# Practice Problems (Final Exam) – ISA 225

Consider the following Tinder stats for Crammer Nation University's Alpha Berry Pi fraternity chapter:

Question #1: Determine the estimated values for the time series with a 3-point moving average.

Answer:

For each of the estimated values...

...all we need to do is take the average of the values in the (1) current period and (2) previous two periods to calculate our 3-point moving average! (You may see 3-point moving average represented as "MA(3)".)

Let's start with the estimated value for 2021. To determine this, we'll take the average of the Tinder Match data for 2019, 2020, and 2021...

...resulting in 1276.67!

To determine estimated value in 2022, we'll take the average of the Tinder Match data for 2020, 2021, and 2022...

...resulting in 1352.67!

Lastly, to determine the estimated value in 2023, we'll take the average of the Tinder Match data for 2021, 2022, and 2023...

...resulting in 1378.33!

This results in the following final answer!

Answer:

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Consider the following regression output for Crammer Nation University's student-run merch company, Steez Inc. and their sales the past 2 years:

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Consider the following: Rolling a six-sided die with two possible events...

• Event A: Roll a number greater than 4
• Event B: Roll an odd number

Question #1: Solve for P(A).

Answer: 1/3 or 0.33

To determine the probability of Event A occurring (a.k.a. "P(A)"), we must first understand that the sample space for all outcomes of a six-sided die roll look like so:

S = {1, 2, 3, 4, 5, 6}

In other words, the outcome of rolling a six-sided die can be 1, 2, 3, 4, 5, or 6.

Now, the sample space for Event A...

• Event A: Roll a number greater than 4
• Event B: Roll an odd number

...looks like so...

SA = {5, 6}

...because we're only looking for rolls greater than 4.

From here, we can solve for P(A) with the following formula...

...which the "# of outcomes in A" equals 2...

SA = {5, 6}

...and the "# of all potential outcomes" equals 6...

S = {1, 2, 3, 4, 5, 6}

...resulting in a P(A) value of 1/3, or 0.33!

This serves as our final answer!

Answer: 1/3 or 0.33

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Consider the following table regarding Delta Blueberry Pi's rush bid breakdown from this year:

Question #1: Identify the marginal probabilities in the table above.

Answer:

The marginal probabilities are the ones on the outskirts...

...as they represent the probability that one event occurs... a.k.a. row / column totals. (Because each row / column represents one given event!)

The rows represent Event #1 (whether or not the student wore a hat)...

...and the columns represent Event #2 (whether or not the student received a bid).

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Consider the following rush stats for Sigma Apple Pi: 75% of students who wore a hat during rush received a bid to Sigma Apple Pi, whereas only 20% of students who didn't wear a hat received a bid. According to brothers, 40% of students wore a hat during rush.

Question #1: Construct a probability tree (Setup: Hat ➡ Bid).

Answer:

To start, we can set up a blank probability tree like so...

...and assign whether or not they wore a hat to the first "round" of legs (since that's what the problem indicates)...

Question #1: Construct a probability tree (Setup: Hat ➡ Bid).

...and then assign whether or not they received a bid to the second "round" of legs (since that's what the problem indicates)...

Question #1: Construct a probability tree (Setup: Hat ➡ Bid).

From here, the prompt tells us that 75% of students who wore a hat received a bid...

Consider the following rush stats for Sigma Apple Pi: 75% of students who wore a hat during rush received a bid to Sigma Apple Pi, whereas only 20% of students who didn't wear a hat received a bid. According to brothers, 40% of students wore a hat during rush.

...meaning that the other 25% did not receive a bid.

We can fill in those values like so:

Then, the prompt tells us that out of the students who didn't wear a hat, 20% received a bid...

Consider the following rush stats for Sigma Apple Pi: 75% of students who wore a hat during rush received a bid to Sigma Apple Pi, whereas only 20% of students who didn't wear a hat received a bid. According to brothers, 40% of students wore a hat during rush.

...meaning that the other 80% did receive a bid.

We can fill in those values like so:

Lastly, the prompt tells us that 40% of students wore a hat during rush...

Consider the following rush stats for Sigma Apple Pi: 75% of students who wore a hat during rush received a bid to Sigma Apple Pi, whereas only 20% of students who didn't wear a hat received a bid. According to brothers, 40% of students wore a hat during rush.

...meaning the other 60% didn't wear a hat.

We can fill in those values like so...

...resulting in our finalized probability tree!

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Consider the following: Crammer Nation University Merch Store has 10,000 transactions. Out of all transactions, 5,000 of them include a hoodie. Of these hoodie transactions, 1,000 of them include a hat.

Question #1: Find the support for "(Hoodie) ➡ (Hat)".

Answer: 0.10

In market basket analysis terms, "support" essentially means "out of all transactions, which contain both a Hoodie & a Hat?".

We can set up the following formula to determine this:

Based on the prompt, we can see that there's 1,000 transactions with both a Hoodie and a Hat...

Consider the following: Crammer Nation University Merch Store has 10,000 transactions. Out of all transactions, 5,000 of them include a hoodie. Of these hoodie transactions, 1,000 of them include a hat.

...which we can plug in like so:

Then, the prompt tells us that there were 10,000 total transactions...

Consider the following: Crammer Nation University Merch Store has 10,000 transactions. Out of all transactions, 5,000 of them include a hoodie. Of these hoodie transactions, 1,000 of them include a hat.

...which we can plug in like so:

When we solve from there...

...we get a support value of 0.10! This serves as our final answer!

Answer: 0.10

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