AP Statistics / Mr. Hansen

1/27/2003

Name: ________________________

 

 

Random Variable Mini-Worksheet—SOLUTION KEY

 

 

1.

Define X = number that appears when die is randomly rolled. [Since X is a numeric outcome resulting from a random process, X is a r.v.]

 

 

 

 

 

 

i

xi

pi

 

 

----------------

----------------

----------------

 

 

1

1

.1

 

 

2

2

.3

 

 

3

3

.1

 

 

4

4

.1

 

 

5

5

.3

 

 

6

6

.1

 

 

 

 

[We usually omit the i column since it adds no information. Note that Spi = 1, as required.]

 

 

 

[If you weren’t able to find the pi values in your head, here is how to show the work:
Let c = most common probability here.
We know the probabilities for 1 through 6 must be (in order) c, 3c, c, c, 3c, and c.
But we also know that these must add up to 1.
Write equation: c + 3c + c + c + 3c + c = 1.
Solve to get c = .1, 3c = .3.]

 

 

2.

mX = Spixi = .1(1) + .3(2) + .1(3) + .1(4) + .3(5) + .1(6) = 3.5. [Note that this exactly the same as for a fair die! Refer to your class notes from Friday, 1/24/2003.]

 

 

 


xi

(dev. from mean)2
= (xi – mX)2

weighted (dev.)2
= pi(xi – mX)2

 

 

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----------------

----------------

 

 

1

(1 – 3.5)2 = 6.25

.1(6.25) = .625

 

 

2

(2 – 3.5)2 = 2.25

.3(2.25) = .675

 

 

3

(3 – 3.5)2 = .25

.1(.25) = .025

 

 

4

(4 – 3.5)2 = .25

.1(.25) = .025

 

 

5

(5 – 3.5)2 = 2.25

.3(2.25) = .675

 

 

6

(6 – 3.5)2 = 6.25

.1(6.25) = .625

 

 

 

 

----------------

 

 

 

TOTAL

2.65

 

 

 

 

total of third column = variance of X = s2X = 2.65

 

sX = Ö2.65 = 1.628

 

 

 

[This is slightly less than the value of 1.708 that we obtained on Friday for a fair die. This makes sense, since the relative frequency histogram for X would show a little less dispersion. More of the “meat” of the distribution stays closer to the mean, which implies that the s.d. will be lower.]

 

 

 

[Do you have to show this much work? Answer: Yes, on free response. The column headings are even more important than the work, since they show that you know how to perform the work. On a multiple-choice question, or to check your work, you can find the mean and s.d. very quickly as follows:

(A) Store xi values in L1, pi values in L2.
(B) Issue command STAT CALC 1-Var Stats L1,L2. That’s all!]

 

 

3.(a)

Not disjoint, since P(A Ç B) = P(prime and even) = P(2) = .3 ¹ 0.

Or, you could simply say that in the sample space {1, 2, 3, 4, 5, 6}, there is a number that is both prime and even, namely 2.

 

 

(b)

Method 1: Check whether prob. of intersection = product of probs.

 

P(A Ç B) = .3 by part (a)

 

P(A) · P(B) = P(prime) · P(even)
 = P(2, 3, or 5) · P(2, 4, or 6)
 = (.3 + .1 + .3) · (.3 + .1 + .1)            [note use of mut. exclusive rule here]
 = .7(.5)
 = .35

 

Since .35 ¹ .3, A and B are not independent.

 

 

 

Method 2: Check whether P(A | B) is unchanged from P(A).

 

P(A) = P(prime) = P(2, 3, or 5) = .3 + .1 + .3 = .7

 

P(A | B) = P(prime | even)
 = P(2, 3, or 5 | 2, 4, or 6)
 = P(2 | 2, 4, or 6)
 = .3/.5
 = .6

 

Since .6 ¹ .7, A and B are not independent.

 

 

(c)

P(A | B) = .6 (already done)

 

 

 

P(B | A) = P(even | prime)
 = P(2, 4, or 6 | 2, 3, or 5)
 = P(2 | 2, 3, or 5)
 = .3/.7
 = .429