Intro to Probability and Statistics

Sample Midterm #2 – Questions And Answers (Answer Key)

Professor Brian Shydlo

brian@shydlo.com

Question 1) (18 points in total) Assume there are 130 MBA students in a given program. 78 are taking an elective on Venture Capital and 40 are taking an elective on Corporate Governance.

Question 1a) (3 points) Assuming Statistical Independence between taking Venture Capital and Corporate Governance, how many students are taking both classes?

Hint: Intersection

Define A = Taking Venture Capital

Define B = Taking Corporate Governance

P(A) = 78/120 = 60%

P(B) = 40/120 = 30.7692%

There are independent so you can use this formula:

P(A ∩ B) = P(A) * P(B) = 60% x 30.7692% = 18.4615385%

18.4615385% * 130 students = 24

The answer is 24 students are taking both.

Question 1b) (3 points) Again, assuming Statistical Independence between taking Venture Capital and Corporate Governance, how many students are taking at least one of the two courses (either one class or the other or taking both)?

Hint: Union

Note that P(A ∩ B) was computed in part A to be 18.4615385%.

You can use this formula:

P(A U B) = P(A) + P(B) - P(A ∩ B)

P(A U B) = 60% + 30.7692% - 18.4615385%

P(A U B) = 72.3076923%

72.3076923 * 130 students = 94 students

The answer is 94 students are taking at least one of the two courses.

Question 1c) (3 points) Now assume that these classes are not Statistically Independent. It turns out, if a student is taking Venture Capital, then it is less likely that they are taking Corporate Governance.

130 MBA students in a given program. (Same as before.)

78 are taking an elective on Venture Capital. (Same as before.)

40 are taking an elective on Corporate Governance. (Same as before.)

Assume 10% are taking both. (new)

How many students are taking exactly one of the two courses?

You can do this using a probability box:

Define A = Taking Venture Capital

Define B = Taking Corporate Governance

P(A) = 78/130 = 60%

P(B) = 40/130 = 30.7692%

P(A ∩ B) = 10% (given in problem)

P(A ∩ B') = P(A) - P(A ∩ B) = 60% - 10% = 50%

P(A' ∩ B) = P(B) - P(A ∩ B) = 30.77% - 10% = 20.77%

The odds that exactly one of them happen are P(A' ∩ B) + P(A ∩ B')

P(A' ∩ B) + P(A ∩ B') = 20.77% + 50.00% = 70.77%

	A	A'
B	10.00%	20.77%	30.77%
B'	50.00%	19.23%	69.23%
	60.00%	40.00%	100.00%

The answer is 70.77% * 130 = 92

Same chart, but with numbers instead of probabilities

	A	A'
B	13	27	40
B'	65	25	90
	78	52	130

The answer is 65 + 27 = 92

Question 1d) (3 points) How many students are not taking either class? Please use the same assumptions and numbers as part c.

P(A' ∩ B') is the middle cell of the probability box above.

P(A ∩ B) + P(A ∩ B') + P(A' ∩ B) + P(A' ∩ B') = 100%

10% (given) + 50% (from in part c) + 20.77% (from in part c) + ? = 1

The answer is 19.23% * 130 = 25

Question 1e) (3 points) What is the probability that a student is taking Venture Capital given that the student is taking Corporate Governance?

Hint: What is P(A | B)?

Please use the same assumptions and numbers as part c.

P(A | B) = P(A ∩ B) / P(B)

P(A | B) = 10% / 30.77%

P(A | B) = 32.50%

The answer is 32.50%

Question 1f) (3 points) Now assume you have 130 students in an MBA program and 78 are taking Venture Capital and 68 are taking Credit Risk Management. Are these classes Mutually Exclusive? Why or why not?

Two classes would be Mutually Exclusive if taking one means you could not take the other (maybe they are electives offered at the same time). Since the total adds up to 78 + 68 = 146 these classes must not be Mutually Exclusive. With only 130 Students in the program, at least 16 of them must be taking 2 of these classes.

Question 2) (15 points in total) You own 1 stock with the following distribution of returns:

Kind of Year	Probability Of This Kind Of Year Occurring	Stock Return
Great	10%	35%
Good	25%	20%
Fair	45%	15%
Poor	15%	5%
Worst	5%	-30%

Question 2a) (6 points) What is the Expected Value of your stock return?

Year	Probability	Return	Probability * Return
Great	10%	35%	3.50%
Good	25%	20%	5.00%
Fair	45%	15%	6.75%
Poor	15%	5%	0.75%
Worst	5%	-30%	-1.50%
	Sum = 100%		Sum = 14.5%

The Expected Value is the probability weighted average of the returns.

The answer is 14.5%

Question 2b) (6 points) What is the Variance of your stock return?

Year	Probability	Return	Expected Value (see Part A)	Return - Expected Value	(Return - Expected Value)^2	(Return - Expected Value) x Probability
Great	10%	35%	14.5%	20.5%	4.20%^2	0.4203%^2
Good	25%	20%	14.5%	5.5%	0.30%^2	0.0756%^2
Fair	45%	15%	14.5%	0.5%	0.00%^2	0.0011%^2
Poor	15%	5%	14.5%	-9.5%	0.90%^2	0.1354%^2
Worst	5%	-30%	14.5%	-44.5%	19.80%^2	0.9901%^2
						Sum = 1.6250%^2

The Variance is the probability weighted average of square in the average of the difference of the returns versus the expected return. The point of using the %² rather than just a % was to show that the Variance is in the units of %-squared (as opposed to the Standard Deviation, which is in units of %). You didn't need to put a unit to get credit for this and I didn't expect anyone to use %².

The answer is 1.6225%²

Question 2c) (3 points) What is the Standard Deviation of your stock return?

The Standard Deviation is the square root of the Variance. The Variance was computed in the problem above.

Standard Deviation = Sqrt(Variance)

Standard Deviation = Sqrt(1.6225%²)

Standard Deviation = 12.7377%

The answer is 12.7475%

Question 3) (31 points in total) You manage Za House, a trendy new pizza place (Za as in pizZa). Your restaurant delivers and it has a guarantee regarding how long it takes to deliver a pizza. Your policy is that you will deliver a pizza in 30 minutes or less or it is free.

If you deliver the Pizza in 30 minutes or less then you make 2 dollars

If it takes more than 30 minutes you lose 10 dollars

You deliver exactly 10 pizzas every day. These are all to the same place, to the MBA students at NYU, who like their pizza. Each pizza is delivered as a separate trip, so you make exactly 10 trips per day. Assume each pizza delivery is independent of the other ones (which normally would not be the case, unless you had 10 pizza delivery people, but we have simplified things for this question).

You have carefully mapped out the shorted (in distance) route, but you are concerned because you are still giving out free pizzas from time to time. You are considering either stopping the free deliveries or switching to a different route. You are not sure what to do, as any other route you take would be longer in distance. You carefully examine two possible routes. Route A is the one you are taking now. Route B is another way that is longer in distance.

Please assume for this problem that Route A is normally distributed with a mean of 26 minutes and a standard deviation of 3 minutes. (Ignore the fact that delivery times can't be negative, so the Normal Distribution is not a perfect fit. Please assume it is Normally Distributed for the purposes of this problem.)

Please assume for this problem that Route B is normally distributed with a mean of 27 minutes and a standard deviation of 2 minutes. (Ignore the fact that delivery times can't be negative, so the Normal Distribution is not a perfect fit. Please assume it is Normally Distributed for the purposes of this problem.)

Route	μ (the mean)	σ (the Standard Deviation)
A	26 minutes	3 minutes
B	27 minutes	2 minutes

Question 3a) (4 points) Why might the Standard Deviation of Route A be higher than the Standard Deviation of Route B?

Basically any reasonable answer will be accepted for this. For example, there may be traffic on Route A that sometimes causes delays.

Question 3b) (4 points) What is the probability that a single delivery of pizza using Route A will result in a free delivery? (Or what is the probability that the time will take more than 30 minutes)?

Step one: Do Z transformation to figure out standard deviations from the mean.

(30 - 26) / 3 = 1.333

What is the probability that a delivery time is greater than 1.333 standard deviations from the mean?

Look up on the chart of standard normal distributions (which assume a mean of zero and a standard deviation of 1).

You see 40.8789%

This means that from 0 to 1.333 standard deviations there is a 40.8789% chance.

We know that from -infinity to 0 there is a 50% chance.

So total to the left of 1.333 standard deviations is 90.8789%.

We want greater than 1.333 standard deviations so its 100% - 90.8789% = 9.1211%

The answer is 9.1211%.

Question 3c) (4 points) What is the probability that a single delivery of pizza using Route B will result in a free delivery? (Or what is the probability that the time will take more than 30 minutes)?

Step one: Do Z transformation to figure out standard deviations from the mean.

(30 - 27) / 2 = 1.5

What is the probability that a delivery time is greater than 1.5 standard deviations from the mean?

Look up on the chart of standard normal distributions (which assume a mean of zero and a standard deviation of 1).

You see 43.3193%

This means that from 0 to 1.5 standard deviations there is a 43.3193% chance.

We know that from -infinity to 0 there is a 50% chance.

So total to the left of 1.5 standard deviations is 93.3193%

We want greater than 1.5 standard deviations so its 100% - 93.3193% = 6.6807%

The answer is 6.6807%

Question 3d) (3 points) What is the expected value of the profit of Route A for a given day (remember there are 10 deliveries in a day)?

Profit	Probability	Profit * Probability
2	90.88%	1.817577
-10	9.12%	-0.91211
		Sum = 0.905465
		Times 10 Deliveries/Day
		$ 9.05

The Expected Value of the profit of a single delivery is $0.905.

There are 10 deliveries in a day so the total Expected Profit is: $0.905 x 10.

The answer is $9.05

Question 3e) (3 points) What is the expected value of the profit of Route B for a given day (remember there are 10 deliveries in a day)?

Profit	Probability	Profit * Probability
2	93.32%	1.866386
-10	6.68%	-0.66807
		Sum = 1.198313
		Times 10 Deliveries/Day
		$ 11.98

The Expected Value of the profit of a single delivery is $1.198313.

There are 10 deliveries in a day so the total Expected Profit is: 1.198313 x 10.

The answer is $11.98

Question 3f) (3 points) Which route would you choose to use and why? Would you choose Route A or Route B?

The answer is: Choose Route B as it has the higher Expected Value of the profit.

Question 3g) (4 points) Looking back at Route A, please write down the distribution of possible outcomes and total payoffs for the day (without regard to probabilities).

Hint: There are 11 possible outcomes for the day. With 0 trials over 30 minutes you make 10 * 2 = $20.

Total Trials Over 30 Minutes	Total Trials Under 30 Minutes	Per Trial Payoff if Under 30 minutes	Total Under 30 minute trials * per trial payoff	Per Trial Payoff if Over 30 minutes	Total Over 30 minute trials * per trial payoff	Total Payoff (summing Column 4 + 6)
Col 1	Col 2	Col 3	Col 4	Col 5	Col 6	Col 7
0	10	2	20	-10	0	20
1	9	2	18	-10	-10	8
2	8	2	16	-10	-20	-4
3	7	2	14	-10	-30	-16
4	6	2	12	-10	-40	-28
5	5	2	10	-10	-50	-40
6	4	2	8	-10	-60	-52
7	3	2	6	-10	-70	-64
8	2	2	4	-10	-80	-76
9	1	2	2	-10	-90	-88
10	0	2	0	-10	-100	-100

Question 3h) (6 points) What are the odds of having a profitable day using Route A? (Or what are the odds of having a day with a profit of more than 0 using Route A?)

Hint: Each trial is assumed to have a normal distribution as given in the intro to this problem. Each trial is independent from every other trial and there are only two possible outcomes for each trial, either you are over or under 30 minutes. If you didn't figure out the probability of the time being more than 30 minutes for Route A, then just use any number for the probability between 5% and 25% for the odds of going over 30 minutes on one trial and if you get a logically consistent answer you'll get full credit.

This should be modeled using the Binomial Distribution.

Define "Success" to be delivering the pizza in less than 30 minutes (you could have defined this as a failure instead). The probability of Failure is called "q" and was the answer to part b of this question. It is 9.12%. The probability of success, "p", is 90.88%, which is 100% - 9.12%

There are only two outcomes that get a profit for the day. They are getting 0 times over 30 minutes and getting exactly one time over 30 minutes. See part g for the possible outcomes and payoffs. So the question can be rephrased as what are the odds of having 9 or more successes?

Successes	Trials	p	Probability
0	10	90.88%	0.00%
1	10	90.88%	0.00%
2	10	90.88%	0.00%
3	10	90.88%	0.00%
4	10	90.88%	0.01%
5	10	90.88%	0.10%
6	10	90.88%	0.82%
7	10	90.88%	4.66%
8	10	90.88%	17.42%
9	10	90.88%	38.57%
10	10	90.88%	38.43%

The probabilities from above came from using the formula. To answer this question, you only need to use the formula for 9 successes and 10 successes.

Number of Trials = 10

Number of successes = 9

(10 choose 9) x 0.9088⁹ x 0.0912¹ = 10 x 0.4228 x 0.0912 = 38.57%

Number of Trials = 10

Number of successes = 10

(10 choose 10) x 0.9088¹⁰ x 0.0912⁰ = 1 x 0.3843 x 1 = 38.43%

P(X>=9) = P( x = 9 ) + P( x = 10 )

P(X>=9) = 38.57% + 38.43% = 77%

The answer is 77%

Question 4) (6 points) A random variable X is Normally distributed with a mean of 10. The probability that x>12 is 15.87%. What is the odds that x is between 9 and 11?

X is N(10, ???): X is normally distributed with a mean of 10 and a standard deviation you need to figure out.

p( 9 < X <= 11)?

Step one: Do reverse Z transformation to figure out that for a standard normal distribution, the area to the right of 1standard deviation is 15.87%.

i.e. P(z>1) = 15.87%

Step Two: If 12 is 1 Standard Deviation away from 10, then the magnitude of the Standard Deviation must be 12 - 10 = 2.

Step Three: Do Z transformation to figure out standard deviations from the mean for left side.

(9 - 10) / 2 = -0.5

What is the probability that a value is less than -0.5 standard deviations from the mean?

From chart it is: 30.85%

Step Four: Do Z transformation to figure out standard deviations from the mean for right side.

(11 - 10) / 2 = 0.5

What is the probability that a value is less than 0.5 standard deviations from the mean?

From chart it is: 69.15%

Step Five: Subtract the two numbers to get the answer:

69.146% - 30.854% = 38.29%

(or you could have looked up the odds that z is between 0 and .5 and got 19.146%, then looked up the odds that z is between -.5 and 0 and got 19.146% and added them up to get the same number, 38.29%)

The answer is 38.29%.

Question 5) (10 points in total) The average American adult man has a height that is normally distributed with a mean of 67 inches and a Standard Deviation of 6 inches.

The average American adult woman has a height that is normally distributed with a mean of 63 inches and a Standard Deviation of 4 inches.

Question 5a) (5 points) How big does a bed maker need to make a bed (in inches) to accommodate 90% of all Adult Male Americans?

The answer is Normsinv(.90) = 1.2910 Standard Deviations to the right of the mean. That is 1.2910 * 6 inches = 7.7461 inches to the right of the mean. That is 67 inches (the mean) + 7.7461 = 74.7461 inches.

The answer is 74.7461 inches.

Question 5b) (5 points) What percentage of American Females fits on a bed the size of the answer to part "a"?

The answer to part "a" is 74.7461 inches. This is 11.746 inches or 2.94 Standard Deviations away from the mean.

To go from inches to Standard Deviations, do a Z transformation:

Z = (74.7461 - 63 ) / 4 = 2.94

Look up 2.94 on the chart and see that 99.83% of the area is to the left of it (or 49.83% of the area is between 0 and 2.94 Standard Deviations).

The answer is 99.83% of all American Females fit on a bed 74.7461 inches long.

Question 6) (20 points in total) A particular car has 6 sub systems. Each system must be working for the car to run (for the car to be operable). The probability of a system failing is 10%. The probability of a system working (not failing) is 90%. Each system is completely independent of the other systems.

Question 6a) (4 points) What is the expected number of failures in the car? Meaning for all 6 systems, how many are expected to fail on average?

This is a Binomial Distribution.

Define n as the number of trials = 6

Define "Success" as system failure.

Define p as the probability of "Success" = 10%

The formula for Expected Value of Binomial Distribution is np.

6 * 10% = 0.6

The answer is .6 expected failures.

Question 6b) (4 points) What are the odds that the car will run? (Or what are the odds of zero subsystem failures of the 6 subsystems)?

This is a Binomial Distribution.

Define n as the number of trials = 6

Define "Success" as system failure.

Define p as the probability of "Success" = 10%

We want to have 0 failures. The odds of this occurring are computed by Binomial Distribution.

Number of Trials = 6

Number of Successes = 0

(6 choose 0) x 0.10⁰ x 0.90⁶ = 1 x 1 x 1 = 53.14%

# Successes	Total Trials	Prob of Success	Prob of x Successes
0	6	10%	53.14%
1	6	10%	35.43%
2	6	10%	9.84%
3	6	10%	1.46%
4	6	10%	0.12%
5	6	10%	0.01%
6	6	10%	0.00%

The answer is 53.14% expected failures.

Question 6c) (4 points) What would the per sub-system working rate need to be in order to have the total success rate for the car to be operational be 99%? Said a different way, what would the odds of working need to be for each sub-system to expect that the odds for all 6 system working (no failures of any subsystems) at the same time to be 99%?

Hint: This relates to question 6b. The answer involves doing a n-square-root. For those without a calculator, I have included the table below. You would read it by taking the first column to the power of the top row and the intersecting cell is the answer. For example, the cube root of .01 is 0.215443 (or .01^1/3 = .215443).

	1/2	1/3	1/4	1/5	1/6
0.01	0.1	0.215443	0.316228	0.398107	0.464159
0.05	0.223607	0.368403	0.472871	0.54928	0.606962
0.1	0.316228	0.464159	0.562341	0.630957	0.681292
0.9	0.948683	0.965489	0.974004	0.979148	0.982593
0.95	0.974679	0.983048	0.987259	0.989794	0.991488
0.99	0.994987	0.996655	0.997491	0.997992	0.998326

The answer is .99^1/6 or 99.8326%

Question 6d) (3 points) Now assume that failures are not Binomially Distributed. Assume the following distributions for numbers of defect in total.

# defects	Prob of this many defects
0	40%
1	40%
2	10%
3	10%
4	0%
5	0%
6	0%

What is the expected value of the number of failures?

# defects	Prob of this many defects	# * Prob
0	40%	0
1	40%	0.4
2	10%	0.2
3	10%	0.3
4	0%	0
5	0%	0
6	0%	0
		Sum = 0.9

The answer is 0.9

Question 6e) (3 points) Assume the same distribution as part d. What is the Standard Deviation of the distribution?

# defects	Prob of this many defects	E(x)	x - E(x)	(x - E(x))^2	p(x) * (x - E(x))^2
0	40%	0.90	(0.90)	0.81	0.324
1	40%	0.90	0.10	0.01	0.004
2	10%	0.90	1.10	1.21	0.121
3	10%	0.90	2.10	4.41	0.441
4	0%	0.90	3.10	9.61	0
5	0%	0.90	4.10	16.81	0
6	0%	0.90	5.10	26.01	0
					0.890

The Variance is 0.89.

The Standard Deviation is the square root of the Variance = Sqrt(0.89).

The Standard Deviation is 0.9434

The Answer is 0.9434

Question 6f) (2 points) Is the odds of 2 failures (as per the chart of probabilities in section d) Statistically Independent with the odds of 3 failures?

e.g. let A = 2 failures

Let B = 3 failures

Are A and B Statistically Independent?

The answer is No. They are Mutually Exclusive. If you got 2 failures you didn't get 3 failure and vice-versa.