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<Y<28]
  • For X=20, determine the mean and standard deviation of the conditional distribution of Y given X=20.
  • Determine P[12<Y<28 \lvert X=20], the probability that 12<Y<28 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.

  1. Compute the probability P[3<Y<5]
  2. Determine E[Y \lvert X=x], the mean of the conditional distribution of Y given X=x.
  3. 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.
  4. 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.
  5. For each of the x values 6, 8, 10 and 12, determine P[3<Y<5 \ \lvert X=x]. 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<X+Y<140]
  • Determine the 5 parameters of the bivariate normal random variables L=X+Y and M=X-Y.
  • Calculate P[100<X+Y<140 \ \biggl \lvert \ X-Y=5]

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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<Y<81.76 \lvert X=70]=0.95. 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<Y<55 \lvert X=60]
  • 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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Problem ………..Answer
5-A
  1. P[12<Y<28]=0.8904 (table), 0.8904014212 (TI84+)
  2. E[Y \lvert X=20]=15.625, Var[Y \lvert X=20]=12.75
  3. P[12<Y<28 \lvert X=20]=0.8458 (table), 0.8447309876 (TI84+)
5-B
  • P[3<Y<5]=0.5934 (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<Y<5 \lvert X=6]=0.8354 (table), 0.8351333522 (TI84+)
    • P[3<Y<5 \lvert X=8]=0.3886 (table), 0.3894682472 (TI84+)
    • P[3<Y<5 \lvert X=10]=0.0262 (table), 0.025919702 (TI84+)
    • P[3<Y<5 \lvert X=12]=0.0002 (table), 0.0001524802646 (TI84+)
5-C
  • P[100<X+Y<140]=0.7568 (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<X+Y<140 \lvert X-Y=5]=0.8967 (table), 0.8966089617 (TI84+)
5-D
  • Normal with mean 163.8 and standard deviation \sqrt{279}.
  • P[170<Y<200 \lvert X=71]=0.3407 (table), 0.3401418637 (TI84+)
  • 90th percentile = 185.18 (table), 185.2061314 (TI84+)
5-E
  • 0.8
5-F
  • P[45<Y<55 \lvert X=60]=0.5123 (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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2 thoughts on “Practice Problem Set 5 – bivariate normal distribution

  1. […] Practice problems to reinforce these concepts are available here. […]

  2. […] The next post is a further discussion on bivariate normal distribution. Practice problems on bivariate normal distribution are available here. […]

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