MTH 180, Fall 2015, Test 2
NAME:........................................... Section ...........
Part I: Show work for credit. If you get your answer from a table state it.
1. At Scott’s Pick-a-Red marble game, you give me $2 to play. There is a box with ten marbles in it;
two red, four green, and four blue. You close your eyes and pick one marble. If you pick red, you
get $4. If you pick green, you get $2. If you pick blue, you get nothing.
What is the expected value of this game to you (the player)?
(5 pts)
Outcomes
Red
Green
Blue
Sum (Σ)
value = x
2
0
-2
P (x)
2/10 = 0.20
4/10 = 0.40
4/10 = 0.40
x · P (x)
0.4
0.0
-0.8
E = -.40
expected value:
-0.40
2. Suppose that 10% of all M&M candies are blue.
Consider a random sample of 15 M&M’s from a very large bag of M&M’s.
(10 pts)
(a) In all such samples of size 15, what is the mean number of blue M&M’s?
µ = 1.5
µ = n · p = 15(.10) = 1.5
(b) What is the probability that you get exactly 4 blue M&M’s.
your answer
0.043
P (x = 4 | n = 15, p = .10) = .043 from Tables 1 with n = 15.
(c) Would 4 blue M&M’s be considered an unusually high number of blue M&M’s and why?
Use the criteria that a number (x) is unusually large if P(x or more) ≤ 0.05.
Is 4 unusually high?
(yes or no) no
P (x ≥ 4) = 0.043 + 0.010 + 0.002 + ∗ + . . . + ∗ = 0.055.
This is greater than 0.05 so 4 is not an unusually large number of blue M&M’s.
3. The weights of Shaw’s potatoes are normally distributed with a mean of 9.0 ounces and a standard
deviation of 0.8 ounces. What is the probability that a single potato weighs more than 11 ounces?
(10 pts)
your answer
0.0062
11−9
If x = 11, then z = x−µ
σ = 0.8 = 2.50
P (x > 11) = P (z > 2.50) =
= 1 − P (z < 2.50) = 1 − 0.9938 = 0.0062
4. The weights of Shaw’s potatoes are normally distributed with a mean of 9.0 ounces and a standard
deviation of 0.8 ounces.
(20 pts)
(a) What weight delineates the heaviest 5% of the potatoes from the others?
Round your answer to two decimal places.
your answer
10.32 ounces
Here we’re looking for P95
Look for 0.95 INSIDE the z-table.
The corresponding z-value is z = 1.645
Now x = µ + zσ = 9 + 1.645(0.8) = 10.316 →
10.32
(b) What is the probability that a bag of 6 potatoes has a mean potato-weight below 8.5 ounces?
Give your answer to four decimal places.
your answer
0.0630
This is for sample data!
x−µ
√
If x = 8.5, then z = σ/
n
8.5−9
√ = −1.53093 → −1.53
z = 0.8/
6
P (x < 8.5) = P (z < −1.53) = 0.0630
5. It is known that 20% of all U.S. college students do not graduate in four years. At State College, the
incoming class of 2008 had 800 students and 135 of them didn’t graduate in four years (about %17).
• Assuming the 20% non-graduation rate is accurate, convert the non-graduating 135 students out
of 800 at State College into a z-score. Round your answer to two decimal places.
(4 pts)
z = -2.21
p
√
µ = n · p = 800(0.20) = 160, σ = npq = 800(.2)(.8) = 11.3
z = x−µ
= 135−160
= −2.2124 → −2.21
σ
11.3
• Use the normal approximation to the binomial distribution to estimate the probability of getting
135 or fewer non-graduating students in a randomly selected group of 800 students.
(4 pts)
your answer
0.0136
P (x < 130) ≈ P (z < −2.21) = 0.0136
Exact Answer with a TI Calculator:
P (x ≤ 135) = binomcdf(800, .2, 135) = 0.0139
• Which statement best describes the situation at State College?
(a) Since the non-graduating rate at State College is only 3% below the national average,
there is no reason to believe anything unusual is happening.
(b) Since the probability of getting 135 or fewer non-graduating students is so small, something unusual is happening.
(c) Since the probability of getting 135 or fewer non-graduating students is bigger than
10%, there is no reason to believe that anything unusual is happening.
(d) There is no chance that the below-average non-graduation rate at State College is due
to random variation.
(2 pts)
answer:
b
6. In a random sample of 150 dog owners, 18 of them said they do not and will not pick up the poop left
behind by their dog in public places.
(15 pts)
(a) What is the best point estimate for the proportion of all dog owners who refuse to pick up after
their dog? Round your answers to three significant digits.
point estimate for p:
0.12
p̂ is the point estimate =
x
n
=
18
160
= 0.12
(b) What is the margin of error (E) for a 99% confidence interval estimate?
Round your answers to three significant digits.
E=
0.0683
E = zα/2
q
p̂q̂
n
q
= 2.575 (.12)(.88)
= 0.068322 → 0.0683
150
(c) What is the 99% confidence interval for the proportion of all dog owners who refuse to pick up
the poop? Round your answers to three decimal places.
99% confidence interval:
0.052 < p < 0.188
p̂ − E = .12 − 0.0683 = 0.0517
p̂ + E = .12 + 0.0683 = 0.1883
(d) Write an understandable concluding statement that explains what the confidence interval from
part (c) means.
Concluding Statement:
I am 99% confident that the proportion of all dog owners who
refuse to pick up poop is between 5.2% and 18.8%.
7. You want to estimate the proportion of dog owners who would go to a proposed dog-park in
Burlington. How many dog owners should you sample in order to be 95% confident that the sample
estimate was in error by no more than three percentage points.
(5 pts)
your answer:
1068
Since we don’t have an estimate for p̂, we use the formula
z2
(.25)
2
(.25)
n = α/2E 2 = (1.96)
= 1067.1
(.03)2
Since this is the minimum sample size, you need a sample with at least 1068 dog owners.
8. In a random sample of 34 cars driving down a given stretch of I-89, the mean speed was 74.2 miles per
hour (mph) with a standard deviation of 3.8 mph.
(15 pts)
(a) What is the point estimate for the mean car speed on this stretch of highway?
your answer:
74.2 mph
x̄ (sample mean) is the point estimate for the mean car speed = 74.2 mph.
(b) What is the margin of error (E) for a 95% confidence interval estimate for the mean speed of all
cars traveling this stretch of highway? Round your answer to two decimal places.
E = 1.33
You don’t have the population standard deviation (σ) so you use the t distribution.
The critical value is found in Table 3, with d.f. = 33 and confidence level of 95%.
= 1.3262 → 1.33
E = tα/2 √sn = 2.035 √3.8
34
(c) What is the 95% confidence interval estimate for the mean speed of all cars traveling this stretch
of highway? Round your answers to one decimal place.
95% confidence interval:
72.9 < µ < 75.5
x̄ − E = 74.2 − 1.33 = 72.87
x̄ + E = 74.2 + 1.33 = 75.53
(d) Can you be 95% confident that the mean speed on this stretch of highway is greater than 70
mph? Why or why not?
Yes, because 70 mph is below the lower entire 95% confidence interval for the mean.
Or
Yes, because 70 mph is below the lower limit of the 95% confidence interval.
Part II: Multiple Choice
(10 pts)
1. Suppose your expected value of a particular gambling game is negative.
Which of the following statements is accurate? There is only one.
(a) You will lose money every time you play the game.
(b) You will win money every time you play the game.
answer:
d
(c) You won’t ever win money playing this game.
(d) If you play this game many times, you can expect to lose money in the long run.
2. It is alright to use small samples (n < 30) in constructing confidence intervals for a population mean
provided
(a) the population is normally distributed.
(b) the standard deviation is less than 10.
answer:
a
answer:
b
(c) the population standard deviation is known.
(d) the population standard deviation is not known.
3. The symbol σ represents
(a) the population mean.
(b) the population standard deviation.
(c) the sample mean.
(d) the sample standard deviation.
4. The margin of error (E) in a confidence interval can be decreased by increasing the sample size or
(a) decreasing the sample size.
(b) moderating the kurtosis.
answer:
d
(c) increasing the confidence level.
(d) decreasing the confidence level.
5. Which of the following is NOT a property of the standard normal distribution (z-distribution)?
(a) It has a mean of 0 and standard deviation of 1.
(b) It has a mean of 1 and standard deviation of 0.
(c) The distribution is symmetric.
(d) It is bell-shaped and centered at 0.
answer:
b
Survey Questions – NO CREDIT.
• A flipped classroom is one where you (the students) learn most of the new content outside of class
by reading or watching videos and most of the class time is spent doing problems.
How do you feel about the flipped classroom approach?
0)
1)
2)
3)
4)
I
I
I
I
I
love it.
like it.
truly don’t care.
dislike it.
hate it.
answer:
• What was your primary source of out-of-class learning?
0) Watched Videos
1) Read Textbook
2) I didn’t read or watch anything outside of class
answer:
• Since your last test or quiz, you’ve had several assignments to read a chapter of the book or watch the
associated lecture before class. How many times did you do this?
0)
1)
2)
3)
4)
zero
more than zero but less than half
about half of them
more than half but not all of them
all of them
answer: