Solution

If you look at the graph of the function (above and to the right) of \(Y=X^2\), you might note that (1) the function is an increasing function of \(X\), and (2) \(0<y<1\). That noted, let's now use the distribution function technique to find the p.d.f. of \(Y\). First, we find the cumulative distribution function of \(Y\):

Having shown that the cumulative distribution function of \(Y\) is:

\(F_Y(y)=y^{3/2}\)

for \(0<y<1\), we now just need to differentiate \(F(y)\) to get the probability density function \(f(y)\). Doing so, we get:

\(f_Y(y)=F'_Y(y)=\dfrac{3}{2} y^{1/2}\)

for \(0<y<1\). Our calculation is complete! We have successfully used the distribution function technique to find the p.d.f of \(Y\), when \(Y\) was an increasing function of \(X\). (By the way, you might find it reassuring to verify that \(f(y)\) does indeed integrate to 1 over the support of \(y\). In general, that's not a bad thing to check.)

Solution

If you look at the graph of the function (above and to the right) of:

\(Y=(1-X)^3\)

you might note that the function is a decreasing function of \(X\), and  \(0<y<1\). That noted, let's now use the distribution function technique to find the p.d.f. of \(Y\). First, we find the cumulative distribution function of \(Y\):

Having shown that the cumulative distribution function of \(Y\) is:

\(F_Y(y)=y\)

for \(0<y<1\), we now just need to differentiate \(F(y)\) to get the probability density function \(f(y)\). Doing so, we get:

\(f_Y(y)=F'_Y(y)=1\)

for \(0<y<1\). That is, \(Y\) is a \(U(0,1)\) random variable. (Again, you might find it reassuring to verify that \(f(y)\) does indeed integrate to 1 over the support of \(y\).)

Solution

Note that the function:

\(Y=X^2\)

defined over the interval \(0<x<1\) is an invertible function. The inverse function is:

\(x=v(y)=\sqrt{y}=y^{1/2}\)

for \(0<y<1\). (That range is because, when \(x=0, y=0\); and when \(x=1, y=1\)). Now, taking the derivative of \(v(y)\), we get:

\(v'(y)=\dfrac{1}{2} y^{-1/2}\)

Therefore, the change-of-variable technique:

\(f_Y(y)=f_X(v(y))\times |v'(y)|\)

tells us that the probability density function of \(Y\) is:

\(f_Y(y)=3[y^{1/2}]^2\cdot \dfrac{1}{2} y^{-1/2}\)

And, simplifying we get that the probability density function of \(Y\) is:

\(f_Y(y)=\dfrac{3}{2} y^{1/2}\)

for \(0<y<1\). We shouldn't be surprised by this result, as it is the same result that we obtained using the distribution function technique.

Solution

Note that the function:

\(Y=(1-X)^3\)

defined over the interval \(0<x<1\) is an invertible function. The inverse function is:

\(x=v(y)=1-y^{1/3}\)

for \(0<y<1\). (That range is because, when \(x=0, y=1\); and when \(x=1, y=0\)). Now, taking the derivative of \(v(y)\), we get:

\(v'(y)=-\dfrac{1}{3} y^{-2/3}\)

Therefore, the change-of-variable technique:

\(f_Y(y)=f_X(v(y))\times |v'(y)|\)

tells us that the probability density function of \(Y\) is:

\(f_Y(y)=3[1-(1-y^{1/3})]^2\cdot |-\dfrac{1}{3} y^{-2/3}|=3y^{2/3}\cdot \dfrac{1}{3} y^{-2/3} \)

And, simplifying we get that the probability density function of Y is:

\(f_Y(y)=1\)

for \(0<y<1\). Again, we shouldn't be surprised by this result, as it is the same result that we obtained using the distribution function technique.

Solution

First, note that the transformation:

\(Y=X^2\)

is not one-to-one over the interval \(-1<x<2\):

For example, in the interval \(-1<x<1\), if we take the inverse of \(Y=X^2\), we get:

\(X_1=-\sqrt{Y}=v_1(Y)\)

for \(-1<x<0\), and:

\(X_2=+\sqrt{Y}=v_2(Y)\)

for \(0<x<1\).

As the graph suggests, the transformation is two-to-one between when \(0<y<1\), and one-to-one when \(1<y<4\). So, let's use the distribution function technique, separately, over each of these ranges. First, consider when \(0<y<1\). In that case:

\(F_Y(y)=P(Y\leq y)=P(X^2 \leq y)=P(-\sqrt{y}\leq X \leq \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})\)

The first equality holds by the definition of the cumulative distribution function. The second equality holds because the transformation of interest is \(Y=X^2\). The third equality holds, because when \(X^2\le y\), the random variable \(X\) is between the positive and negative square roots of \(y\). And, the last equality holds again by the definition of the cumulative distribution function. Now, taking the derivative of the cumulative distribution function \(F(y)\), we get (from the Fundamental Theorem of Calculus and the Chain Rule) the probability density function \(f(y)\):

\(f_Y(y)=F'_Y(y)=f_X(\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2} + f_X(-\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2}\)

Using what we know about the probability density function of \(X\):

\(f(x)=\dfrac{x^2}{3}\)

we get:

\(f_Y(y)=\dfrac{(\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}+\dfrac{(-\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}\)

And, simplifying, we get:

\(f_Y(y)=\dfrac{1}{6}y^{1/2}+\dfrac{1}{6}y^{1/2}=\dfrac{\sqrt{y}}{3}\)

for \(0<y<1\). Note that it readily becomes apparent that in the case of a two-to-one transformation, we need to sum two terms, each of which arises from a one-to-one transformation.

So, we've found the p.d.f. of \(Y\) when \(0<y<1\). Now, we have to find the p.d.f. of \(Y\) when \(1<y<4\). In that case:

\(F_Y(y)=P(Y\leq y)=P(X^2 \leq y)=P(X\leq \sqrt{y})=F_X(\sqrt{y})\)

The first equality holds by the definition of the cumulative distribution function. The second equality holds because \(Y=X^2\). The third equality holds, because when \(X^2\le y\), the random variable \(X \le \sqrt{y}\). And, the last equality holds again by the definition of the cumulative distribution function. Now, taking the derivative of the cumulative distribution function \(F(y)\), we get (from the Fundamental Theorem of Calculus and the Chain Rule) the probability density function \(f(y)\):

\(f_Y(y)=F'_Y(y)=f_X(\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2}\)

Again, using what we know about the probability density function of \(X\), and simplifying, we get:

\(f_Y(y)=\dfrac{(\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}=\dfrac{\sqrt{y}}{6}\)

for \(1<y<4\).

Solution

The transformation \(Y=X^2\) is two-to-one over the entire support \(-\infty<x<\infty\):

That is, when \(-\infty<x<0\), we have:

\(X_1=-\sqrt{Y}=v_1(Y)\)

and when \(0<x<\infty\), we have:

\(X_2=+\sqrt{Y}=v_2(Y)\)

Then, the change of variable technique tells us that, over the two-to-one portion of the transformation, that is, when \(0<y<\infty\):

\(f_Y(y)=f_X(\sqrt{y})\cdot \left |\dfrac{1}{2} y^{-1/2}\right|+f_X(-\sqrt{y})\cdot \left|-\dfrac{1}{2} y^{-1/2}\right|\)

Recalling the p.d.f. of the standard normal distribution:

\(f_X(x)=\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{x^2}{2}\right]\)

the p.d.f. of \(Y\) is then:

\(f_Y(y)=\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{(\sqrt{y})^2}{2}\right]\cdot \left|\dfrac{1}{2} y^{-1/2}\right|+\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{(\sqrt{y})^2}{2}\right]\cdot \left|-\dfrac{1}{2} y^{-1/2}\right|\)

Adding the terms together, and simplifying a bit, we get:

\(f_Y(y)=2 \dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{y}{2}\right]\cdot \dfrac{1}{2} y^{-1/2}\)

Crossing out the 2s, recalling that \(\Gamma(1/2)=\sqrt{\pi}\), and rewriting things just a bit, we should be able to recognize that, with \(0<y<\infty\), the probability density function of \(Y\):

\(f_Y(y)=\dfrac{1}{\Gamma(1/2) 2^{1/2}} e^{-y/2} y^{-1/2}\)

is indeed the p.d.f. of a chi-square random variable with 1 degree of freedom!

Solution

The cumulative distribution function of an exponential random variable with a mean of 5 is:

\(y=F(x)=1-e^{-x/5}\)

for \(0\le x<\infty\). We need to invert the cumulative distribution function, that is, solve for \(x\), in order to be able to determine the exponential(5) random numbers. Manipulating the above equation a bit, we get:

\(1-y=e^{-x/5}\)

Then, taking the natural log of both sides, we get:

\(\text{log}(1-y)=-\dfrac{x}{5}\)

And, multiplying both sides by −5, we get:

\(x=-5\text{log}(1-y)\)

for \(0<y<1\). Now, it's just a matter of inserting the student's three random U(0,1) numbers into the above equation to get our three exponential(5) random numbers:

• If \(y=0.2\), we get \(x=1.1\)
• If \(y=0.5\), we get \(x=3.5\)
• If \(y=0.9\), we get \(x=11.5\)

We would simply continue the same process — that is, generating \(y\), a random U(0,1) number, inserting y into the above equation, and solving for \(x\) — 997 more times if we wanted to generate 1000 exponential(5) random numbers. Of course, we wouldn't really do it by hand, but rather let statistical software do it for us. At least we now understand how random number generation works!

Solution

We can calculate the mean and variance of \(Y\) in three different ways.

1. By recognizing that \(Y\) is a binomial random variable with \(n=5\) and \(p=\frac{1}{2}\), we can use what know about the mean and variance of a binomial random variable, namely that the mean of \(Y\) is:

   \(E(Y)=np=5(\frac{1}{2})=\frac{5}{2}\)

   and the variance of \(Y\) is:

   \(Var(Y)=np(1-p)=5(\frac{1}{2})(\frac{1}{2})=\frac{5}{4}\)

   Since sums of independent random variables are not always going to be binomial, this approach won't always work, of course. It would be good to have alternative methods in hand!

2. We could use the linear operator property of expectation. Before doing so, it would be helpful to note that the mean of \(X_1\) is:

   \(E(X_1)=np=3(\frac{1}{2})=\frac{3}{2}\)

   and the mean of \(X_2\) is:

   \(E(X_2)=np=2(\frac{1}{2})=1\)

   Now, using the property, we get that the mean of \(Y\) is (thankfully) again \(\frac{5}{2}\):

   \(E(Y)=E(X_1+X_2)=E(X_1)+E(X_2)=\dfrac{3}{2}+1=\dfrac{5}{2}\)

   Recall that the second equality comes from the linear operator property of expectation. Now, using the linear operator property of expectation to find the variance of \(Y\) takes a bit more work. First, we should note that the variance of \(X_1\) is:

   \(Var(X_1)=np(1-p)=3(\frac{1}{2})(\frac{1}{2})=\frac{3}{4}\)

   and the variance of \(X_2\) is:

   \(Var(X_2)=np(1-p)=2(\frac{1}{2})(\frac{1}{2})=\frac{1}{2}\)

   Now, we can (thankfully) show again that the variance of \(Y\) is \(\frac{5}{4}\):

   Okay, as if two methods aren't enough, we still have one more method we could use.

3. We could use the independence of the two random variables \(X_1\) and \(X_2\), in conjunction with the definition of expected value of \(Y\) as we know it. First, using the binomial formula, note that we can present the probability mass function of \(X_1\) in tabular form as:

   \(x_1\)	0	1	2	3
   \(f(x_1)\)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)

   And, we can present the probability mass function of \(X_2\) in tabular form as well:

   \(x_2\)	0	1	2
   \(f(x_2)\)	\(\frac{1}{4}\)	\(\frac{2}{4}\)	\(\frac{1}{4}\)

   Now, recall that if \(X_1\) and \(X_2\) are independent random variables, then:

   \(f(x_1,x_2)=f(x_1)\cdot f(x_2)\)

   We can use this result to help determine \(g(y)\), the probability mass function of \(Y\). First note that, since \(Y\) is the sum of \(X_1\) and \(X_2\), the support of \(Y\) is {0, 1, 2, 3, 4 and 5}. Now, by brute force, we get:

   \(g(0)=P(Y=0)=P(X_1=0,X_2=0)=f(0,0)=f_{X_1}(0) \cdot f_{X_2}(0)=\dfrac{1}{8} \cdot \dfrac{1}{4}=\dfrac{1}{32}\)

   The second equality comes from the fact that the only way that \(Y\) can equal 0 is if \(X_1=0\) and \(X_2=0\), and the fourth equality comes from the independence of \(X_1\)and \(X_2\). We can make a similar calculation to find the probability that \(Y=1\):

   \(g(1)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=f_{X_1}(0) \cdot f_{X_2}(1)+f_{X_1}(1) \cdot f_{X_2}(0)=\dfrac{1}{8} \cdot \dfrac{2}{4}+\dfrac{3}{8} \cdot \dfrac{1}{4}=\dfrac{5}{32}\)

   The first equality comes from the fact that there are two (mutually exclusive) ways that \(Y\) can equal 1, namely if \(X_1=0\) and \(X_2=1\) or if \(X_1=1\) and \(X_2=0\). The second equality comes from the independence of \(X_1\) and \(X_2\). We can make similar calculations to find \(g(2), g(3), g(4)\), and \(g(5)\). Once we've done that, we can present the p.m.f. of \(Y\) in tabular form as:

   \(y = x_1 + x_2\)	0	1	2	3	4	5
   \(g(y)\)	\(\frac{1}{32}\)	\(\frac{5}{32}\)	\(\frac{10}{32}\)	\(\frac{10}{32}\)	\(\frac{5}{32}\)	\(\frac{1}{32}\)

   Then, it is a straightforward calculation to use the definition of the expected value of a discrete random variable to determine that (again!) the expected value of \(Y\) is \(\frac{5}{2}\): 

   \(E(Y)=0(\frac{1}{32})+1(\frac{5}{32})+2(\frac{10}{32})+\cdots+5(\frac{1}{32})=\frac{80}{32}=\frac{5}{2}\)

   The variance of \(Y\) can be calculated similarly. (Do you want to calculate it one more time?!)

   The following summarizes the method we've used here in calculating the expected value of \(Y\):

   \begin{align} E(Y)=E(X_1+X_2) &= \sum\limits_{x_1 \in S_1}\sum\limits_{x_2 \in S_2} (x_1+x_2)f(x_1,x_2)\\ &= \sum\limits_{x_1 \in S_1}\sum\limits_{x_2 \in S_2} (x_1+x_2)f(x_1) f(x_2)\\ &= \sum\limits_{y \in S} yg(y)\\ \end{align}

   The first equality comes, of course, from the definition of \(Y\). The second equality comes from the definition of the expectation of a function of discrete random variables. The third equality comes from the independence of the random variables \(X_1\) and \(X_2\). And, the fourth equality comes from the definition of the expected value of \(Y\), as well as the fact that \(g(y)\) can be determined by summing the appropriate joint probabilities of \(X_1\) and \(X_2\).

Solution

We'll use the fact that the expectation of the product is the product of the expectations:

Solution

The mean of the sum is:

\(E(X_1+X_2)=E(X_1)+E(X_2)=2+3=5\)

and the variance of the sum is:

\(Var(X_1+X_2)=(1)^2Var(X_1)+(1)^2Var(X_2)=4+5=9\)

Solution

The mean of the difference is:

\(E(X_1-X_2)=E(X_1)-E(X_2)=2-3=-1\)

and the variance of the difference is:

\(Var(X_1-X_2)=Var(X_1+(-1)X_2)=(1)^2Var(X_1)+(-1)^2Var(X_2)=4+5=9\)

That is, the variance of the difference in the two random variables is the same as the variance of the sum of the two random variables.

Solution

The mean of the linear combination is:

\(E(3X_1+4X_2)=3E(X_1)+4E(X_2)=3(2)+4(3)=18\)

and the variance of the linear combination is:

\(Var(3X_1+4X_2)=(3)^2Var(X_1)+(4)^2Var(X_2)=9(4)+16(5)=116\)

Solution

The fact that \(X_1,X_2,\ldots, X_n\) constitute a random sample tells us that (1) \(X_i\) is independent of \(X_j\), for all \(i\ne j\), and (2) the \(X_i\) are identically distributed. Now, we know from our previous work that if \(X_i\) is independent of \(X_j\), for \(i\ne j\), then the covariance between \(X_i\) is independent of \(X_j\) is 0. That is:

\(E[(X_i-\mu)(X_j-\mu)]=Cov(X_i,X_j)=0\)

Solution

The only way that the maximum of the \(X_i\) will be less than or equal to 2 is if all of the \(X_i\) are less than or equal to 2. That is:

\(P(\max X_i\leq 2)=P(X_1\leq 2,X_2\leq 2,X_3\leq 2)\)

Now, because \(X_1,X_2,X_3\) are a random sample, we know that (1) \(X_i\) is independent of \(X_j\), for all \(i\ne j\), and (2) the \(X_i\) are identically distributed. Therefore:

\(P(\max X_i\leq 2)=P(X_1\leq 2)P(X_2\leq 2)P(X_3\leq 2)=[P(X_1\leq 2)]^3\)

The first equality comes from the independence of the \(X_i\), and the second equality comes from the fact that the \(X_i\) are identically distributed. Now, the probability that \(X_1\) is less than or equal to 2 is:

\(P(X\leq 2)=P(X=1)+P(X=2)=\left(\dfrac{3}{4}\right) \left(\dfrac{1}{4}\right)^{1-1}+\left(\dfrac{3}{4}\right) \left(\dfrac{1}{4}\right)^{2-1}=\dfrac{3}{4}+\dfrac{3}{16}=\dfrac{15}{16}\)

Therefore, the probability that the maximum of the \(X_i\) is less than or equal to 2 is:

\(P(\max X_i\leq 2)=[P(X_1\leq 2)]^3=\left(\dfrac{15}{16}\right)^3=0.824\)

Solution

We know that:

• \(X_1\) is a binomial random variable with \(n=3\) and \(p=\frac{1}{2}\)
• \(X_2\) is a binomial random variable with \(n=2\) and \(p=\frac{1}{2}\)

Therefore, based on what we know of the moment-generating function of a binomial random variable, the moment-generating function of \(X_1\) is:

\(M_{X_1}(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^3\)

And, similarly, the moment-generating function of \(X_2\) is:

\(M_{X_2}(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^2\)

Now, because \(X_1\) and \(X_2\) are independent random variables, the random variable \(Y\) is the sum of independent random variables. Therefore, the moment-generating function of \(Y\) is:

\(M_Y(t)=E(e^{tY})=E(e^{t(X_1+X_2)})=E(e^{tX_1} \cdot e^{tX_2} )=E(e^{tX_1}) \cdot E(e^{tX_2} )\)

The first equality comes from the definition of the moment-generating function of the random variable \(Y\). The second equality comes from the definition of \(Y\). The third equality comes from the properties of exponents. And, the fourth equality comes from the expectation of the product of functions of independent random variables. Now, substituting in the known moment-generating functions of \(X_1\) and \(X_2\), we get:

\(M_Y(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^3 \cdot \left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^2=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^5\)

That is, \(Y\) has the same moment-generating function as a binomial random variable with \(n=5\) and \(p=\frac{1}{2}\). Therefore, by the uniqueness properties of moment-generating functions, \(Y\) must be a binomial random variable with \(n=5\) and \(p=\frac{1}{2}\). (Of course, we already knew that!)

Solution

The moment-generating function of a gamma random variable \(X\) with \(\alpha=7\) and \(\theta=5\) is:

\(M_X(t)=\dfrac{1}{(1-5t)^7}\)

for \(t<\frac{1}{5}\). Therefore, the corollary tells us that the moment-generating function of \(Y\) is:

\(M_Y(t)=[M_{X_1}(t)]^3=\left(\dfrac{1}{(1-5t)^7}\right)^3=\dfrac{1}{(1-5t)^{21}}\)

for \(t<\frac{1}{5}\), which is the moment-generating function of a gamma random variable with \(\alpha=21\) and \(\theta=5\). Therefore, \(Y\) must follow a gamma distribution with \(\alpha=21\) and \(\theta=5\).

Solution

Again, the moment-generating function of a gamma random variable \(X\) with \(\alpha=7\) and \(\theta=5\) is:

\(M_X(t)=\dfrac{1}{(1-5t)^7}\)

for \(t<\frac{1}{5}\). Therefore, the corollary tells us that the moment-generating function of \(\bar{X}\) is:

\(M_{\bar{X}}(t)=\left[M_{X_1}\left(\dfrac{t}{3}\right)\right]^3=\left(\dfrac{1}{(1-5(t/3))^7}\right)^3=\dfrac{1}{(1-(5/3)t)^{21}}\)

for \(t<\frac{3}{5}\), which is the moment-generating function of a gamma random variable with \(\alpha=21\) and \(\theta=\frac{5}{3}\). Therefore, \(\bar{X}\) must follow a gamma distribution with \(\alpha=21\) and \(\theta=\frac{5}{3}\).

Solution

The previous theorem tells us that \(Y\) is normally distributed with mean 7 and variance 48 as the following calculation illustrates:

\((2X_1+3X_2)\sim N(2(2)+3(1),2^2(3)+3^2(4))=N(7,48)\)

Solution

The previous theorem tells us that \(Y\) is normally distributed with mean 1 and variance 7 as the following calculation illustrates:

\((X_1-X_2)\sim N(2-1,(1)^2(3)+(-1)^2(4))=N(1,7)\)

Solution

We can find the requested probability by noting that \(P(X>Y)=P(X-Y>0)\), and then taking advantage of what we know about the distribution of \(X-Y\). That is, \(X-Y\) is normally distributed with a mean of 55 and variance of 12100 as the following calculation illustrates:

\((X-Y)\sim N(529-474,(1)^2(5732)+(-1)^2(6368))=N(55,12100)\)

Then, finding the probability that \(X\) is greater than \(Y\) reduces to a normal probability calculation:

\begin{align} P(X>Y) &=P(X-Y>0)\\ &= P\left(Z>\dfrac{0-55}{\sqrt{12100}}\right)\\ &= P\left(Z>-\dfrac{1}{2}\right)=P\left(Z<\dfrac{1}{2}\right)=0.6915\\ \end{align}

That is, the probability that the first student's Math score is greater than the second student's Verbal score is 0.6915.

Solution

Because the bags are selected at random, we can assume that \(X_1, X_2, X_3\) and \(W\) are mutually independent. The theorem helps us determine the distribution of \(Y\), the sum of three one-pound bags:

\(Y=(X_1+X_2+X_3) \sim N(1.18+1.18+1.18, 0.07^2+0.07^2+0.07^2)=N(3.54,0.0147)\)

That is, \(Y\) is normally distributed with a mean of 3.54 pounds and a variance of 0.0147. Now, \(Y-W\), the difference in the weight of three one-pound bags and one three-pound bag is normally distributed with a mean of 0.32 and a variance of 0.0228, as the following calculation suggests:

\((Y-W) \sim N(3.54-3.22,(1)^2(0.0147)+(-1)^2(0.09^2))=N(0.32,0.0228)\)

Therefore, finding the probability that \(Y\) is greater than \(W\) reduces to a normal probability calculation:

\begin{align} P(Y>W) &=P(Y-W>0)\\ &= P\left(Z>\dfrac{0-0.32}{\sqrt{0.0228}}\right)\\ &= P(Z>-2.12)=P(Z<2.12)=0.9830\\ \end{align}

That is, the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag is 0.9830. Hey, if you want more bang for your buck, it looks like you should buy multiple one-pound bags of carrots, as opposed to one three-pound bag!

Anwser

In general, the variance of the sample mean is:

\(Var(\bar{X})=\dfrac{\sigma^2}{n}\)

Therefore, the variance of the sample mean of the first sample is:

\(Var(\bar{X}_4)=\dfrac{16^2}{4}=64\)

(The subscript 4 is there just to remind us that the sample mean is based on a sample of size 4.) And, the variance of the sample mean of the second sample is:

\(Var(\bar{Y}_8=\dfrac{16^2}{8}=32\)

(The subscript 8 is there just to remind us that the sample mean is based on a sample of size 8.) Now, the corollary therefore tells us that the sample mean of the first sample is normally distributed with mean 100 and variance 64. That is:

\(\bar{X}_4 \sim N(100,64)\)

And, the sample mean of the second sample is normally distributed with mean 100 and variance 32. That is:

\(\bar{Y}_8 \sim N(100,32)\)

So, we have two, no actually, three normal random variables with the same mean, but difference variances:

• We have \(X_i\), an IQ of a random individual. It is normally distributed with mean 100 and variance 256.
• We have \(\bar{X}_4\), the average IQ of 4 random individuals. It is normally distributed with mean 100 and variance 64.
• We have \(\bar{Y}_8\), the average IQ of 8 random individuals. It is normally distributed with mean 100 and variance 32.

It is quite informative to graph these three distributions on the same plot. Doing so, we get:

As the plot suggests, an individual \(X_i\), the mean (\bar{X}_4\) and the mean \(\bar{Y}_8\) all provide valid, "unbiased" estimates of the population mean \(\mu\). But, our intuition coincides with reality... that is, the sample mean \(\bar{Y}_8\) will be the most precise estimate of \(\mu\).

All the work that we have done so far concerning this example has been theoretical in nature. That is, what we have learned is based on probability theory. Would we see the same kind of result if we were take to a large number of samples, say 1000, of size 4 and 8, and calculate the sample mean of each sample? That is, would the distribution of the 1000 sample means based on a sample of size 4 look like a normal distribution with mean 100 and variance 64? And would the distribution of the 1000 sample means based on a sample of size 8 look like a normal distribution with mean 100 and variance 32? Well, the only way to answer these questions is to try it out!

I did just that for us. I used Minitab to generate 1000 samples of eight random numbers from a normal distribution with mean 100 and variance 256. Here's a subset of the resulting random numbers:

As you can see, the second last column, titled Mean4, is the average of the first four columns X1 X2, X3, and X4. The last column, titled Mean8, is the average of the first eight columns X1, X2, X3, X4, X5, X6, X7, and X8. Now, all we have to do is create a histogram of the sample means appearing in the Mean4 column:

Ahhhh! The histogram sure looks fairly bell-shaped, making the normal distribution a real possibility. Now, recall that the Empirical Rule tells us that we should expect, if the sample means are normally distributed, that almost all of the sample means would fall within three standard deviations of the population mean. That is, in the case of Mean4, we should expect almost all of the data to fall between 76 (from 100−3(8)) and 124 (from 100+3(8)). It sure looks like that's the case!

Let's do the same thing for the Mean8 column. That is, let's create a histogram of the sample means appearing in the Mean8 column. Doing so, we get:

Again, the histogram sure looks fairly bell-shaped, making the normal distribution a real possibility. In this case, the Empirical Rule tells us that, in the case of Mean8, we should expect almost all of the data to fall between 83 (from 100−3(square root of 32)) and 117 (from 100+3(square root of 32)). It too looks pretty good on both sides, although it seems that there were two really extreme sample means of size 8. (If you look back at the data, you can see one of them in the eighth row.)

In summary, the whole point of this exercise was to use the theory to help us derive the distribution of the sample mean of IQs, and then to use real simulated normal data to see if our theory worked in practice. I think we can conclude that it does!

Solution

Because the sample size is \(n=8\), the above theorem tells us that:

\(\dfrac{(8-1)S^2}{\sigma^2}=\dfrac{7S^2}{\sigma^2}=\dfrac{\sum\limits_{i=1}^8 (X_i-\bar{X})^2}{\sigma^2}\)

follows a chi-square distribution with 7 degrees of freedom. Here's what the theoretical density function would look like:

Again, all the work that we have done so far concerning this example has been theoretical in nature. That is, what we have learned is based on probability theory. Would we see the same kind of result if we were take to a large number of samples, say 1000, of size 8, and calculate:

\(\dfrac{\sum\limits_{i=1}^8 (X_i-\bar{X})^2}{256}\)

for each sample? That is, would the distribution of the 1000 resulting values of the above function look like a chi-square(7) distribution? Again, the only way to answer this question is to try it out! I did just that for us. I used Minitab to generate 1000 samples of eight random numbers from a normal distribution with mean 100 and variance 256. Here's a subset of the resulting random numbers:

click to enlarge

As you can see, the last column, titled FnofSsq (for function of sums of squares), contains the calculated value of:

\(\dfrac{\sum\limits_{i=1}^8 (X_i-\bar{X})^2}{256}\)

based on the random numbers generated in columns X1 X2, X3, X4, X5, X6, X7, and X8. For example, given that the average of the eight numbers in the first row is 98.625, the value of FnofSsq in the first row is:

\(\dfrac{1}{256}[(98-98.625)^2+(77-98.625)^2+\cdots+(91-98.625)^2]=5.7651\)

Now, all we have to do is create a histogram of the values appearing in the FnofSsq column. Doing so, we get:

Hmm! The histogram sure looks eerily similar to that of the density curve of a chi-square random variable with 7 degrees of freedom. It looks like the practice is meshing with the theory!

Solution

The probability calculation is quite similar to a calculation we'd have to make for a normal random variable. First, rewriting the probability in terms of \(T\) instead of the absolute value of \(T\), we get:

\(P(|T|<2.306)=P(-2.306<T<2.306)\)

Then, we have to rewrite the probability in terms of cumulative probabilities that we can actually find, that is:

\(P(|T|<2.306)=P(T<2.306)-P(T<-2.306)\)

Pictorially, the probability we are looking for looks something like this:

But the \(t\)-table doesn't contain negative \(t\)-values, so we'll have to take advantage of the symmetry of the \(T\) distribution. That is:

 >\(P(|T|<2.306)=P(T<2.306)-P(T>2.306)\) 

The \(t\)-table tells us that \(P(T<2.306)=0.975\) and \(P(T>2.306)=0.025\).  Therefore:

\(P(|T|>2.306)=0.975-0.025=0.95\)

What is \(t_{0.05}(8)\)?

Solution

Our previous work on the continuous Uniform(0, 1) random variable tells us that the mean of a \(U(0,1)\) random variable is:

\(\mu=E(X_i)=\dfrac{0+1}{2}=\dfrac{1}{2}\)

while the variance of a \(U(0,1)\) random variable is:

\(\sigma^2=Var(X_i)=\dfrac{(1-0)^2}{12}=\dfrac{1}{12}\)

The Central Limit Theorem, therefore, tells us that the sample mean \(\bar{X}\) is approximately normally distributed with mean:

\(\mu_{\bar{X}}=\mu=\dfrac{1}{2}\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{1/12}{n}=\dfrac{1}{12n}\)

Now, our end goal is to compare the normal distribution, as defined by the CLT, to the actual distribution of the sample mean. Now, we could do a lot of theoretical work to find the exact distribution of \(\bar{X}\) for various sample sizes \(n\). Instead, we'll use simulation to give us a ballpark idea of the shape of the distribution of \(\bar{X}\). Here's an outline of the general strategy that we'll follow:

1. Specify the sample size \(n\).
2. Randomly generate 1000 samples of size \(n\) from the Uniform (0,1) distribution.
3. Use the 1000 generated samples to calculate 1000 sample means from the Uniform (0,1) distribution.
4. Create a histogram of the 1000 sample means.
5. Compare the histogram to the normal distribution, as defined by the Central Limit Theorem, in order to see how well the Central Limit Theorem works for the given sample size \(n\).

Let's start with a sample size of \(n=1\). That is, randomly sample 1000 numbers from a Uniform (0,1) distribution, and create a histogram of the 1000 generated numbers. Of course, the histogram should look roughly flat like a Uniform(0,1) distribution. If you're willing to ignore the artifacts of sampling, you can see that our histogram is roughly flat:

Okay, now let's tackle the more interesting sample sizes. Let \(n=2\). Generating 1000 samples of size \(n=2\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

It can actually be shown that the exact distribution of the sample mean of 2 numbers drawn from the Uniform(0, 1) distribution is the triangular distribution. The histogram does look a bit triangular, doesn't it? The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=\dfrac{1}{2}\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{1}{12n}=\dfrac{1}{12(2)}=\dfrac{1}{24}\)

As you can see, already at \(n=2\), the normal curve wouldn't do too bad of a job of approximating the exact probabilities. Let's increase the sample size to \(n=4\). Generating 1000 samples of size \(n=4\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=\dfrac{1}{2}\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{1}{12n}=\dfrac{1}{12(4)}=\dfrac{1}{48}\)

Again, at \(n=4\), the normal curve does a very good job of approximating the exact probabilities. In fact, it does such a good job, that we could probably stop this exercise already. But let's increase the sample size to \(n=9\). Generating 1000 samples of size \(n=9\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=\dfrac{1}{2}\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{1}{12n}=\dfrac{1}{12(9)}=\dfrac{1}{108}\)

And not surprisingly, at \(n=9\), the normal curve does a very good job of approximating the exact probabilities. There is another interesting thing worth noting though, too. As you can see, as the sample size increases, the variance of the sample mean decreases. That's a good thing, as it doesn't seem that it should be any other way. If you think about it, if it were possible to increase the sample size \(n\) to something close to the size of the population, you would expect that the resulting sample means would not vary much, and would be close to the population mean. Of course, the trade-off here is that large sample sizes typically cost lots more money than small sample sizes.

Well, just for the heck of it, let's increase our sample size one more time to \(n=16\). Generating 1000 samples of size \(n=16\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=\dfrac{1}{2}\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{1}{12n}=\dfrac{1}{12(16)}=\dfrac{1}{192}\)

Again, at \(n=16\), the normal curve does a very good job of approximating the exact probabilities. Okay, uncle! That's enough of this example! Let's summarize the two take-away messages from this example:

1. If the underlying distribution is symmetric, then you don't need a very large sample size for the normal distribution, as defined by the Central Limit Theorem, to do a decent job of approximating the probability distribution of the sample mean.
2. The larger the sample size \(n\), the smaller the variance of the sample mean.

Solution

We are going to do exactly what we did in the previous example. The only difference is that our underlying distribution here, that is, the chi-square(3) distribution, is highly-skewed. Now, our previous work on the chi-square distribution tells us that the mean of a chi-square random variable with three degrees of freedom is:

\(\mu=E(X_i)=r=3\)

while the variance of a chi-square random variable with three degrees of freedom is:

\(\sigma^2=Var(X_i)=2r=2(3)=6\)

The Central Limit Theorem, therefore, tells us that the sample mean \(\bar{X}\) is approximately normally distributed with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{n}\)

Again, we'll follow a strategy similar to that in the above example, namely:

1. Specify the sample size \(n\).
2. Randomly generate 1000 samples of size \(n\) from the chi-square(3) distribution.
3. Use the 1000 generated samples to calculate 1000 sample means from the chi-square(3) distribution.
4. Create a histogram of the 1000 sample means.
5. Compare the histogram to the normal distribution, as defined by the Central Limit Theorem, in order to see how well the Central Limit Theorem works for the given sample size \(n\).

Again, starting with a sample size of \(n=1\), we randomly sample 1000 numbers from a chi-square(3) distribution, and create a histogram of the 1000 generated numbers. Of course, the histogram should look like a (skewed) chi-square(3) distribution, as the blue curve suggests it does:

Now, let's consider samples of size \(n=2\). Generating 1000 samples of size \(n=2\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{2}=3\)

As you can see, at \(n=2\), the normal curve wouldn't do a very job of approximating the exact probabilities. The probability distribution of the sample mean still appears to be quite skewed. Let's increase the sample size to \(n=4\). Generating 1000 samples of size \(n=4\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{4}=1.5\)

Although, at \(n=4\), the normal curve is doing a better job of approximating the probability distribution of the sample mean, there is still much room for improvement. Let's try \(n=9\). Generating 1000 samples of size \(n=9\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{9}=0.667\)

We're getting closer, but let's really jump up the sample size to, say, \(n=25\). Generating 1000 samples of size \(n=25\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{25}=0.24\)

1. Okay, now we're talking! There's still just a teeny tiny bit of skewness in the sampling distribution. Let's increase the sample size just one more time to, say, \(n=36\). Generating 1000 samples of size \(n=36\), calculating the 1000 sample means, and creating a histogram of the 1000 sample means, we get:

The blue curve overlaid on the histogram is the normal distribution, as defined by the Central Limit Theorem. That is, the blue curve is the normal distribution with mean:

\(\mu_{\bar{X}}=\mu=3\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{6}{36}=0.167\)

Okay, now, I'm perfectly happy! It appears that, at \(n=36\), the normal curve does a very good job of approximating the exact probabilities. Let's summarize the two take-away messages from this example:

1. Again, the larger the sample size \(n\), the smaller the variance of the sample mean. Nothing new there.
2. If the underlying distribution is skewed, then you need a larger sample size, typically \(n>30\), for the normal distribution, as defined by the Central Limit Theorem, to do a decent job of approximating the probability distribution of the sample mean.

Solution

The expected value of the random variable \(X\) is 0, as the following calculation illustrates:

\(\mu=E(X)=\int^1_{-1} x \cdot \dfrac{3}{2} x^2dx=\dfrac{3}{2} \int^1_{-1}x^3dx=\dfrac{3}{2} \left[\dfrac{x^4}{4}\right]^{x=1}_{x=-1}=\dfrac{3}{2} \left(\dfrac{1}{4}-\dfrac{1}{4} \right)=0\)

The variance of the random variable \(X\) is \(\frac{3}{5}\), as the following calculation illustrates:

\(\sigma^2=E(X-\mu)^2=\int^1_{-1} (x-0)^2 \dfrac{3}{2} x^2dx=\dfrac{3}{2} \int^1_{-1}x^4dx=\dfrac{3}{2} \left[\dfrac{x^5}{5}\right]^{x=1}_{x=-1}=\dfrac{3}{2} \left(\dfrac{1}{5}+\dfrac{1}{5} \right)=\dfrac{3}{5}\)

Therefore, the CLT tells us that the sample mean \(\bar{X}\) is approximately normal with mean:

\(E(\bar{X})=\mu_{\bar{X}}=\mu=0\)

and variance:

\(Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{3/5}{15}=\dfrac{3}{75}=\dfrac{1}{25}\)

Therefore the standard deviation of \(\bar{X}\) is \(\frac{1}{5}\). Drawing a picture of the desired probability:

we see that:

\(P(-2/5<\bar{X}<1/5)=P(-2<Z<1)\)

Therefore, using the standard normal table, we get:

\(P(-2/5<\bar{X}<1/5)=P(Z<1)-P(Z<-2)=0.8413-0.0228=0.8185\)

That is, there is an 81.85% chance that a random sample of size 15 from the given distribution will yield a sample mean between \(-\frac{2}{5}\) and \(\frac{1}{5}\).

Solution

It is reasonable to assume that \(X_i\) is an exponential random variable. And, based on the assistant manager's claim, the mean of \(X_i\) is:

\(\mu=\theta=2\).

Therefore, knowing what we know about exponential random variables, the variance of \(X_i\) is:

\(\sigma^2=\theta^2=2^2=4\).

Now, we need to know, if the mean \(\mu\) really is 2, as the assistant manager claims, what is the probability that the manager would obtain a sample mean as large as (or larger than) 3.2 minutes? Well, the Central Limit Theorem tells us that the sample mean \(\bar{X}\) is approximately normally distributed with mean:

\(\mu_{\bar{X}}=2\)

and variance:

\(\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{4}{36}=\dfrac{1}{9}\)

Here's a picture, then, of the normal probability that we need to determine:

\(z = \dfrac{3.2 - 2}{\sqrt{\frac{1}{9}}} = 3.6\)

That is:

\(P(\bar{X}>3.2)=P(Z>3.6)\)

The \(Z\) value in this case is so extreme that the table in the back of our text book can't help us find the desired probability. But, using statistical software, such as Minitab, we can determine that:

\(P(\bar{X}>3.2)=P(Z>3.6)=0.00016\)

That is, if the population mean \(\mu\) really is 2, then there is only a 16/100,000 chance (0.016%) of getting such a large sample mean. It would be quite reasonable, therefore, for the manager to reject his assistant's claim that the mean \(\mu\) is 2. The manager should feel comfortable concluding that the population mean \(\mu\) really is greater than 2. We will leave it up to him to decide whether or not he should fire his assistant!

By the way, this is the kind of example that we'll see when we study hypothesis testing in Stat 415. In general, in the process of performing a hypothesis test, someone makes a claim (the assistant, in this case), and someone collects and uses the data (the manager, in this case) to make a decision about the validity of the claim. It just so happens to be that we used the CLT in this example to help us make a decision about the assistant's claim.

Solution

There is really nothing new here. We can calculate the exact probability using the binomial table in the back of the book with \(n=10\) and \(p=\frac{1}{2}\). Doing so, we get:

\begin{align} P(Y=5)&= P(Y \leq 5)-P(Y \leq 4)\\ &= 0.6230-0.3770\\ &= 0.2460\\ \end{align}

That is, there is a 24.6% chance that exactly five of the ten people selected approve of the job the President is doing.

Solution

First, recognize in our case that the mean is:

\(\mu=np=10\left(\dfrac{1}{2}\right)=5\)

and the variance is:

\(\sigma^2=np(1-p)=10\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=2.5\)

Now, if we look at a graph of the binomial distribution with the rectangle corresponding to \(Y=5\) shaded in red:

we should see that we would benefit from making some kind of correction for the fact that we are using a continuous distribution to approximate a discrete distribution. Specifically, it seems that the rectangle \(Y=5\) really includes any \(Y\) greater than 4.5 but less than 5.5. That is:

\(P(Y=5)=P(4.5< Y < 5.5)\)

Such an adjustment is called a "continuity correction." Once we've made the continuity correction, the calculation reduces to a normal probability calculation:

Now, recall that we previous used the binomial distribution to determine that the probability that \(Y=5\) is exactly 0.246. Here, we used the normal distribution to determine that the probability that \(Y=5\) is approximately 0.251. That's not too shabby of an approximation, in light of the fact that we are dealing with a relative small sample size of \(n=10\)!

Solution

If we look at a graph of the binomial distribution with the area corresponding to \(7<Y\le 9\) shaded in red:

we should see that we'll want to make the following continuity correction:

\(P(7<Y \leq 9)=P(7.5< Y < 9.5)\)

Now again, once we've made the continuity correction, the calculation reduces to a normal probability calculation:

By the way, you might find it interesting to note that the approximate normal probability is quite close to the exact binomial probability. We showed that the approximate probability is 0.0549, whereas the following calculation shows that the exact probability (using the binomial table with \(n=10\) and \(p=\frac{1}{2}\) is 0.0537:

\(P(7<Y \leq 9)=P(Y\leq 9)-P(Y\leq 7)=0.9990-0.9453=0.0537\)

Solution

If we look at a graph of the binomial distribution with the area corresponding to \(2\le Y<4\) shaded in red:

we should see that we'll want to make the following continuity correction:

\(P(2 \leq Y <4)=P(1.5< Y < 3.5)\)

Again, once we've made the continuity correction, the calculation reduces to a normal probability calculation:

\begin{align} P(2 \leq Y <4)=P(1.5< Y < 3.5) &= P(\dfrac{1.5-5}{\sqrt{2.5}}<Z<\dfrac{3.5-5}{\sqrt{2.5}})\\ &= P(-2.21<Z<-0.95)\\ &= P(Z>0.95)-P(Z>2.21)\\ &= 0.1711-0.0136=0.1575\\ \end{align}

By the way, the exact binomial probability is 0.1612, as the following calculation illustrates:

\(P(2 \leq Y <4)=P(Y\leq 3)-P(Y\leq 1)=0.1719-0.0107=0.1612\)

Solution.

We can, of course use the Poisson distribution to calculate the exact probability. Using the Poisson table with \(\lambda=6.5\), we get:

\(P(Y\geq 9)=1-P(Y\leq 8)=1-0.792=0.208\)

Now, let's use the normal approximation to the Poisson to calculate an approximate probability. First, we have to make a continuity correction. Doing so, we get:

\(P(Y\geq 9)=P(Y>8.5)\)

Once we've made the continuity correction, the calculation again reduces to a normal probability calculation:

\begin{align} P(Y\geq 9)=P(Y>8.5)&= P(Z>\dfrac{8.5-6.5}{\sqrt{6.5}})\\ &= P(Z>0.78)=0.218\\ \end{align}

So, in summary, we used the Poisson distribution to determine the probability that \(Y\) is at least 9 is exactly 0.208, and we used the normal distribution to determine the probability that \(Y\) is at least 9 is approximately 0.218. Not too shabby of an approximation!