Final Exam Review

Use the following problems to prepare for the exam. There will be in-class review on Thursday, April 23. Your recitation this week will also be exam review.

Allowed Materials

You will be allowed to use a scientific calculator (not a graphing calculator, not a calculator app on your phone). You may not share a calculator with another student; you must use your own calculator.

You may bring a standard 3 in x 5 in index card with prepared notes. You may use both sides of the notecard. You must put your full name in the top right corner of the card, and turn it in along with your exam.

Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, B = \{2, 4, 9, 10, 12, 14, 19\}, and C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of \Omega = \{1, 2, 3, \dotsc, 20\}.

Find A - (B \cap C).

\{1, 2, 3, 4, 5, 6, 7, 8\}

Find |A|, |B|, |C|, |A\cup B|, |A \cap B|, |B\cap C|, |A\cap C|, and |A\cup B\cup C|. Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets?

|A| = 10, |B| = 7, |C| = 7, |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, |A\cap C| = 2, |A\cup B\cup C| = 17. In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.

Find A^c and (A\cup B)^c.

A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}.

Suppose we have a 6-sided die that's weighted to roll a 6 half of the time. We roll the die two times. List the set of all possible results. [Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result (4, 2)---rolling a 4 and then a 2.]

\Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} Note that \Omega simply lists outcomes with no reference to the probabilities. So the answer here is the same as in .

Suppose we flip a coin two times. List the set of all possible results. What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be?

For two flips: \Omega = \{ HH, HT, TH, TT \}.

For three flips: \Omega = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}.

For four flips: \Omega = \{ \amp HHHH, HHHT, HHTH, HTHH, \amp THHH, HHTT, HTHT, HTTH, \amp THHT, THTH, TTHH, HTTT, \amp THTT, TTHT, TTTH, TTTT \}.

Each additional flip doubles the number of outcomes. So, with ten flips, we'll have |\Omega| = 2^{10} = 1024.

If we roll a 6-sided die ten times, how many possible results will there be?

Each additional roll will multiply the number of outcomes by 6. So, with 10 rolls, we'll have |\Omega| = 6^{10}.

Consider the sample space \Omega = \{1, 2, 3, 4, 5, 6, 7, 8\} with probability distribution below. Calculate the probabilities of A = \{1, 3, 7, 8\}, B = \{2, 3, 6, 7\}, A\cup B, and A \cap B.

x \Pr(x) 1 0.1 2 0.05 3 0.2 4 0.15 5 0.15 6 0.1 7 0.05 8 0.1 9 0.1

\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, the probabilities scale by the same amount as the face values. For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than a result of 1; and so on. Write a probability distribution table for this die.

Probability Distribution for a Linearly Scaled Die x \Pr(x) 1 1/21 2 2/21 3 3/21 4 4/21 5 5/21 6 6/21

Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but the die is not fair. Instead, each even value has an equal probability, each odd value has an equal probability, and the even values are each twice as likely as the odd values to appear on a roll. Write a probability distribution table for this die.

Probability Distribution for an Even-biased Die x \Pr(x) 1 1/9 2 2/9 3 1/9 4 2/9 5 1/9 6 2/9

A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period. For each value of n = 1, 2, 3, \dotsc, find the probability of the toxin molecule leaving the cell during the nth minute. What is the probability of the molecule leaving the cell during the first 3 minutes?

For short, write \Pr(n) to mean the probability of the toxin molecule leaving during the nth minute. Then \Pr(n) = (0.7)^{n - 1} (0.3).

The probability of leaving during the first 3 minutes is \Pr(1) + \Pr(2) + \Pr(3) = 0.657.

In each of the following scenarios with given events A and B, alculate \Pr(A), \Pr(B), \Pr(A\cap B), \Pr(A \mid B), and \Pr(B \mid A).

An experiment consists of rolling a fair die two times. Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first.

A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3}

An experiment consists of flipping a fair coin three times. Let A be the event that the first and second flips match. Let B be the event that there are at least two heads.

A \amp = \{ HHH, HHT, TTH, TTT \} B \amp = \{ HHH, HHT, HTH, THH \} A\cap B \amp = \{HHH, HHT\} So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2}

A diagnostic test is developed to detect a disease present in 3.2% of the population. For a patient who has the disease, the test will accurately give a positive result 65% of the time. When the patient does not have the disease, the test will accurately give a negative result 99.9% of the time.

For a patient who receives a positive test, what is the probability they have the disease?

Let P be the event of testing positive and D the event of having the disease. Then the prevalence \Pr(D) is given as 3.2%, or 0.032. The sensitivity is \Pr(P\mid D) = 0.65, and the specificity is \Pr(P^c\mid D^c) = 0.999. So, according to Bayes' Theorem: \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} \amp \approx 0.96

For a patient who receives a negative test, what is the probability they do not have the disease?

\Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} \amp \approx 0.99

An experiment consists of rolling a fair die two times. Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first. Are A and B independent?

A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), so A and B are not independent.

An experiment consists of flipping a fair coin three times. Let A be the event that the first and second flips match. Let B be the event that there are at least two heads. Are A and B independent?

A \amp = \{ HHH, HHT, TTH, TTT \} B \amp = \{ HHH, HHT, HTH, THH \} A\cap B \amp = \{HHH, HHT\} So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), so A and B are independent.

Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}. Create a probability distribution for \Omega so that A, B are independent.

Example Distribution x \Pr(x) 1 0.1 2 0.2 3 0.2 4 0.1 5 0.1 6 0.3

Now \Pr(A) = 0.5, \Pr(B) = 0.4, and \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B), so A and B are independent.

An experiment consists of flipping a biased coin 20 times. If the coin comes up heads with probability p = 0.3, find the probability of seeing 5 heads. Find the probability of seeing up to (and including) 3 heads.

Let S be the number of heads. Then S \sim \Bin(20, 0.3), so: \Pr(S = 5) \amp = {20 \choose 5} (0.3)^5 (0.7)^{20 - 5} \amp = \frac{20!}{(5!)(15!)} (0.3)^5 (0.7)^{15} \amp = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} (0.3)^5 (0.7)^{15} \amp = (19 \times 3 \times 17 \times 16) (0.3)^5 (0.7)^{15} \amp \approx 0.179

An experiment consists of flipping a coin repeatedly until we first see heads.

If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip?

Let T be the number of flips until we see heads. Then T is geometric with parameter p = 0.4, so: \Pr(T = k) \amp = (1-0.4)^{k-1}(0.4) = 0.6^{k-1} \cdot 0.4

Which flip has the highest chance of being the first flip to come up heads?

A particular store has an average of 20 customers each hour. During a 4-hour afternoon shift, what is the probability of serving 80 customers.

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = k(x - \sqrt{x}) for some constant k.

What is the value of k?

k = \frac{6}{17}.

Find \Pr(2 \leq X \leq 3).

\Pr(2 \leq X \leq 3) \approx 0.325

A continuous random variable X taking values in [1, 2] has p.d.f. \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. Find the c.d.f. F(x). Use your c.d.f. to find \Pr\left(1 \leq X \leq \frac{3}{2}\right).

\frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}. \Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.

A continuous random variable X taking values in [2, 3] has c.d.f. F(x) = \frac{x^3}{3} - x^2 + 4. Find the p.d.f. f(x).

f(x) = x^2 - 2x.

Consider X, Y with the joint distribution table below. Are X, Y independent?

Joint distribution for <m>X, Y</m> X = 0 X = 1 Y = 0 0.2 0.3 Y = 1 0.4 0.1

No. For example, \Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).

Suppose X, Y have the distributions: \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 \Pr(X = 2) \amp = 0.5 Assuming X, Y are independent, write a joint distribution table.

Joint Distribution X = 0 X = 1 X = 2 Y = 0 0.04 0.16 0.2 Y = 1 0.06 0.24 0.3

Consider a random variable X with probability distribution below. Find \E(X).

x \Pr(X = x) 1 0.1 2 0.05 3 0.2 4 0.15 5 0.15 6 0.1 7 0.05 8 0.1 9 0.1

4.8.

Suppose we flip a coin n = 100 times, and let N count the number of heads.

If the coin comes up heads on a flip with probability p = 0.4, what is \E(N)?

40.

What if n = 80 and p = 0.6?

48.

What if n = 200 and p = 0.5?

100.

If \E(X) = 3, \E(Y) = -2, and \E(Z) = 1, what is \E(4X + 5Y - Z + 3)?

4.

A continuous random variable X taking values in [0, 1] has p.d.f. f(x) = 2x. What is \E(X)?

2/3.

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = \frac{4}{3x^2}. What is \E(X)?

\frac{4}{3}\ln(4) \approx 1.85.

A continuous random variable X taking values in [1, 2] has p.d.f. \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. Find \E(X).

\frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513.

Consider a random variable X with probability distribution below. Find \Var(X).

x \Pr(X = x) 1 0.1 2 0.05 3 0.2 4 0.15 5 0.15 6 0.1 7 0.05 8 0.1 9 0.1

Suppose we flip a coin n = 100 times, and let N count the number of heads. If the coin comes up heads on a flip with probability p = 0.4, what is \Var(N)? What if n = 80 and p = 0.6? What if n = 200 and p = 0.5?

If \E(X) = 3, \Var(X) = 2, what is \E(X^2)?

A continuous random variable X taking values in [0, 1] has p.d.f. f(x) = 2x. What is \Var(X)?

A continuous random variable X taking values in [1, 4] has p.d.f. f(x) = \frac{4}{3x^2}. What is \Var(X)?

A continuous random variable X taking values in [1, 2] has p.d.f. \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. Find \Var(X).

Suppose a coin has an unknown probability of coming up heads. We perform the experiment in n independent trials, during which it takes k_1, k_2, \dotsc, k_n flips to see our first heads in each trial. Find a "common sense" MLE formula for the geometric distribution.

A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate?

A radioactive material emits particles at an unknown probabilistic rate \lambda particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function \mathcal{L}(\lambda) based on this data. What is the maximum likelihood estimation for \lambda?

Suppose a parameter \theta takes values in [0, 1] with likelihood function \mathcal{L}(\theta) = \sqrt{\theta} - \theta^2. Find the maximum likelihood estimation of \theta.

Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times. Let S be the number of heads. Estimate the probability that 34 \leq S \leq 44.

\Pr(34 \leq S \leq 44) \approx 0.8212 - 0.0918 = 0.7294.

Suppose a fair die is rolled 100 times, and let m be the average value of the rolls. Estimate the probability that 3.45 \leq m \leq 3.55.

\Pr(3.45 \leq m \leq 3.55) \approx 0.6255 - 0.3745 = 0.2510.

The heights of men in the US have a mean of 69 in and a variance of about 9 in^2, and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in^2.

Suppose the heights of 30 men are sampled, and a sample mean m is taken. Find the expected value and variance of m.

\E(m) = 69 and \Var(m) = 0.3.

Estimate the probability that m \geq 69.5.

\Pr(m \geq 69.5) \approx 1 - 0.8186 = 0.1814.

If the sampled group was women, estimate the probability that m \geq 64.

\E(m) = 63.5 and \Var(m) \approx 0.208. \Pr(m \geq 64) \approx 1 - 0.8643 = 0.1357.

Suppose we flip a coin 100 times and count 60 heads. Let p be the (unknown) probability that the coin comes up heads on a flip. Give an approximate 95% confidence interval for the value of p.

[0.504, 0.696].

Suppose in a sample of 100 people, 12 are left-handed.

Give a 95% confidence interval for the proportion p of left-handed people.

[0.056, 0.184].

Give a 90% confidence interval.

[0.066, 0.174].

The weights of five mice are measured and recorded below. Give a 95% confidence interval for the sample mean weight of mice. (Pretend 5 measurements is large enough for the CLT to apply.)

mouse i weight W_i (g) 1 26 2 32 3 33 4 20 5 29

[23.39, 32.61].

The heights of five plants are measured and recorded below. Give a 95% confidence interval around the sample mean for the heights of the plants. (Pretend 5 measurements is large enough for the CLT to apply.)

plant i height H_i (in) 1 15 2 14 3 18 4 21 5 17

[14.61, 19.39].

In each of the following scenarios, determine whether we should use a 1-tailed test or a 2-tailed test.

We have a coin which we've flipped many times, seeing an above-average number of heads. We suspect the coin comes up heads more often than a fair coin would.

We find a coin on the street and wonder whether or not it's a fair coin.

We suspect there will be a difference in average weight of mice caught during the summer versus during the winter.

In a medical study, a group of patients are gathered and the proportion experiencing particular symptoms is measured. A new drug intended to eliminate these symptoms is administered, after which the proportion experiencing symptoms is measured again.

Suppose we find a coin and wonder whether it's fair. As a first test, we decide to flip the coin 200 times and count the number of heads. If we see 112 heads, should we accept or reject the null hypothesis of a fair coin at a significance level of 0.05?

Suppose we have a coin which we suspect comes up heads more often than a fair coin would. As a first test, we decide to flip the coin 200 times and count the number of heads. If we see 112 heads, should we accept or reject the null hypothesis of a fair coin at a significance level of 0.05?

Suppose we find a six-sided die and wonder whether it's fair. As a first test, we decide to roll the die 100 times and count the number of times it comes up 1. If we roll 22 1's, should we accept or reject the null hypothesis of a fair die at a significance level of 0.05?

Suppose a particular plant when grown outdoors has an average height of 39 in with a variance of 20 in^2. We suspect that growing this plant in a greenhouse will increase its height. A sample of 50 plants grown in a greenhouse has an average height of 40 in. Is this significant enough data to reject the null hypothesis of equal means at the p = 0.05 significance level?

A farmer is testing an experimental new plant fertilizer that is supposed to increase the weight of a particular apple variety. A control sample of 25 apples grown using the usual fertilizer have a mean weight of 75 grams and a sample variance of 90 grams^2 (for an individual apple). An experimental sample of 25 apples grown using the new fertilizer have a mean weight of 79 grams and a sample variance of 90 grams^2.

We have an established factory which produces coins that are close to fair. We're opening up a second factory, and we'd like to ensure the machines are calibrated to produce coins which behave similarly to the ones produced in the established factory. We pick one sample coin from each factory, and flip each sample coin 100 times. The coin from the established factory flips 52 heads in 100 flips. The coin from the new factory flips 62 heads in 100 flips. Is this strong enough evidence to reject the null hypothesis that the two factories produce similar coins at a 0.05 significance level?

Suppose we find a coin and wonder whether it's fair. As a first test, we decide to flip the coin 200 times and count the number of heads, S. What values of S would be extreme enough to reject the null hypothesis of a fair coin? If the coin actually has a 0.6 probability of coming up heads, what is the power of this test?

Suppose we have a coin which we suspect comes up heads more often than a fair coin would. As a first test, we decide to flip the coin 200 times and count the number of heads, S. What values of S would be extreme enough to reject the null hypothesis of a fair coin? If the coin actually has a 0.6 probability of coming up heads, what is the power of this test?

Suppose we find a six-sided die and wonder whether it's fair. As a first test, we decide to roll the die 100 times and count the number of times it comes up 1. The expected number of 1's is 50/3, with a variance of 125/9.

Using a normal approximation, what is the smallest number of 1's greater than 50/3 that would be extreme enough to reject the null hypothesis of a fair die?

Using a normal approximation, what is the greatest number of 1's less than 50/3 that would be extreme enough to reject the null hypothesis of a fair die?

Suppose that this die is weighted so that it rolls a 1 with probability 0.2. What would be the power of our test?

Suppose we roll the die 100 times and see 23 1's. Use the maximum likelihood value for the probability of rolling a 1 to calculate the power of the test.

Suppose a particular plant when grown outdoors has an average height of 39 in with a variance of 20 in^2. We suspect that growing this plant in a greenhouse will increase its height. We take the average height of a sample of 50 plants grown in a greenhouse. What is the minimum average height of this sample that would be extreme enough to reject the null hypothesis of equal means at the p = 0.05 significance level? If the plants, when grown in a greenhouse, would truly have an average height of 41 in, what is the power of our test?

A store owner wants to determine how much shelf space to allocate to each of the drinks that they sell. They survey their customers about their favorite drinks. Is the data below consistent with the null hypothesis that each type of drink will be equally preferred?

drink type water soda tea coffee energy drinks favorite 28 17 15 26 14

A particular drug is administered in 100 independent trials. In each trial, the drug is administered to four people, and we count how many respond to the drug. The table below shows how many trials have each different count of people who respond to the drug.

# who respond to drug 0 1 2 3 4 # of trials 3 11 31 34 21

Is the data consistent with a binomial distribution with parameter \theta = 0.7?

What is the total number of people who have been administered the drug? What is the total number who have responded to it? What is the maximum likelihood estimation \widehat{\theta} for the probability that a person will respond to the drug?

Is the data consistent with a binomial distribution with the MLE value of \widehat{\theta}?

Calculate the covariance of X and Y:

X = 1 X = 2 Y = 0 0.12 0.24 Y = 1 0.3 34

Calculate the covariance of X and Y:

X = 0 X = 1 X = 2 Y = 0 0.08 0.16 0.1 Y = 1 0.14 0.2 0.32

Suppose we roll a fair, 4-sided die two times. Let X be the sum of the rolls, and let Y be the product of the rolls. Find the covariance of X and Y.

[Note: Somewhat tedious.]

Calculate the correlation of X and Y:

X = 1 X = 2 Y = 0 0.12 0.24 Y = 1 0.3 34

Calculate the correlation of X and Y:

X = 0 X = 1 X = 2 Y = 0 0.08 0.16 0.1 Y = 1 0.14 0.2 0.32

Suppose we roll a fair, 4-sided die two times. Let X be the sum of the rolls, and let Y be the product of the rolls. Find the correlation of X and Y.

[Note: Somewhat tedious.]

A survey of local companies collects information about marketing budgets X and revenue Y (each measures in thousands of dollars), shown below. A linear regression gives the best linear fit as Y = 18.28 X + 29.69. What is the coefficient of determination r^2?

x_i y_i (y_i - \text{avg})^2 \text{pred } y_i \text{res}^2 200 4300 2073600 3686 377377 420 7700 3841600 7707 53 270 4500 1537600 4965 216495 380 7000 1587600 6976 572 300 5200 291600 5514 98401 sum: 28700 9332000 28848 692898

A sample of 100 measurements are taken and a best fit line is calculated, resulting in the data below (the final line shows the sums for each column). Find the coefficient of determination.

x_i y_i \text{pred } y_i (y - \text{avg})^2 \text{res}^2 38.00 121.00 93.17 11.83 774.33 87.00 241.00 238.95 13586.23 4.22 30.00 61.00 69.37 4024.63 70.12 35.00 85.00 84.25 1555.51 0.57 ⋮ ⋮ ⋮ ⋮ ⋮ 33.00 108.00 78.30 270.27 882.19 26.00 32.00 57.47 8545.15 648.91 sum: 12444.00 12444.00 515206.64 29789.80