## Practice Problem Set 5 – bivariate normal distribution

This post provides practice problems to reinforce the concept of bivariate normal distribution discussed in two posts – one is a detailed introduction to bivariate normal distribution and the other is a further discussion that brings out more mathematical properties of the bivariate normal distribution. The properties discussed in these two posts form the basis for the calculation behind the practice problems presented here.

The practice problems presented here are mostly on calculating probabilities. The normal probabilities can be obtained using a normal table or a calculator that has a function for normal distribution (such as TI84+). The answers for normal probabilities given at the end of the post have two versions – one using a normal table (found here) and the other one using TI84+.

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 Practice Problem 5-A Suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=15$, $\sigma_X=4$, $\mu_Y=20$, $\sigma_Y=5$ and $\rho=-0.7$. Determine the following. Compute the probability $P[12 For $X=20$, determine the mean and standard deviation of the conditional distribution of $Y$ given $X=20$. Determine $P[12, the probability that $12 given $X=20$.

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 Practice Problem 5-B Suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=6$, $\sigma_X=1.6$, $\mu_Y=4$, $\sigma_Y=1.2$ and $\rho=0.8$. Determine the following. Compute the probability $P[3 Determine $E[Y \lvert X=x]$, the mean of the conditional distribution of $Y$ given $X=x$. Determine $\sigma_{Y \lvert x}^2=Var[Y \lvert X=x]$ and $\sigma_{Y \lvert x}$, the variance and the standard deviation of the conditional distribution of $Y$ given $X=x$. For each of the $x$ values 6, 8, 10 and 12, determine the 99.7% interval $(a,b)$ for the conditional distribution of $Y$ given $x$, i.e. $a$ is three standard deviations below the mean and $b$ is 3 standard deviations above the mean. For each of the $x$ values 6, 8, 10 and 12, determine $P[3. Explain the magnitude of each of these probabilities based on the intervals in 6.

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 Practice Problem 5-C Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=50$, $\sigma_X=10$, $\mu_Y=60$, $\sigma_Y=5$ and $\rho=0.6$. Determine the following. Calculate $P[100 Determine the 5 parameters of the bivariate normal random variables $L=X+Y$ and $M=X-Y$. Calculate $P[100

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 Practice Problem 5-D Suppose $X$ is the height (in inches) and $Y$ is the weight (in pounds) of a male student in a large university. Furthermore suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=69$, $\mu_Y=155$, $\sigma_X=2.5$, $\sigma_Y=20$ and $\rho=0.55$. What is the distribution of the weights of all male students what are 5 feet 11 inches tall (71 inches)? For a randomly chosen male student who is 71 inches tall, what is the probability that his weight is between 170 and 200 pounds? For male students who are 71 inches tall, what is the 90th percentile of weight?

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 Practice Problem 5-E Suppose that $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=70$, $\sigma_X=5$, $\sigma_Y=10$ and $\rho>0$. Further suppose that $P[58.24. Determine $\rho$.

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 Practice Problem 5-F Suppose that $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=60$, $\sigma_X=10$, $\sigma_Y=12$ and $\rho=0.8$. Compute $P[45 When $X=60$, 4 values of $Y$ are observed. Compute $P[45<\overline{Y}<55 \lvert X=60]$ where $\overline{Y}$ is the mean of the sample of size 4.

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 Practice Problem 5-G Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=50$, $\sigma_X=10$, $\sigma_Y=12$ and $\rho=-0.65$. Determine the following. $P[X-Y<50]$ $\displaystyle P[55<\frac{X+Y}{2}<65]$

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 Practice Problem 5-H Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\sigma_X=5$, $\mu_Y=50$, $\sigma_Y=10$ and $\rho=0.75$. Determine the following probabilities. $P \biggl[ \frac{X+Y}{2}<68 \biggr]$ $P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert Y=60 \biggr]$

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 Practice Problem 5-I For a couple from a large population of married couples, let $X$ be the height (in inches) of the husband and let $Y$ be the height (in inches) of the wife. Suppose that $X$ and $Y$ have a bivriate normal distribution with parameters $\mu_X=68$, $\mu_Y=65$, $\sigma_X=2.2$, $\sigma_Y=2.5$ and $\rho=0.5$. For a randomly selected wife from this population, determine the probability that her height is between 68 inches and 72 inches. For a randomly selected wife from this population whose husband is 72 inches tall, determine the probability that her height is between 68 inches and 72 inches. For a randomly selected couple from this population, determine the probability that the wife is taller than the husband.

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 Practice Problem 5-J The annual revenues of Company X and Company Y are positively correlated since the correlation coefficient between the two revenues is 0.65. The annual revenue of Company X is, on average, 4,500 with standard deviation 1,500. The annual revenue of Company Y is, on average, 5,500 with standard deviation 2,000. Calculate the probability that annual revenue of Company X is less than 6,800 given that the annual revenue of Company Y is 6,800. Calculate the probability that the annual revenue of Company X is greater than that of Company Y given that their total revenue is 12,000.

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5-A
1. $P[12 (table), 0.8904014212 (TI84+)
2. $E[Y \lvert X=20]=15.625$, $Var[Y \lvert X=20]=12.75$
3. $P[12 (table), 0.8447309876 (TI84+)
5-B
• $P[3 (table),0.5953433508 (TI84+)
• $\displaystyle E[Y \lvert x]=0.4+0.6 \ x$
• $\displaystyle Var[Y \lvert x]=0.5184$, standard deviation = 0.72
• For x = 6, (1.84, 6.16)
• For x = 8, (3.04, 7.36)
• For x = 10, (4.24, 8.56)
• For x = 12, (5.44, 9.76)
• $P[3 (table), 0.8351333522 (TI84+)
• $P[3 (table), 0.3894682472 (TI84+)
• $P[3 (table), 0.025919702 (TI84+)
• $P[3 (table), 0.0001524802646 (TI84+)
5-C
• $P[100 (table), 0.7551912515 (TI84+)
• $\displaystyle \mu_L=110 \ \ \ \sigma_L=\sqrt{185} \ \ \ \mu_M=-10 \ \ \ \sigma_M=\sqrt{65} \ \ \ \rho_{L,M}=\frac{75}{\sqrt{185} \sqrt{65}}$
• $P[100 (table), 0.8966089617 (TI84+)
5-D
• Normal with mean 163.8 and standard deviation $\sqrt{279}$.
• $P[170 (table), 0.3401418637 (TI84+)
• 90th percentile = 185.18 (table), 185.2061314 (TI84+)
5-E
• 0.8
5-F
• $P[45 (table), 0.5119251771 (TI84+)
• $P[45<\overline{Y}<55 \lvert X=60]=0.8329$ (table), 0.8325288097 (TI84+)
5-G
• $P[X-Y<50]=0.9332$ (table), 0.9331927713 (TI84+)
• $\displaystyle P[55<\frac{X+Y}{2}<65]=0.7154$ (table), 0.7135779177 (TI84+)
5-H
• $P \biggl[ \frac{X+Y}{2}<68 \biggr]=0.8708$ (table), 0.8710504336 (TI84+)
• $P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert Y=60 \biggr]=0.7517$ (table), 0.7518542213 (TI84+)
5-I
• 0.1125 (table), 0.1125145409 (TI84+)
• 0.3523 (table), 0.3539664536 (TI84+)
• 0.1020 (table), 0.1022447094 (TI84+)
5-J
• 0.9279 (table), 0.9280950079 (TI84+)
• 0.1736 (table), 0.1736950626 (TI84+)

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## Calculating bivariate normal probabilities

This post extends the discussion of the bivariate normal distribution started in this post from a companion blog. Practice problems are given in the next post.

Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X$, $\sigma_X$, $\mu_Y$, $\sigma_Y$ and $\rho$. What to make of these five parameters? According to the previous post, we know that

• $\mu_X$ and $\sigma_X$ are the mean and standard deviation of the marginal distribution of $X$,
• $\mu_Y$ and $\sigma_Y$ are the mean and standard deviation of the marginal distribution of $Y$,
• and finally $\rho$ is the correlation coefficient of $X$ and $Y$.

So the five parameters of a bivariate normal distribution are the means and standard deviations of the two marginal distributions and the fifth parameter is the correlation coefficient that serves to connect $X$ and $Y$. If $\rho=0$, then $X$ and $Y$ are simply two independent normal distributions.

When calculating probabilities involving a bivariate normal distribution, keep in mind that both marginal distributions are normal. Furthermore, the conditional distribution of one variable given a value of the other is also normal. Much more can be said about the conditional distributions.

The conditional distribution of $Y$ given $X=x$ is usually denoted by $Y \lvert X=x$ or $Y \lvert x$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant (it does not matter what $x$ is, the variance is always the same). The linear conditional mean and constant variance are given by the following:

$\displaystyle E[Y \lvert X=x]=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X)$

$\displaystyle Var[Y \lvert X=x]=\sigma_Y^2 \ (1-\rho^2)$

Similarly, the conditional distribution of $X$ given $Y=y$ is usually denoted by $X \lvert Y=y$ or $X \lvert y$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant. The linear conditional mean and constant variance are given by the following:

$\displaystyle E[X \lvert Y=y]=\mu_X+\rho \ \frac{\sigma_X}{\sigma_Y} \ (y-\mu_Y)$

$\displaystyle Var[X \lvert Y=y]=\sigma_X^2 \ (1-\rho^2)$

The information about the conditional distribution of $Y$ on $X=x$ is identical to the information about the conditional distribution of $X$ on $Y=y$, except for the switching of $X$ and $Y$. An example is helpful.

Example 1
Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=10$, $\sigma_X=10$, $\mu_Y=20$, $\sigma_Y=5$ and $\rho=0.6$. The first two parameters are the mean and standard deviation of the marginal distribution of $X$. The next two parameters are the mean and standard deviation of the marginal distribution of $Y$. The parameter $\rho$ is the correlation coefficient of $X$ and $Y$. Both marginal distributions are normal.

Let’s focus on the conditional distribution of $Y$ given $X=x$. It is normally distributed. Its mean and variance are:

\displaystyle \begin{aligned} E[Y \lvert X=x]&=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X) \\&=20+0.6 \ \frac{5}{10} \ (x-10) \\&=20+0.3 \ (x-10) \\&=17+0.3 \ x \end{aligned}

$\displaystyle \sigma_{Y \lvert x}^2=Var[Y \lvert X=x]=\sigma_Y^2 (1-\rho^2)=25 \ (1-0.6^2)=16$

$\displaystyle \sigma_{Y \lvert x}=4$

The line $y=17+0.3 \ x$ is also called the least squares regression line. It gives the mean of the conditional distribution of $Y$ given $x$. Because $X$ and $Y$ are positively correlated, the least squares line has positive slope. In this case, the larger the $x$, the larger is the mean of $Y$. The standard deviation of $Y$ given $x$ is constant across all possible $x$ values.

With mean and standard deviation known, we can now compute normal probabilities. Suppose the realized value of $X$ is 25. Then the mean of $Y \lvert 25$ is $E[Y \lvert 25]=24.5$. The standard deviation, as indicated above, is 4. In fact, for any other $x$, the standard deviation of $Y \lvert x$ is also 4. Now calculate the probability $P[20. We first calculate it using a normal table found here.

\displaystyle \begin{aligned} P[20

Using a TI84+ calculator, $P[20. In contrast, the probability $P[20 is (using the table found here):

\displaystyle \begin{aligned} P[20

Using a TI84+ calculator, $P[20. Note that $P[20 is for the marginal distribution of $Y$. It is not conditioned on any realized value of $X$.

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