &\doteq 1 - 2\cdot P( Z \ge 0.2\cdot \sqrt{40}/0.75) \\ &\doteq 1 - 2\cdot P( Z \ge 1.69 ) \\ &\doteq 1 - 2\cdot 0.0455 \\ &\doteq 0.909. \end{align*}

## Problem 7.36 (2 points)

We are given a random sample $$Y_1, Y_2, \dotsc, Y_{50}$$, where $$E[Y_i]=\mu$$ and $$V[Y_i] = \sigma^2 \doteq 4$$. We want to find $$\mu$$ so that $$P(\sum_{i=1}^{50} Y_i > 200) \doteq 0.95$$. This is equivalent to the condition $$P(\overline{Y} > 4) \doteq 0.95$$ since $$\sum_{i=1}^{50} Y_i = 50\cdot \overline{Y}$$. If we suppose that $$\overline{Y}$$ is approximately normal, then $$\sqrt{50}\cdot (\overline{Y} - \mu)/2$$ is approximated by the standard normal random variable $$Z$$. Thus, $$P(\overline{Y} > 4)$$ is estimated by $$P(Z > \sqrt{50} \cdot (4 - \mu)/2)$$. Since $$P(Z > -1.645) \doteq 0.95$$, we see that the desired estimate for $$\mu$$ is obtained by solving the equation $$(4 - \mu)/2 \doteq -1.645$$. This yields $$\mu \doteq 4 + 2\cdot 1.645/\sqrt{50} \doteq 4.465$$.

## Problem 7.38 (2 points)

We are given independent random samples $$X_1, \dotsc, X_n$$ and $$Y_1, \dotsc, Y_n$$ from populations with means $$\mu_1$$ and $$\mu_2$$, and variances $$\sigma_1^2$$ and $$\sigma_2^2$$, respectively. Define the random variables $$W_1, \dotsc, W_n$$ by $$W_i = X_i - Y_i$$, $$i=1, \dotsc, n$$. The $$W_i$$ are clearly independent and identically distributed, $$E[W_i] = \mu_1 - \mu_2$$, $$V[W_i] = \sigma_1^2 + \sigma_2^2$$, $$i=1, \dotsc, n$$. The hypotheses of Theorem 7.4 are satisfied, so the distribution function for $U_n = \frac{\overline{W} - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2 + \sigma_2^2)/n}} = \frac{(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2 + \sigma_2^2)/n}}$ converges, as $$n\rightarrow\infty$$, to the distribution function for the standard normal random variable.

## Problem 7.40 (2 points)

\begin{align*} P( |(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)| \le 0.05 ) &= P\left(\left|\frac{(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2/n_1) + (\sigma_2^2/n_2)}}\right| \le \frac{0.05}{\sqrt{(0.01/50) + (0.02/100)}}\right) \\ &\doteq P( |Z| \le 2.50 ) = 1 - 2\cdot P(Z \ge 2.50) \\ &\doteq 1 - 2\cdot 0.0062 \\ &\doteq 0.9876. \end{align*}

## Problem 7.42 (2 points)

We are given a population mean $$\mu=2.5$$ minutes and standard deviation $$\sigma = 2$$ minutes. In a random sample of size $$n=100$$, we want to estimate \begin{align*} P\left(\sum_{i=1}^{100} Y_i > 240\right) &= P\left(\overline{Y} > 2.4\right) \\ &= P\left(\sqrt{n}\cdot \frac{\overline{Y} - \mu}{\sigma} > \sqrt{100}\cdot\frac{2.4 - 2.5}{2} \right) \\ &\doteq P(Z > -0.5) = 1 - P(Z > 0.5) \\ &\doteq 1 - 0.3085 = 0.6915. \end{align*}
---
title: "Solutions to Homework Assignment 2"
output: html_notebook
---

## Problem 7.4 (2 points)
We are given an observable random variable that is normally distributed with standard deviation \(\sigma = 4\) square inches. Thus
\begin{align*}
  P(-1 \le \overline{Y} - \mu \le 1) 
  &= P\left(
  -\frac{\sqrt{n}}{4}
  \le \sqrt{n}\frac{\overline{Y} - \mu}{\sigma}
  \le \frac{\sqrt{n}}{4}
  \right) \\
  &= P\left(
  -\frac{\sqrt{n}}{4}
  \le Z
  \le \frac{\sqrt{n}}{4}
  \right) \\
  &= 0.90
\end{align*}
is satisfied for \(\sqrt{n}/4 = z_{0.05} \doteq 1.645\), or \(n \doteq [(4)(1.645)]^2\doteq 43.30\). Since \(n\) must be an integer, we may take \(n=44\) to guarantee that the probability is at least \(0.90\).

## Problem 7.8 (2 points)
Let \(X = \overline{Y}_A - \overline{Y}_B\). We are given \(E[Y_A] = E[Y_B]\), and \(V[Y_A] = 0.4\), \(V[Y_B] = 0.8\), and \(n_A = n_B = 10\). Then \(E[X] = 0\) and \(V[X] = (0.4/10) + (0.8/10) = 0.12\). Thus,
\begin{align*}
  P(\overline{Y}_A > \overline{Y}_B + 1)
  &= P(X > 1) \\
  &= P(X/\sqrt{0.12} > 1/\sqrt{0.12}) \\
  &\doteq P(Z > 2.89) \\
  &\doteq 0.0019.
\end{align*}

## Problem 7.10 (4 points)
<ol type="a">
<li>
Let \(U\) have a \(\chi^2\) distribution with \(\nu\) degrees of freedom. Then \(U\) has a Gamma distribution with parameters \(\alpha=\nu/2\) and \(\beta=2\). Thus (see Theorem 4.8)
\begin{align*}
    E[U] &= \alpha \beta = \nu, \\
    V[U] &= \alpha \beta^2 = 2\nu.
\end{align*}
</li>
<li>
\(U = (n-1) S^2 / \sigma^2\) has a \(\chi^2\) distribution with \((n-1)\) degrees of freedom. Thus,
\begin{align*}
  E[S^2] &= 
  E\left[\frac{\sigma^2}{(n-1)} U \right] =
  \frac{\sigma^2}{(n-1)} E[U] =
  \frac{\sigma^2}{(n-1)} (n - 1) =
  \sigma^2, \\
  V[S^2] &= 
  V\left[\frac{\sigma^2}{(n-1)} U \right] =
  \frac{\sigma^4}{(n-1)^2} V[U] \\
  &=
  \frac{\sigma^4}{(n-1)^2} 2 (n - 1) =
  \frac{2\sigma^4}{(n-1)}.
\end{align*}
</li>
</ol>

##Problem 7.12 (2 points)
We are given \(n=9\), thus
\[
T = \sqrt{9}\cdot\frac{\overline{Y} - \mu}{S}
\]
has a \(t\)-distribution with \(\nu=8\) degrees of freedom. Since
\[
P( -t_{0.05,8} \le T \le t_{0.05,8}) = 0.90
\]
and \(t_{0.05,8} \doteq `r round(qt(.05,df=8,lower.tail=FALSE),3)`\) we see that
\[
P\left( -\frac{t_{0.05,8}}{\sqrt{9}}\cdot S \le \overline{Y} - \mu \le \frac{t_{0.05,8}}{\sqrt{9}}\cdot S\right) = 0.90,
\]
so we may take
\begin{align*}
g_1 &= -\frac{t_{0.05,8}}{\sqrt{9}}\cdot S \doteq -0.620\cdot S \\
g_2 &= \frac{t_{0.05,8}}{\sqrt{9}}\cdot S \doteq 0.620\cdot S.
\end{align*}

##Problem 7.14 (4 points)
<ol type="a">
<li>\(E[Z]=0\), \(E[Z^2] = V[Z] + E[Z]^2 = 1 + 0^2 = 1\).
</li>
<li>
\(T = \sqrt{\nu}\cdot Z\cdot Y^{-1/2}\). By problem 4.90, if \(\nu>1\) then \(E[Y^{-1/2}]\) exists. Since \(Z\) and \(Y\) are independent, if \(\nu>1\) then 
\[
E[T] = \sqrt{\nu}\cdot E[Z]\cdot E[Y^{-1/2}] = \sqrt{\nu}\cdot 0\cdot E[Y^{-1/2}] = 0.
\]
By problem 4.90, if \(\nu > 2\) then \(E[Y^{-1}] = 1/(\nu - 2)\). Thus, for \(\nu > 2\)
\begin{align*}
  V[T] &= E[T^2] - E[T]^2 = E[T^2] - 0^2 = E[T^2] \\
    &= E[\nu\cdot Z^2\cdot Y^{-1}] \\
    &= \nu\cdot E[Z^2]\cdot E[Y^{-1}] \qquad\text{(since \(Z\) and \(Y\) are independent)} \\
    &= \nu\cdot 1 \cdot \frac{1}{\nu-2} \\
    &= \frac{\nu}{\nu -2}.
\end{align*}
Therefore, if \(\nu>2\) then a \(t\)-distributed random variable with \(\nu\) degrees of freedom has mean 0 and variance \(\nu/(\nu-2)\).
</li>
</ol>

##Problem 7.16 (4 points)
Let \(W_1\) and \(W_2\) be independent \(\chi^2\) distributed random variables with \(\nu_1\) and \(\nu_2\) degrees of freedom, respectively. Let
\[
F = \frac{W_1/\nu_1}{W_2/\nu_2} = \nu_2\cdot \nu_1^{-1}\cdot W_1\cdot W_2^{-1}.
\]
<ol type="a">
<li>
By problem 4.90, if \(\nu_2 > 2\) then \(E[W_2^{-1}] = 1/(\nu_2 - 2)\). Since expected value is linear and since \(W_1\) and \(W_2\) are independent random variables,
\begin{align*}
  E[F] &= E[\nu_2\cdot \nu_1^{-1}\cdot W_1\cdot W_2^{-1}] \\
    &= \nu_2\cdot \nu_1^{-1}\cdot E[W_1]\cdot E[W_2^{-1}] \\
    &= \nu_2\cdot \nu_1^{-1}\cdot \nu_1\cdot \frac{1}{\nu_2 - 2} \\
    &= \frac{\nu_2}{\nu_2 -2}.
\end{align*}
</li>
<li>
By problem 4.90, \(E[W_1^2]=\nu_1\cdot(\nu_1+2)\), and if \(\nu_2 > 4\) then \(E[W_2^{-2}] = 1/[(\nu_2 - 2)(\nu_2-4)]\). Since expected value is linear and since \(W_1\) and \(W_2\) are independent random variables,
\begin{align*}
  E[F^2] &= E[\nu_2^2\cdot \nu_1^{-2}\cdot W_1^2\cdot W_2^{-2}] \\
    &= \nu_2^2\cdot \nu_1^{-2}\cdot E[W_1^2]\cdot E[W_2^{-2}] \\
    &= \nu_2^2\cdot \nu_1^{-2}\cdot \nu_1\cdot(\nu_1+2)\cdot \frac{1}{(\nu_2 - 2)\cdot(\nu_2 - 4)} \\
    &= \frac{\nu_2^2\cdot(\nu_1 + 2)}{\nu_1\cdot(\nu_2 -2)\cdot(\nu_2 - 4)}.
\end{align*}
Finally, the variance of \(F\) is \(V[F] = E[F^2] - E[F]^2\) which is, after simplification,
\[
V[F] = \frac{2\cdot\nu_2^2\cdot(\nu_1 + \nu_2 -2)}{\nu_1\cdot (\nu_2 - 2)^2\cdot(\nu_2 - 4)}.
\]
</li>
</ol>

##Problem 7.24 (2 points)
We are given that the population standard deviation is \(\sigma = 2.5\) inches. We are trying to find a sample size \(n\) so that \(P( |\overline{Y} - \mu | \le 0.4) \ge 0.95\). This is equivalent to the condition
\[
P\left(-\frac{0.4\cdot \sqrt{n}}{2.5}
\le \sqrt{n}\cdot\frac{\overline{Y} - \mu}{\sigma} 
\le \frac{0.4\cdot \sqrt{n}}{2.5}\right) \ge 0.95.
\]
For a large enough sample size, \(\sqrt{n}\cdot (\overline{Y}-\mu)/\sigma\) is well approximated by the standard normal random variable \(Z\). Recall that \(z_{\alpha}\) is the number defined by the property \(P(Z \ge z_{\alpha}) = \alpha\), where \(0<\alpha<1\).
Since \(P(-z_{0.025} \le Z \le z_{0.025}) = 0.95\), and since \(z_{0.025}\doteq 1.96\), our estimated \(n\) should satisfy \(0.4\cdot\sqrt{n}/2.5 \ge 1.96\), so it suffices to take 
\[
n \ge (1.96\cdot 2.5 / 0.4)^2 \doteq (12.25)^2 \doteq 150.06.
\]
We will take \(n=151\).

##Problem 7.26 (2 points)
We are given a sample size of \(n=40\) and we are given that the range for pH is \(8-5=3\), therefore we can estimate the standard deviation by \(\sigma \doteq 3/4 = 0.75\). Using the estimate provided by the central limit theorem, we calculate
\begin{align*}
  P(| \overline{Y} - \mu | \le 0.2) &=
  P\left( -\frac{0.2\cdot \sqrt{40}}{\sigma} \le
  \sqrt{40}\cdot \frac{\overline{Y} - \mu}{\sigma}
  \le \frac{0.2\cdot \sqrt{40}}{\sigma} \right) \\
  &\doteq 1 - 2\cdot P( Z \ge 0.2\cdot \sqrt{40}/\sigma) \\
  &\doteq 1 - 2\cdot P( Z \ge 0.2\cdot \sqrt{40}/0.75) \\
  &\doteq 1 - 2\cdot P( Z \ge 1.69 ) \\
  &\doteq 1 - 2\cdot 0.0455 \\
  &\doteq 0.909.
\end{align*}

##Problem 7.36 (2 points)
We are given a random sample \(Y_1, Y_2, \dotsc, Y_{50}\), where \(E[Y_i]=\mu\) and 
\(V[Y_i] = \sigma^2 \doteq 4\). We want to find \(\mu\) so that
\(P(\sum_{i=1}^{50} Y_i > 200) \doteq 0.95\). This is equivalent to the condition
\(P(\overline{Y} > 4) \doteq 0.95\) since \(\sum_{i=1}^{50} Y_i = 50\cdot \overline{Y}\).
If we suppose that \(\overline{Y}\) is approximately normal, then 
\(\sqrt{50}\cdot (\overline{Y} - \mu)/2\) is approximated by the standard normal random variable \(Z\). 
Thus, \(P(\overline{Y} > 4)\) is estimated by \(P(Z > \sqrt{50} \cdot (4 - \mu)/2)\). Since 
\(P(Z > -1.645) \doteq 0.95\), we see that the desired estimate for \(\mu\) is obtained by solving the equation
\((4 - \mu)/2 \doteq -1.645\). This yields \(\mu \doteq 4 + 2\cdot 1.645/\sqrt{50} \doteq 4.465\).

##Problem 7.38 (2 points)
We are given independent random samples \(X_1, \dotsc, X_n\) and \(Y_1, \dotsc, Y_n\) from populations with means \(\mu_1\) and \(\mu_2\), and variances \(\sigma_1^2\) and \(\sigma_2^2\), respectively. Define the random variables \(W_1, \dotsc, W_n\) by \(W_i = X_i - Y_i\), \(i=1, \dotsc, n\). The \(W_i\) are clearly independent and identically distributed,
\(E[W_i] = \mu_1 - \mu_2\), \(V[W_i] = \sigma_1^2 + \sigma_2^2\), \(i=1, \dotsc, n\). The hypotheses of Theorem 7.4 are satisfied, so the distribution function for
\[
U_n = \frac{\overline{W} - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2 + \sigma_2^2)/n}}
  = \frac{(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2 + \sigma_2^2)/n}}
\]
converges, as \(n\rightarrow\infty\), to the distribution function for the standard normal random variable.

##Problem 7.40 (2 points)
\begin{align*}
  P( |(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)| \le 0.05 )
  &= P\left(\left|\frac{(\overline{X} - \overline{Y}) - (\mu_1 - \mu_2)}{\sqrt{(\sigma_1^2/n_1) + (\sigma_2^2/n_2)}}\right|
  \le \frac{0.05}{\sqrt{(0.01/50) + (0.02/100)}}\right) \\
  &\doteq P( |Z| \le 2.50 ) = 1 - 2\cdot P(Z \ge 2.50) \\
  &\doteq 1 - 2\cdot 0.0062 \\
  &\doteq 0.9876.
\end{align*}

##Problem 7.42 (2 points)
We are given a population mean \(\mu=2.5\) minutes and standard deviation \(\sigma = 2\) minutes. In a random sample of size \(n=100\), we want to estimate
\begin{align*}
P\left(\sum_{i=1}^{100} Y_i > 240\right) 
  &= P\left(\overline{Y} > 2.4\right) \\
  &= P\left(\sqrt{n}\cdot \frac{\overline{Y} - \mu}{\sigma} > \sqrt{100}\cdot\frac{2.4 - 2.5}{2} \right) \\
  &\doteq P(Z > -0.5) = 1 - P(Z > 0.5) \\
  &\doteq 1 - 0.3085 = 0.6915.
\end{align*}










