Download Midterm Result

Midterm Results
Points /200
% Correct
Freq
192-199
96-99.5
181-190
91-95
17 / 62
172-180
86-90
16 / 62
168-170
83-85
8 / 62
155-162
75-81
6 / 62
<145
<75
8 / 62
7 / 62
1. Three fair coins are tossed in the air. If the first coin
lands as tails, what is the probability that the other two
coins will both land as tails? Explain.
• Because of independence: (1/2) x (1/2) = 1/4
• Binomial: 2C2 (1/2)2(1/2)0 =(1)(1/4)(1) = ¼
• Conditional probability:
P (T & TT ) (1 / 2)(1 / 4)
P (TT | T ) 
>90%
0
0
86-90%
0
0
75-85%
0
0
<75%
3
-30
P (T )

1/ 2
 1/ 4
2. A city has 50,000 voters. A random sample of 100
voters are asked for whom they would vote. 60 of the
respondents said that would vote for Candidate A and
40 said that they would vote for Candidate B. If in fact
the voting public in that city was equally divided
between the candidates, what is the probability that
the sample would turn out just like it did?
• Binomial:
100C60
(1/2)60(1/2)40
>90%
0
0
86-90%
2
-10
75-85%
2
-15
<75%
3
-20
3. If X is distributed normally with mean = 50
and variance = 16, between what two values
at equal distances from the mean will 95% of
the area under the curve lie?
  1.96  50  1.96(4)
= 42 to 58 approx
>90%
7
-25
86-90%
8
-28
75-85%
9
-37
<75%
8
-61
4. A die has three of its sides painted red, two painted
yellow, and one painted black? If the die is rolled
twice, what is the probability of two yellows?
Sample space: 36 points (6x6)
Event A: Y1Y1, Y1Y2, Y2Y1, Y2Y2 = 4 points
4/36 = 1/9
Independence of rolls: P(Y, 1st) x P(Y, 2nd) =
2/6 x 2/6 = 4/36 = 1/9
>90%
0
0
86-90%
0
0
75-85%
0
0
<75%
0
0
5. Assume that the ratio of male:female children is ½. In
a family where 4 children are desired, what is the
probability (a) that all 4 children will be of the same sex,
and (b) that exactly 2 will be boys and 2 girls?
• From the binomial:
• (a) 4C4 (1/2)4(1/2)0 = 1/16 = P(all boys)
4(1/2)0 = 1/16 = P(all girls)
C
(1/2)
4 4
1/8 = P(all same sex)
• (b) 4C2 (1/2)2(1/2)2 = 6(1/16) = 3/8 = P(2 b and 2 g)
(p=1/3 boys & 2/3 girls
>90%
1
-3
also OK)
86-90% 2
-6
75-85% 0
0
<75%
2
-7
6. Compare the chances of rolling a 3 with 1 die
and rolling a total of 6 with 2 dice.
• P(3 w/1 die) = 1/6
• P(sum to 6 w/ 2 dice) = 5/36
(5+1, 4+2, 3+3, 2+4, 1+5)
>90%
0
0
86-90%
75-85%
0
0
0
0
<75%
2
-12
7. The probability = 3/4 that a putt on the 6th
hole will go in. What is the probability that 3
out of 5 putts on the 6th hole will go in?
3
2
• Binomial:
5C3 (3/4) (1/4)
= (10)(27/64)(1/16)
= 270/1024 = 135/512 = .264
>90%
86-90%
0
0
0
0
75-85%
0
0
<75%
0
0
8. A card is drawn from an ordinary deck.
What is the probability that it is a jack, given
that it is a black card?
• Conditional probability:
P(J|black) = P(J & black) = 2/52 = 2 = 1
P(black)
26/52 26 13
>90%
0
0
86-90%
2
-14
75-85%
4
-19
<75%
6
-51
9. Two balls are drawn, without replacement,
from an urn containing 1 white ball, 2 black
balls, and 3 green balls. What is the
probability that the first is white and the
second is black?
Sample space: P(WB) = 2/30 = 1/15
Joint probability: P(WB) = P(W)P(B|W)
= 1/6 x 2/5
>90%
0
0
= 2/30 = 1/15
86-90% 2
-15
75-85% 0
0
<75%
2
-15
10. Consider the joint distribution of suit and
denomination in an ordinary deck of playing
cards, as shown below. Find (a) the marginal
distribution of suit and (b) the marginal
distribution of face vs non-face cards.
♣
♦
♥
♠
Face cards
(KQJA)
4
4
4
4
16/52
Non-face
cards
(2-10)
9
9
9
9
36/52
Marginal
suit
13/52
13/52
13/52
13/52
of
Marginal of
denomina
tion
52
>90%
1
-2
86-90% 3
75-85% 2
-22
-20
<75%
-39
5
11. Given that a binomial distribution has mean
= 15 and variance = 6, find its parameters.
• np = 15
npq = 6
• q = 6/15 = 2/5
• p = 1-q = 3/5
• n = 15 / p = 15 x 5/3 = 25
>90%
0
0
86-90%
0
0
75-85%
0
0
<75%
2
-24
12. A company sells three brands of cameras in four different
stores, which use the same advertising. Because there is a
problem with cameras being returned quite often, the company
wants to know which brand has the most defective returns. So
the company decides to audit camera returns in the largest
store for each of the three camera brands at that store.
In order to audit camera returns, 100 return receipts for one
particular week are selected at random for each brand and
examined for whether or not the returned cameras were
defective. If 20 defectives were returned for brand A, 21 for
brand B, and 18 for brand C, what exactly can the company
conclude about its defective camera returns and brands? Why?
• The company can conclude that for the week in question at their
largest store, the brands have almost equal numbers of
defective returns because that week at that store was the
sampling frame.
• The company cannot conclude anything about the three brands
at the other stores or in other weeks.
>90%
18
-66
86-90%
15
-74
75-85%
12
-68
<75%
8
-50
13. A test of the breaking strength in pounds of 10’
PVC pipes in a sample of 100 pipes produced an
estimated mean of 300 and an estimated standard
deviation of 50. Find a 95% confidence interval for
the mean, assuming that the distribution of
breaking strength is normal. Interpret this interval.
300  1.96
50
100
 300  1.96(5)  300  10
This interval means that if we did infinitely many
studies of samples of 100 of these pipes, 95% of
the intervals constructed this way would capture the
true mean μ.
Only 4 students got this interpretation right!!!
>90%
1
-5
>90%
19
-83
86-90%
1
-2
86-90%
16
-77
75-85%
2
-4
75-85%
14
-70
<75%
2 -10
<75%
8
-40
14. A city has received some complaints about
the driving ability of its senior citizens, so
they decide to test a sample of seniors to
evaluate their driving ability.
Clearly the city can’t test every senior, so
they decide to test the first 20 seniors who
show up for license renewals in the first week
of May. If they do this, what statement can
they make about the driving ability of seniors?
Why?
• The city cannot conclude anything about the
driving ability of seniors, except for the 20 in
the sample, because the city didn’t use a
random sample for their study.
>90%
7
-33
86-90% 10
-63
75-85% 10
-85
<75%
-50
7
15. The height in feet of 56 pine trees in a forest
was measured with the following results, where
X = height and freq(X) = the frequency with
which the value X occurred.
X
5
10
15
20 25
30
35 40 45
Freq(x)
1
3
5
10 18
10
5
3
1
Draw the histogram and find the mean, the
median, and the mode of this distribution.
• Distribution is symmetric so mean = median =
mode = 25
20
18
16
14
12
10
8
6
4
2
0
5
10
15
20
25
30
35
40
45
>90%
4
-16
86-90%
1
-4
75-85%
3
-11
<75%
4
-15
16. If two points on a binomial cdf are
(4, 0.38) and (3, 0.18) what is the
probability of X > 3 ?
• P(X > 3) = 1 - .18 = .82
>90%
0
0
86-90%
1
-10
75-85%
0
0
<75%
1
-10
17. If two points on a normal cdf are (1.20,
0.38) and (1.21, .40), what is the probability
of X ≥ 1.20 ?
• P(X ≥ 1.20) = 1- .38 = .62
>90%
0
0
86-90%
0
0
75-85%
1
-5
<75%
4
-40
18. In men’s tennis matches, a player must win 3 out of
5 sets to win the match. Roger and David are fairly
well matched and the odds do not favor either player.
What is the probability that the match will end in 3 sets?
• P(Roger wins in 3) = 3C3(1/2)3(1/2)0 = 1/8
• P(David wins in 3) = 3C3(1/2)3(1/2)0 = 1/8
•
P(match ends in 3) = 1/4
>90%
1
-5
86-90%
3
-15
75-85%
4
-36
<75%
5
-44
19. What is the expected value of mX + b,
where X is a random variable and m and b
are constants? What is the variance of mX
+ b?
• E(mX + b) = mE(X) + b = mμ + b
• Var(mX + b) = m2Var(X) + 0
>90%
4
-21
86-90%
3
-11
75-85%
5
-40
<75%
8
-80
20. X is normally distributed with mean = 20 and
standard deviation = 10. Y is also normally
distributed with mean = 15 and standard deviation =
5. If X is transformed to standard normal form, and
Y is transformed to standard normal form, what will
be the difference between their standardized means?
• Both means = 0, so
their difference = 0.
>90%
1
-10
86-90% 4
-31
75-85% 9
-85
<75%
-65
7