AP Statistics / Mr. Hansen
2/7/2003

Name: _________________________

Key to Practice Test from December 1998

Note: In most cases, work is omitted here. For full credit on your HW assignment, you must show work and circle your answers.

1.(a)

P(B) = P(B | A)
In other words, events A and B are independent iff knowing that event A has occurred does not change the likelihood of B.

(b)

P(A Ç B) = P(A) P(B)
In other words, events are independent iff their probabilities can be multiplied for an intersection. (We called this the “IMI” rule, which is a nonstandard abbreviation.)

(c)

The middle choice is the best of the three.

(d)

Proof that (a) Þ (b):

 

Statement

Reason

1. P(B) = P(B | A)

1. Given (provided that (a) is true)

2. P(B | A) = P(B Ç A) / P(A)

2. Conditional prob. formula

3. P(B) = P(B Ç A) / P(A)

3. Transitive prop. (steps 1, 2)

4. P(A) P(B) = P(B Ç A)

4. Mult. prop. (mult. both sides by P(A))

5. P(B Ç A) = P(A Ç B)

5. Intersection is commutative

6. P(A Ç B) = P(A) P(B)

6. Transitive prop. (steps 5, 4)

Q.E.D.

 

(e)

Proof that (b) Þ (a):

 

Statement

Reason

1. P(A Ç B) = P(A) P(B)

1. Given (provided that (b) is true)

2. P(B Ç A) = P(B) P(A)

2. By subst., since both intersection and mult. are commutative

3. P(A) ¹ 0

3. Given in (c) if you answered correctly

4. P(B Ç A) / P(A) = P(B)

4. Div. prop. (divide eqn. 2 by P(A))

5. P(B | A) = P(B Ç A) / P(A)

5. Conditional prob. formula

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

6. Transitive prop. (steps, 5, 4)

Q.E.D.

 

(d)

Short version: Assume (a) is true, i.e., that P(B) = P(B | A). Then P(B) = P(B Ç A) / P(A). Mult. both sides by P(A) to get (b). Q.E.D.

(e)

Short version: Assume (b) is true, i.e., that P(A Ç B) = P(A) P(B). If P(A) ¹ 0 [and P(A) certainly would be nonzero in order for cond. prob. involving A to have any meaning], then we can div. both sides by P(A) to get P(A Ç B) / P(A) = P(B). LHS equals P(B | A), so we have (a). Q.E.D.

2.

E
A
B
F
X (The answer is not G. Do you understand why?)
D
C
H

3.(a)

.9740 = .296

(b)

1 – .9740 = .704, or you could say P(X ³ 1) = .704 where X follows B(40, .03)

Note: The notation B(40, .03) means a binomial distribution with parameters n = 40 and p = .03. Do you understand why X does not follow B(100, .03) for the purposes of this question?

Note: You cannot write 1 – binompdf(40,0.03,0) or 1 – binomcdf(40,0.03,0), even though they would give the correct answer. Calculator notation is forbidden.

(c)

.03 since trials are independent

4.(a)

 

Denominator is 52C4 since 4 cards are dealt from a 52-card deck. (This is not poker. Although poker questions, even simplified poker question like this one, should not appear on the AP, you are expected to be able to perform analysis to the depth shown here.)

Numerator signifies a choice of 2 values from among 13, then twice choosing 2 of the 4 suits within each chosen value. By multiplication rule, total number of ways is the product as shown.

(b)

1/(52C4) = .00000369

There is only one way to get 4 kings, and that is to get all 4 of them. Again, note that since this is not poker, there is no need to account for an extra junk card.

(c)

0/(52C4) = 0

In a free-response problem, you would also write, “It is now clearly impossible to get 4 kings.”

(d)

Method 1
    It is as if we wished to draw 3 of the 3 remaining kings from a 51-card deck. Since there is only one way to do this (namely, to draw all 3 of them), the answer is 1/(51C3) = .000048.

Method 2
    Use cond. prob. formula where K = event of drawing first card as king, L = event of drawing all 4 cards as kings. P(L | K) = P(L Ç K) / P(K). Numerator is the same as the answer from part (b), and denominator is clearly 4/52. Final answer is .000048.

5.

There were two versions of this question to discourage cheating during the test. Some students had a prize of $5000, others a prize of $6000.

Suggestion: Work the problem twice, once with a $5000 prize and once with a $6000 prize. That way, you will get more practice.

Let X = net winnings after adjusting for the cost of a ticket

$5000 prize
    P(X = 4999) = .0001, P(X = –1) = .9999
    mX = –.5, sX2 = 2499.75, sX = 49.997

$6000 prize
    P(X = 5999) = .0001, P(X = –1) = .9999
    mX = –.4, sX2 = 3599.64, sX = 59.997

Units for expected value and s.d. are dollars. Units for variance are irrelevant.

Note: You cannot read variance directly from your calculator. Double-check your answers to make sure you did not take an invalid shortcut. Variance is the square of the s.d., and after you have punched in 1-Var Stats L1,L2, you can access the s.d. with VARS 5 4. (That avoids the errors associated with rey-keying the s.d.)

Remember that you must show all your work for full credit. (See questions 1 and 2 on the Random Variable Mini-Worksheet Solution Key for examples of how to show your work.)

6.

Assume P(A) = .15, P(B) = .01. However, on a rainy day, umbrella usage is much more common, perhaps giving P(B | A) = .25. Since P(B) ¹ P(B | A), the events are not indep.

7.(a)

26 · 26 · 26 · 26 · 10 · 10 · 10 = 456,976,000

(b)

Yes, since there are fewer than 300 million residents. Although there are more registered vehicles nationwide than there are licensed drivers, not everyone has a driver’s license.

Of the 50 states plus D.C., which one has the highest ratio of registered vehicles to licensed drivers? The lowest? Answers are at the end of the handout.

(c)

Some arrangements of 4 letters would not be acceptable. For example, SHED might cause some people to become distracted from the task of driving.

 

 

 

 

 

 

 

Answers to 7(b) challenge questions:

Iowa has the highest ratio of registered vehicles to licensed drivers (more than 3 million vehicles for fewer than 2 million drivers), probably because of the large number of farm vehicles and the small number of unlicensed vehicles.

The District of Columbia has the lowest ratio (235,000 vehicles, 350,000 licensed drivers), which is understandable since the entire jurisdiction is urban. The 50 states all have a mixture of both urban and rural populations, and rural populations use more cars per capita. D.C. is the only place where there is no rural population to boost the number of cars.

In 46 of the 50 states, vehicle registrations outnumber driver's licenses. In fact, only D.C., Nevada, Florida, Arkansas, and Hawaii have fewer registered vehicles than drivers. Most of these are plausible: Florida has many elderly people who no longer drive, Arkansas is a poor state with possibly a high number of unregistered vehicles, and D.C. and Hawaii have overwhelmingly urban populations. However, Nevada is second only to D.C. in having a low ratio of vehicles to drivers (1.2 million registered vehicles, somewhere between 1.3 million and 1.4 licensed drivers), and I am not sure why. Perhaps you can come up with some reasons. Submit your ideas by e-mail, and the best explanation will qualify for an extra-credit bonus.