Download Math 137 Review Unit 7 KEY(1)

Math 137 Unit 7 Review KEY
1. The following data show the favorite color and pet for a class of five year old children.
Gender Color
Pet
M
Blue
Dog
M
Green
Dog
F
Pink
Cat
M
Blue
Cat
F
Purple
Fish
F
Pink
Cat
answer the following:
F
Blue
Dog
c. P(Male l Pink) = 0/4 = 0
M
Purple
Cat
d. P(Female l Pink) = 4/4 = 1
F
Pink
Dog
e. P( dog or cat) = 11/12 = .917
F
Pink
Dog
f. P(green) = 2/12 = .167
M
Blue
Dog
g. P (female and blue) = 1/12 = 0.083
M
Green
Cat
a. Create a two way table from
the data set that relates gender
to favorite color.
b. Create a two way table from
the data set that relates gender
to favorite pet.
Use your two way tables to
a.
Male
Female
totals
Blue
3
1
4
Green
2
0
2
Pink
0
4
4
Purple
1
1
2
totals
6
6
12
b.
Dog
3
3
6
Male
Female
totals
Cat
3
2
5
Fish
0
1
1
Totals
6
6
12
2. The numbers of endangered species for several groups are listed here.
Location
Mammals
Birds
Reptiles
Amphibians
Total
USA
63
78
14
10
165
Foreign
251
175
64
8
498
Total
314
253
78
18
663
If one endangered species is selected at random, find the probability that it is
(a) Found in the USA and is a bird 78/663= .118
(b) Foreign or a mammal (251+175+64+8+63)/663 = 561/663 = .846
(c) Warm-blooded (mammals or birds) (314 + 253)/663 = 567/663 = .855
3. A box has 5 red balls and 2 white balls. If two balls are randomly selected (one after the
other), what is the probability that they both are red?
(a) With replacement P(1st red) and P(2nd red) = 7 ∙ 7 =
5 5
25
49
(b) Without replacement P(1st red) and P(2nd red) =
=
5 4
∙
7 6
= .510 or 51.0%
20
42
= .476 = 47.6%
4. Consider the following Probability Distribution for the probability of rainy days in
November for a particular city.
(a)Fill in the missing probability.
X (total rainy days in Nov)
P(X)
P(X) ∙ 𝑥
(𝑥 − 𝜇)2 ∙ 𝑝(𝑥)
0
.10
0
.576
1
.30
. 30
.588
2
.10
.20
.016
3
.20
.60
.072
4
.20
.80
.512
5
.10
.50
.676
(b) Find P(3) and interpret the meaning. The probability that it will rain 3 days in
November is 20%.
(c) Calculate the population mean. 𝜇 = ∑ 𝑥 ∙ 𝑝(𝑥) = 0+.30+.20+.60+.80+.50 = 2.4 days
(d) Calculate the population standard deviation. 𝜎 = √∑(𝑥 − 𝜇)2 ∙ 𝑝(𝑥) =
√. 576 + .588 + .016 + .072 + .512 + .676 = √2.44 ≈ 1.6
(e) It typically rains between __0.8__ and ___4.0___ days in November. (Use the mean
and standard deviation). 2.4±1.6
(f) What is P(x< 2)? Write a sentence in context. P(0)+P(1) = .10+.30 = .40. The
probability that it will rain less than two days in November is 40%.
(g) What is the probability that it will rain a least 4 days? P(4) + P(5) = .20 + .10 =
.30.The probability that it will rain at least four days in November is 30%.
5. A casino in Las Vegas offers the following gambling game. To play you must pay $5.
You role a die one time. If you role a 1, 2 or 3, you lose your $5. If you role a 4 or a 5,
you win $3. (You get your $5 back plus an additional $3). If you role a 6, you win $7.
(You get your $5 back plus an additional $7.)
Let X represent the random variable describing the amount of money won or lost and
P(x) represent the probability of winning or losing that money.
a. Fill out the following probability table with all the missing probabilities.
X, Money Lost or Won
-$5
+$3
+$7
P(x)
3/6
2/6
1/6
P(x)∙ 𝑥
-15/6
6/6
7/6
b. Find the expected value for the probability distribution. -15/6+6/6+7/6= -2/6= -1/3=
−.33
c. Write a sentence explaining the meaning of the expected value in this context. In the
long run, on average you can expect to lose 33 cents per game played.
6. A box contains three $1 bills, two $5 bills, five $10 bills, and one $20 bill. Construct a
Probability distribution for the data if x represents the value of a single bill drawn at
random and then replaced.
X
P(x)
$1 bill
3/11
$5 bill
2/11
$10 bill
5/11
$20 bill
1/11
7. This data reflects Honda sales at a car dealership in Northern California.
Is the probability that someone purchases a Hybrid Honda independent of whether the
buyer is female? Which two proportions are NOT a useful comparison for addressing this
question?
a. 117 / 311 and 34 / 111
b. 111 / 311 and 34 / 117
c. 117 / 311 and 111 / 311
8. State the mean and the best approximation of the standard deviation of the normal curve
shown below.
Mean: 20 , SD: 5
9. The mean of each probability distribution pictured is 5. Which has the largest standard
deviation?
GRAPH A
10.
I. 60/995 = 0.060 answer C
II. 60/505 = 0.119 answer A
III. 170/490 and 60/505 answer D
11. A college intramural sports program requires all students to take a fitness test. The time to
run one mile is recorded for 200 male students in the program. These times are
approximately normal with mean 8 minutes and standard deviation 1 minute.
Mike’s time for one mile is 9 minutes. What is the probability that a randomly selected
student takes longer than Mike to run a mile?
a. 2.5%
b. 16%
c. 32%
d. 68%
9 min. is at the 1st standard deviation which captures the middle 68%. This makes the
area to the right of the mean and the first standard deviation 34% (68/2). The total
area below 9 min. becomes 50% + 34% = 84%. The area to the right is then 100 – 84
= 16%.
12. A college intramural sports program requires all students to take a fitness test. The time to
run one mile is recorded for 200 male students in the program. These times are
approximately normal with mean 8 minutes and standard deviation 1 minute.
Coaches invite the fastest runners to try out for the track team. They invite students with
the lowest 1% of the run times in the distribution. Tom’s time for one mile is 7 minutes and
30 seconds. Will he get an invitation?
a. Yes
b. No
c. Not enough information given to answer this question
Between 7 and 9 captures 68% of the population (1 standard deviation). The area below
7 would be 50 – 34 = 16%. And 7.5 would be to the right. 7.5 is not below 1%.
13. Assume that the distribution of weights of adult men in the United States is normal with
mean 190 pounds and standard deviation 30 pounds. Bill’s weight has a z-score of 1.5.
Which of the following is true?
a.
b.
c.
d.
Bill’s weight is in the upper 2.5% of men’s weights.
Bill weighs less than 220 pounds.
Bill weighs more than 230 pounds.
None of the above.
1 SD is 210 pounds, 2 SD is 240 pounds. So 1.5 SD is halfway which is 225 pounds.
14. An instructor has calculated z-scores for all the grades on the midterm. He just realized
that he used the wrong number for the standard deviation. The correct standard deviation
is twice as large as the one he used.
Which is the quickest way for him to correct the z-scores?
a. double all the original z-scores
b. divide all of the original z-scores by 2
c. there is no quick way to do this, each z-score needs to be separately recalculated
d. none of above
Formula for z score:
𝑥−𝜇
𝜎
should have been
𝑥−𝜇
2𝜎
.
15. Your friend rolls a standard six-sided die 1200 times and you write down the number of
times each outcome comes up (the number of 1’s, the number of 2’s, the number of 3’s,
etc.). Note that the probability that any single outcome comes up when the die is rolled is
1/6. How many times do you expect the number 4 to show up in the 1200 rolls?
a. 600
b. 400
c. 200
d. 800
1200/6 = 200
16. Your friend rolls a standard six-sided die 1200 times and you write down the number of
times each outcome comes up (the number of 1’s, the number of 2’s, the number of 3’s,
etc.). Note that the probability that any single outcome comes up when the die is rolled is
1/6.
I.
Let’s say that after the 1200 rolls are done, you realize that the number 4 came up 600
times. Which of the following statements would be a reasonable conclusion to draw
from this observation?
a. It is not surprising that a fair 6-sided die will come up with the number 4 will occur
600 times in 1200 rolls.
b. This was an unlikely outcome, but is nothing more than bad luck. It is not unlikely
enough to suspect that there is something wrong with the die.
c. The number 4 is only expected to come up around 200 times in 1200 rolls. The fact
that four came up 600 times is strong evidence that the die is not fair.
d. This result is impossible if the die is fair.
II. Suppose you and your friend make a bet before the 1200 rolls take place. You agree that
you will pay your friend $1 for every roll that comes up a 3 or a 4, and your friend will
pay you $3 for every roll that comes up a 5. No money will be exchanged for rolls that
come up 1, 2, or 6.
Who do you expect will make more money on this deal? How much more money will
that person make?
a. You---$200 more than your friend.
b. Your friend---$200 more than you.
c. You---$600 more than your friend.
Expected value: P(3 or 4) + P(5) + P(1 or 2 or 6) = ( - 1 *2/6) + (3*1/6) + (0*3/6) = + 1/6 for
each roll
1200 rolls: $1/6 x 1200 = $+200
17. The ACT exam is used by colleges across the country to make a decision about whether
a student will be admitted to their college. ACT scores are normally distributed with a
mean average of 21 and a standard deviation of 5.
a) Draw a picture of the normal curve with the ACT scores for 1, 2 and 3 standard deviations
above and below the mean.
b) What percent of students score higher than a 31 on the ACT? 2.5% (100-95% = 5% divided by
2 = 2.5%)
c) What are the two ACT scores that the middle 68% of people are in between? 16 and 26 (1 SD
above and below)
d) What percent of people score between a 16 and 21 on the ACT? 68% (1 SD above and
below)
e) Find the ACT score that 84% of people score less than? 50+34 =84% is the cutoff for the first
SD. So 26. (Or look .8400 up in the table and get it more exact. You have to use the z score
formula.)
f) A typical ACT score is between ____16____ and ____26______.
g) An unusual ACT score is above ____31_____ and below ____11______.
h) An extremely unusual ACT score is above _____36____ and below ___6_______.
18. IQ tests are normally distributed with a mean of 100 and a standard deviation of 15.
a) A girl scored a 140 on the IQ test. Find and interpret the z-score for this IQ. Find the
probability of someone scoring higher than 140. Is it unusual for someone to score a
140? Z = 2.47 (using z score formula). It is unusual to have an IQ of 140 since the z
score was greater than 2 SD’s from the mean.
b) A boy scored a 90 on the IQ test. Find the probability of someone scoring lower than 90
Find and interpret the z-score for this IQ. Is it unusual for someone to score a 90?
Z = - 0.67. It is not unusual to have an IQ of 90. It (-0.67) is not below 2SD’s from the
mean. The negative shows that it is below or to the left of the mean.
c) One man scored a 120 on the IQ test. Find the probability of someone scoring higher
than 120. Find and interpret the z-score for this IQ. Is it unusual for someone to score a
120? Z score = 1.33. area from table is 0.9082.
1 - .9082 = 0.0918. The probability of someone scoring higher than 120 is 9.18%.
It is not unusual for someone to score 120 since the z score (1.33) was not greater than
2 SD’s from the mean.
d) Mike scored a 135 on the IQ test and Jake scored a 71 on the IQ test. Find the Z-scores
for each. Which score was more unusual?
Mike: z score = 2.33 Above average and unusual (since greater than 2SD’s)
Jake: z score = -1.93 Below average but not unusual (since it is not below 2SD”s from
mean)