Midterm Exam 1 with Solutions - Applied Statistics I | MATH 3070, Exams of Statistics

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The University of Utah, Spring 2002

Mathematics 3070–5, Midterm 1

Solutions

1. A truth serum has the property that 80% of the guilty suspects are properly judged. Moreover, inno- cent suspects are misjudged 2% of the time. If the suspect was selected from a group of suspects of which only 5% have ever committed a crime, and if the serum indicates that he is guilty, what is the probability that he is innocent? Show your work.

Solution Let G = {guilty}, and J = {judged guilty} to see that G′^ = {innocent} and J′^ = {judged innocent}. Moreover, the problem tells us that P {G} = 0.05, P {J | G} = 0.8, and P {J | G′} = 0 .02. We are after P {G′^ | J}. First, we compute P {J}: by Bayes’ rule,

P {J} = P {J ∩ G} + P {J ∩ G′} = P {J | G}P {G} + P {J | G′}P {G′} = 0. 8 × 0 .05 + 0. 02 × 0. 95 = 0. 059.

Thus,

P {G′^ | J} = P {J | G′} ·

P {G′}

P {J}

= 0. 02 ×

≈

 

0.322 (^)
.

2. Three fair coins are tossed independently from one another. Let X denote the total number of heads, and Y denote the number of tails thus obtained.

(a) Find the joint probability distribution of X and Y. (b) What is P {X ≥ 2 }?

Solution to (a): The joint distribution is given by the following.

x 0 1 2 3 (^0 0 0 0 ) y 1 0 0 38 0 2 0 38 0 0 3 18 0 0 0

Solution to (b): P {X ≥ 2 } = 38 + 18 =

 

1

3. Suppose the joint density function of X and Y is

f (x, y) =

{

4 xy, if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 , 0 , otherwise.

(a) Compute the marginal density functions of X and Y. (b) Are X and Y independent? Explain clearly.

Solution to (a): To find the marginal g(x) of X, integrate f (x, y) against dy, i.e.,

g(x) =







∫ 1

0

4 xy dy = 4x

∫ 1

0

y dy = 2x, if 0 ≤ x ≤ 1 ,

0 , otherwise.

Similarly,

h(y) =

{

2 y, if 0 ≤ y ≤ 1 , 0 , otherwise.

Solution to (b):

 

 Yes , since f (x, y) = g(x)h(y).

4. SAT verbal scores are known to be approximately distributed according to a normal curve. This year, 10,000 people took the test, and the average verbal SAT score was 550 points, and the standard de- viation was 100 points. A graduate program in creative writing will only admit students whose SAT verbal score was in the top 2% of the applicant pool. What is the minimum score needed for admission in to this program? Show your work.

Solution: The z-score for top 2% is between 2.05 and 2.06, which we take as z = 2.055, i.e.,

Min.score − μ σ

= 2. 055.

Plug σ = 100 and μ = 550 to get Min.Score = 2. 055 × 100 + 550 =

 

755. (^5)
     points.

756. Consider the random variables X and Y whose joint probability distribution is

x 1 2 3 1 0.05 0.05 0. y 2 0.05 0.1 0. 3 0 0.2 0.

One can show that E[X] = 2. 45 , E[Y ] = 2. 1 , Var(X) ≈ 0. 5 , and Var(Y ) ≈ 0. 50. Use this informa- tion to compute the correlation between X and Y.

Solution: We need the covariance, which is in turn found by first calculating E[XY ].

E[XY ] = (1 × 1 × 0 .05) + (2 × 1 × 0 .05) + (3 × 1 × 0 .1) + (1 × 2 × 0 .05) + (2 × 2 × 0 .01) +(3 × 2 × 0 .35) + (1 × 3 × 0) + (2 × 3 × 0 .2) + (3 × 3 × 0 .1) = 4. 79.

So the covariance is

σXY = E[XY ] − μX μY = 4. 79 − (2. 45 × 2 .1) = − 0. 355.

Thus, the correlation is

ρXY = σXY σX · σY

=

− 0. 355

√

0. 5 ×

√

≈

 

 − 0. (^71) .