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无代写-MAT 452

时间：2021-03-08

DePaul University

Department of Mathematical Sciences

MAT 452: Probability and Statistics II

Sample Final Test - 2021 Winter

Instructions:

i. This test consists of SEVEN questions for a total of xx points possible. Answer all of

them.

ii. The time allowed for this test is 135 minutes

iii. This test is closed book, closed notes. You may use calculators.

iv. Show all your work to justify your answers. Answers without adequate justification will

not receive credit.

Warning: The sample test can not guarantee full preparation of the real test.

Please also review the materials, in class examples, quiz questions and homework

assignments!

1. Find the correlation coefficient of the random variable X and Y if var(X) = 16,

var(Y ) = 9 and var(X − 2Y ) = 28. (Refer to lecture note 6 and lecture note 2)

Answer : 28 = var(X − 2Y ) = var(X) − 4 cov(X, Y ) + 4 var(Y ) = 16 − 4 cov(X, Y ) +

4(9) = 52 − 4 cov(X, Y ) ⇒ cov(X, Y ) = 6. So, the correlation coefficient of X and Y

is

corr(X, Y ) =

cov(X, Y )√

var(X) var(Y )

=

6√

16(9)

= 0.5

2. Let X1,..., Xn be independent, exponentially distributed random variables with mean

θ. Show that Y1 = min(X1, ..., Xn) has an exponential distribution, with mean

θ

n

.

(Refer to lecture note 5)

Answer : The cumulative distribution function of Y1 = min(X1, ..., Xn) is

FY1(y) = Pr(Y1 ≤ y)

= Pr(min(X1, ..., Xn) ≤ y)

= 1− Pr(X1 ≥ y,X2 ≥ y, ..., Xn ≥ y)

= 1− [Pr(X ≥ y)]n

= 1− [e− yθ ]n

= 1− e−

y

θ

n

1

This is the cdf of an exponential distribution with mean θ

n

. Based on the one-to-one

correspondence between cdf and distribution, Y1 has an exponential distribution with

mean θ

n

.

3. X1 and X2 denoted the proportions of time during which employees I and II actually

performed their assigned tasks during a workday. The joint density of X1 and X2 is

f(x1, x2) =

{

x1 + x2, 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1

0, otherwise

.

(refer to lecture note 1(b) and 3)

(a) Find the marginal density functions for X1 and X2.

(b) Find Pr(X1 ≥ 1/2|X2 ≥ 1/2).

(c) If employee II spends exactly 50% of the day working on assigned duties, find the

probability that employee I spends more than 75% of the day working on similar

duties.

Answer :

(a)

fX1(x1) =

∫ 1

0

(x1 + x2)dx2 = x1 +

1

2

, 0 ≤ x1 ≤ 1.

fX2(x2) =

∫ 1

0

(x1 + x2)dx1 = x2 +

1

2

, 0 ≤ x2 ≤ 1.

(b)

Pr(X1 ≥ 1/2|X2 ≥ 1/2) = Pr(X1 ≥ 1/2, X2 ≥ 1/2)

Pr(X2 ≥ 1/2) =

∫ 1

1/2

∫ 1

1/2

(x1 + x2)dx1dx2∫ 1

1/2

(x2 +

1

2

)dx2

=

3

5

(c) The conditional distribution of X1 given X2 = 0.5 is

fX1|X2(x1|0.5) =

f(x1, 0.5)

fX2(0.5)

=

x1 + 0.5

1

= x1 + 0.5, 0 ≤ x1 ≤ 1.

Thus, Pr(X1 > 0.75|X2 = 0.5) =

∫ 1

0.75

fX1|X2(x1|0.5)dx1 =

∫ 1

0.75

(x1 + 0.5)dx1 =

11

32

.

4. A forester studying the effects of fertilization on certain pine forests in the Southeast

is interested in estimating the average basal area of pine trees. In studying basal

areas of similar trees for many years, he has discovered that these measurements (in

square inches) are normally distributed with standard deviation approximately 4 square

inches. Suppose the forester would like the sample mean to be within 0.5 square inch

of the population mean, with probability .95. How many trees must be measure in

order to ensure this degree of accuracy? (Refer to lecture note 9)

Answer : With confidence level 0.95, the MOE 1.96 σ√

n

≤ 0.5⇒ n = 246.

2

5. Owing to the variability of trade-in allowance, the profit per new car sold by an au-

tomobile dealer varies from car to car. The profits per sale (in hundreds of dollars),

tabulated for the past week, were

2.1, 3.0, 1.2, 6.2, 4.5, 5.1.

Find a 95% confidence interval for the mean profit per sale. What assumptions must

be valid for the technique that you used to be appropriate? (Refer to lecture note 9)

Answer : The sample mean x¯ = 3.68 and the sample standard deviation s = 1.91. So,

the 95% confidence interval for the mean profit per sale is

x¯± t0.025,5 s√

n

= 3.68± 2.571.91√

6

= 3.68± 2.

The population needs to be approximately normal.

6. Suppose that E(θˆ1) = E(θˆ2) = θ, var(θˆ1) = σ

2

1 and var(θˆ2) = σ

2

2. Consider the

estimator

θˆ3 = aθˆ1 + (1− a)θˆ2.

(Refer to lecture note 8(b))

(a) Show that θˆ3 is an unbiased estimator for θ.

(b) If θˆ1 and θˆ2 are independent, how should the constant a be chosen in order to

minimize the variance of θˆ3?

Answer :

(a)

E(θˆ3) = E(aθˆ1 + (1− a)θˆ2)

= aθ + (1− a)θ

= θ

(b) var(θˆ3) = var(aθˆ1 + (1−a)θˆ2) = a2σ21 + (1−a)2σ22. To minimize the variance with

respect to a, we take the derivative of variance with respect to a and set it to 0,

i.e.,

2aσ21 − 2(1− a)σ22 = 0⇒ a =

σ22

σ21 + σ

2

2

7. In a large population of adults, the mean IQ is µ = 112 with standard deviation σ = 20.

Suppose 200 adults are randomly selected for a market research campaign. Applying

the central limit theorem, would you be surprised to see the sample mean to be 105?

(Refer to lecture note 7)

Answer : Based on the central limit theorem, the distribution of the sample mean

IQ is approximately normal with mean equal to 112, and standard deviation equal to

σ√

n

= 20√

200

= 1.414. Using the empirical rule, if we take a random sample of 200 adults,

the observed average IQ is likely to be around 112 plus or minus 2.8 or two standard

deviations. That is

x¯ falls in the interval [109.2, 114.8] with probability approximaltely 0.95.

So, we would be surprising to see a sample mean of 105.

3

学霸联盟

Department of Mathematical Sciences

MAT 452: Probability and Statistics II

Sample Final Test - 2021 Winter

Instructions:

i. This test consists of SEVEN questions for a total of xx points possible. Answer all of

them.

ii. The time allowed for this test is 135 minutes

iii. This test is closed book, closed notes. You may use calculators.

iv. Show all your work to justify your answers. Answers without adequate justification will

not receive credit.

Warning: The sample test can not guarantee full preparation of the real test.

Please also review the materials, in class examples, quiz questions and homework

assignments!

1. Find the correlation coefficient of the random variable X and Y if var(X) = 16,

var(Y ) = 9 and var(X − 2Y ) = 28. (Refer to lecture note 6 and lecture note 2)

Answer : 28 = var(X − 2Y ) = var(X) − 4 cov(X, Y ) + 4 var(Y ) = 16 − 4 cov(X, Y ) +

4(9) = 52 − 4 cov(X, Y ) ⇒ cov(X, Y ) = 6. So, the correlation coefficient of X and Y

is

corr(X, Y ) =

cov(X, Y )√

var(X) var(Y )

=

6√

16(9)

= 0.5

2. Let X1,..., Xn be independent, exponentially distributed random variables with mean

θ. Show that Y1 = min(X1, ..., Xn) has an exponential distribution, with mean

θ

n

.

(Refer to lecture note 5)

Answer : The cumulative distribution function of Y1 = min(X1, ..., Xn) is

FY1(y) = Pr(Y1 ≤ y)

= Pr(min(X1, ..., Xn) ≤ y)

= 1− Pr(X1 ≥ y,X2 ≥ y, ..., Xn ≥ y)

= 1− [Pr(X ≥ y)]n

= 1− [e− yθ ]n

= 1− e−

y

θ

n

1

This is the cdf of an exponential distribution with mean θ

n

. Based on the one-to-one

correspondence between cdf and distribution, Y1 has an exponential distribution with

mean θ

n

.

3. X1 and X2 denoted the proportions of time during which employees I and II actually

performed their assigned tasks during a workday. The joint density of X1 and X2 is

f(x1, x2) =

{

x1 + x2, 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1

0, otherwise

.

(refer to lecture note 1(b) and 3)

(a) Find the marginal density functions for X1 and X2.

(b) Find Pr(X1 ≥ 1/2|X2 ≥ 1/2).

(c) If employee II spends exactly 50% of the day working on assigned duties, find the

probability that employee I spends more than 75% of the day working on similar

duties.

Answer :

(a)

fX1(x1) =

∫ 1

0

(x1 + x2)dx2 = x1 +

1

2

, 0 ≤ x1 ≤ 1.

fX2(x2) =

∫ 1

0

(x1 + x2)dx1 = x2 +

1

2

, 0 ≤ x2 ≤ 1.

(b)

Pr(X1 ≥ 1/2|X2 ≥ 1/2) = Pr(X1 ≥ 1/2, X2 ≥ 1/2)

Pr(X2 ≥ 1/2) =

∫ 1

1/2

∫ 1

1/2

(x1 + x2)dx1dx2∫ 1

1/2

(x2 +

1

2

)dx2

=

3

5

(c) The conditional distribution of X1 given X2 = 0.5 is

fX1|X2(x1|0.5) =

f(x1, 0.5)

fX2(0.5)

=

x1 + 0.5

1

= x1 + 0.5, 0 ≤ x1 ≤ 1.

Thus, Pr(X1 > 0.75|X2 = 0.5) =

∫ 1

0.75

fX1|X2(x1|0.5)dx1 =

∫ 1

0.75

(x1 + 0.5)dx1 =

11

32

.

4. A forester studying the effects of fertilization on certain pine forests in the Southeast

is interested in estimating the average basal area of pine trees. In studying basal

areas of similar trees for many years, he has discovered that these measurements (in

square inches) are normally distributed with standard deviation approximately 4 square

inches. Suppose the forester would like the sample mean to be within 0.5 square inch

of the population mean, with probability .95. How many trees must be measure in

order to ensure this degree of accuracy? (Refer to lecture note 9)

Answer : With confidence level 0.95, the MOE 1.96 σ√

n

≤ 0.5⇒ n = 246.

2

5. Owing to the variability of trade-in allowance, the profit per new car sold by an au-

tomobile dealer varies from car to car. The profits per sale (in hundreds of dollars),

tabulated for the past week, were

2.1, 3.0, 1.2, 6.2, 4.5, 5.1.

Find a 95% confidence interval for the mean profit per sale. What assumptions must

be valid for the technique that you used to be appropriate? (Refer to lecture note 9)

Answer : The sample mean x¯ = 3.68 and the sample standard deviation s = 1.91. So,

the 95% confidence interval for the mean profit per sale is

x¯± t0.025,5 s√

n

= 3.68± 2.571.91√

6

= 3.68± 2.

The population needs to be approximately normal.

6. Suppose that E(θˆ1) = E(θˆ2) = θ, var(θˆ1) = σ

2

1 and var(θˆ2) = σ

2

2. Consider the

estimator

θˆ3 = aθˆ1 + (1− a)θˆ2.

(Refer to lecture note 8(b))

(a) Show that θˆ3 is an unbiased estimator for θ.

(b) If θˆ1 and θˆ2 are independent, how should the constant a be chosen in order to

minimize the variance of θˆ3?

Answer :

(a)

E(θˆ3) = E(aθˆ1 + (1− a)θˆ2)

= aθ + (1− a)θ

= θ

(b) var(θˆ3) = var(aθˆ1 + (1−a)θˆ2) = a2σ21 + (1−a)2σ22. To minimize the variance with

respect to a, we take the derivative of variance with respect to a and set it to 0,

i.e.,

2aσ21 − 2(1− a)σ22 = 0⇒ a =

σ22

σ21 + σ

2

2

7. In a large population of adults, the mean IQ is µ = 112 with standard deviation σ = 20.

Suppose 200 adults are randomly selected for a market research campaign. Applying

the central limit theorem, would you be surprised to see the sample mean to be 105?

(Refer to lecture note 7)

Answer : Based on the central limit theorem, the distribution of the sample mean

IQ is approximately normal with mean equal to 112, and standard deviation equal to

σ√

n

= 20√

200

= 1.414. Using the empirical rule, if we take a random sample of 200 adults,

the observed average IQ is likely to be around 112 plus or minus 2.8 or two standard

deviations. That is

x¯ falls in the interval [109.2, 114.8] with probability approximaltely 0.95.

So, we would be surprising to see a sample mean of 105.

3

学霸联盟