Answer: Bar chart

"family_type" is a categorical variable, and we use bar charts to visualize the distribution of categorical variables.

• A scatter plot visualizes the relationship between two numerical variables, but we are only using one numerical variable here ("avg_housing_cost").
• A line plot is used to visualize the trend between two numerical variables, but we are only using one numerical variable.
• A histogram is used to visualize the distribution of numerical variables, but we want to see the distribution of the categorical variable "family_type".

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: The median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

When we grouped living_cost by "family_type", families is a DataFrame with one row per "family_type". Using the .median() aggregation method takes the median of all numerical columns per "family_type".

sorted_families is the families DataFrame, but sorted in ascending order based on the "avg_housing_cost" column. The first row of sorted_families is the "family_type" with the lowest median "avg_housing_cost", and the last row of sorted_families is the "family_type" with the highest median "avg_housing_cost".

In the last line of code, we’re getting the "avg_childcare_cost" column from the sorted_families DataFrame. We then use iloc to get the first entry in the "avg_childcare_cost" column. Since sorted_families is sorted in ascending order, this means that we’re getting the lowest median in the column. Therefore, result evaluates to the median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: The state with the most counties.

The living_cost DataFrame is being grouped by the "state" column, so there is now one row per "state". By using the .count() aggregation method, the columns in the DataFrame will contain the number of rows in living_count for each "state". All of the columns will also be the same after using .count(), so they will all contain the distribution of "state". Since living_cost has data on every county in the US, the grouped DataFrame represents the number of counties that each state has.

We then sort the DataFrame in descending order, so the state with the most counties is at the top of the DataFrame. The last line of the expression gets a column and uses .index to get the state corresponding to the first entry, which happens to be the state with the most counties and the value that gets assigned to another_result.

Since all the columns are the same, it doesn’t matter which column we pick to use in the .sort_values() method. In this case, we used the "median_income" column, but picking any other column will produce the same result.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Answer: living_cost.groupby("family_type").sum()

Since we can’t take the sum of columns with categorical data, all of the columns in living_cost that contain non-numerical data are dropped after we use the .sum() aggregation method. There are four columns in living_cost that have numerical data ("is_metro", "avg_housing_cost", "avg_childcare_cost", and "median_income"). Since Python can take the sum of these numerical columns, these four columns are kept. Therefore, the resulting DataFrame has exactly four columns.

Although "is_metro" contains Boolean values, Python can still calculate the sum of this column. The Boolean value True corresponds to 1 and False corresponds to 0.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: 31430, the number of rows in the living_cost DataFrame.

The first line of the expression creates a new column in living_cost, called "geo_family" that represents the concatenation of the values in "three_columns". When we group the DataFrame by "geo_family", we create a new DataFrame that contains a row for every unique value in "three_columns". "three_columns" has various combinations of "state", "country", and "family_type". Since it’s given in the DataFrame description that each of the 31430 rows of the DataFrame represents a different combination of "state", "country", and "family_type", this means that the grouped DataFrame has 31430 unique combinations as well. Therefore, when we use .shape[0] to get the number of rows in the grouped DataFrame in the last line of the expression, we get the same value as the number of rows in the living_cost DataFrame, 31430.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer:

1. living_cost[J & K & L]

The first step is to query the rows in the DataFrame that meet our specific criteria. In this case, we want the rows in the DataFrame where the county is "Benton County", the state is "IN", and the family has 1 adult and 2 children. J, K, and L specify these criteria. When used to query the living_cost DataFrame, we are able to obtain a DataFrame with only one row, corresponding this family type in this specific county.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer:

2. get("avg_childcare_cost")

Once we have a DataFrame that contains the row we need, we need to use it to get the average yearly childcare cost. To do that, we get out the "avg_childcare_cost" column and access the entry in row 0 with .iloc[0]. This works because after querying, there is only one row, and it corresponds exactly to the families with one adult and two children in Benton County, IN.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer:

1. living_cost[J]

Since we want to find how many states have a county named "Benton County", we first want to obtain all the rows of the DataFrame where the county is "Benton County". Variable J specifies this condition, so we use it to query and obtain a DataFrame with the rows in living_cost where the county is "Benton County."

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer:

2. shape[0]

Now that we have all the rows in the DataFrame where the county is "Benton County", let’s consider how many rows it has. Each row of this DataFrame represents a unique combination of "state" and "family_type" for counties called "Benton County". We know from the data description that each of the ten family structures is present in each county. This means that for each state with a county called "Benton County", our DataFrame has exactly ten rows. Therefore, the number of states with a county called "Benton County" is the total number of rows in our DataFrame divided by ten. Therefore, we should fill in blank (d) with .shape[0] to extract the number of rows from our DataFrame.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.

Answer: Options 1, 3, 4, 6

Option 1: In order to get the number of people within a family, we can look at the character at position 0 (for the number of adults) and the character at position 2 (for the number of children). Converting each character into an int and adding these ints yields the correct results.

Option 2: This is similar to Option 1, however, the key difference is that the separate strings are concatenated first, then converted into an integer afterwards. Remember that the plus sign between two strings concatenates the strings together, and does not add mathematically. For example, on a family type of "1a2c", "1" and "2" will be extracted and concatenated together as "12", then converted to the int 12. This is returned instead of the value 3 that we are looking for.

Option 3: This option is similar to Option 2, however, it includes an extra step after concatenation. int(x/10) gets the value in the tens place, taking advantage of the fact that the int() function always rounds down. At the same time, x % 10 gets the value in the ones place by calculating the remainder upon division by ten. Looking at the example of "1a2c", the first line will set x = 12 and then int(12/10) will yield 1 while 12 % 10 yields 2. Adding these together achieves the correct answer of 3.

Option 4: This option is similar to Option 1, but includes the initial step of removing "c" from the string and separating by "a". After this, x is a list of two elements, the first of which represents the number of adults in the family, and the second of which represents the number of children in the family. These are separately converted to ints then added up in the last line.

Option 5: This option iterates through the input string, where i represents each individual character in the string. For example, on an input of "1a2c", i is first set to 1, then a, then 2, then c. However, calculating the remainder when we divide by two (i % 2) only makes sense when i is a number, and results in an error when i is a string.

Option 6: This is a similar approach to Option 5, except this time, i represents each of the numbers 0, 1, 2, and 3, since len(fam) is always 4. For each such i, which we can think of as the position number, the code will check if the position number is even (i % 2 == 0). This is only true for position 0 and 2, which are the positions that contain the numbers of adults and children in the family. When this condition is met, we add the value at that position onto our running total, x, which at the end, equals the total number of adults and children in the family.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: ["state", "county"] or ["county", "state"]

We are told that all expensive counties appear in the expensive DataFrame, but some may appear multiple times, for several different "family_type" values. The question we want to answer, however, is about the proportion of expensive counties in each state that are part of a metropolitan area, which has nothing to do with "family_type". In other words, we don’t want or need multiple rows corresponding to the same US county.

To keep just one row for each US county, we can group by both "state" and "county" (in either order). Then the resulting DataFrame will have one row for each unique combination of "state" and "county", or one row for each US county. Notice that the .max() aggregation method keeps the last alphabetical value from the "is_metro" column in each US county. If there are multiple rows in expensive corresponding to the same US county, we are told that they will all have the same value in the "is_metro" column, so taking the maximum just takes any one of these values, which are all the same. We could have just as easily taken the minimum.

Notice the presence of .reset_index() in the provided code. That is a clue that we may need to group by multiple columns in this problem!

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 14%.

Answer: "state"

Now that we have one row for each US county that is considered expensive, we want to proceed by calculating the proportion of expensive counties within each state that are in a metropolitan area. Our goal is to organize the counties by state and create a DataFrame indexed only by "state" so we want to group by "state" to achieve this.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer: mean()

Recall that the "is_metro" column consists of Boolean values, where True equals 1 and False equals 0. Notice that if we take the average of the "is_metro" column for all the counties in a given state, we’ll be computing the sum of these 0s and 1s (or the number of True values) divided by the total number of expensive counties in that state. This gives the proportion of expensive counties in the state that are in a metropolitan area. Thus, when we group the expensive counties according to what state they are in, we can use the .mean() aggregation method to calculate the proportion of expensive counties in each state that are in a metropolitan area.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: Minimum: 0

In a histogram, we do not know how data are distributed within a bin. This means that when we split the bin with range [50, 80) into two smaller bins, we have no way of knowing how the data from the original bin will be distributed. It is possible that all of the data in the [50, 80) bin fell between 50 and 70. In this case, there would be no data in the [70, 80) bin, and as such, the height of this new bar would be 0.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 61%.

Answer: Maximum: 0.072

Similarly, if all of the data in the original [50,80) bin fell between 70 and $80, then all of the data that was originally in the [50, 80) bin would be allocated to the [70, 80) bin. In a density histogram, the area of a bar corresponds to the proportion of the data contained within the bar (for example, a bar with area 0.5 contains 50% of the total data). Since the maximum value of h_2 is achieved when the bin [70, 80) contains all of the data originally contained in the bin [50, 80), this means area of the [70, 80) bar must be the same as the original area of the [50, 80) bar, since it contains the same proportion of data.

The original bar had area 30 * 0.024 = 0.72, which comes from multiplying its base and its height. Since the new bar has a base of 10, its height must be 0.072 to make its area equal to 0.72. Intuitively, if a rectangle is one third as wide as another rectangle and has the same area, it must be three times as tall.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: \frac{r}{2}

The key to solving this problem is recognizing that the number of counties in a given interval is directly related to the area of that interval’s bar in the histogram. This comes from the property of density histograms that the area of a bar corresponds to the proportion of the data contained within the bar.

Given that there are r times the amount of data in the interval [50, 70), in comparison to the interval [70, 80), we know that the area of the bar corresponding to the bin [50, 70) is r times the area of the bar corresponding to the bin [70, 80).

Therefore, if A_1 represents the area of the [50, 70) bar and A_2 represents the area of the [70, 80) bar, we have

A_1 = r \cdot A_2.

Then, since each bar is a rectangle, its area comes from the product of its height and its base. We know the [50, 70) bar has a base of 20 and a height of h_1, and the [70, 80) bar has a base of 10 and a height of h_2. Plugging this in gives

h_1 \cdot 20 = r \cdot h_2 \cdot 10.

From here, we can rearrange terms to get

\frac{h_1}{h_2} = \frac{r}{2}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.

Answer 21

We are told Montana has 5 counties. We don’t know how many counties Nevada has, but let’s call the number of counties in Nevada x and see how many rows the merged DataFrame should have, in terms of x. If Montana has 5 counties, since there are 10 "family_type" values per county, this means the state_A DataFrame has 50 rows.

Similarly, if Nevada has x counties, then state_B has 10x rows. When we merge on "family_type", each of the 5 rows in state_A with a given "family_type" (say "2a3c") will match with each of the x rows in state_B with that same "family_type". This will lead to 5x rows in the output corresponding to each "family_type", and since there are 10 different values for "family_type", this means the final output will have 50x rows.

We are told that the merged DataFrame has 1050 rows, so we can find x by solving 50x = 1050, which leads to x = 21.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 36%.

Answer: \frac{3}{5}

Given that the chosen individual is from a family with one child, we know that they must be from either Family X or Family Z. There are three individuals in Family X, and there are a total of five individuals from these two families. Thus, the probability of choosing any one of the three individuals from Family X out of the five individuals from both families is \frac{3}{5}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: False

If two events are independent, knowledge of one event happening does not change the probability of the other event happening. In this case, events A and B are not independent because knowledge of one event gives complete knowledge of the other.

To see this, note that the probability of choosing a child randomly out of the fifteen individuals is \frac{9}{15}. That is, P(B) = \frac{9}{15}.

Suppose now that we know that the chosen individual is an adult. In this case, the probability that the chosen individual is a child is 0, because nobody is both a child and an adult. That is, P(B \text{ given } A) = 0, which is not the same as P(B) = \frac{9}{15}.

This problem illustrates the difference between mutually exclusive events and independent events. In this case A and B are mutually exclusive, because they cannot both happen. But that forces them to be dependent events, because knowing that someone is an adult completely determines the probability that they are a child (it’s zero!)

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

Answer: True

If two events are independent, the probability of one event happening does not change when we know that the other event happens. In this case, events C and D are indeed independent.

If we know that the chosen individual is a child, the probability that they come from Family Y is \frac{3}{9}, which simplifies to \frac{1}{3}. That is P(D \text{ given } C) = \frac{1}{3}.

On the other hand, without any prior knowledge, when we select someone randomly from all fifteen individuals, the probability they come from Family Y is \frac{5}{15}, which also simplifies to \frac{1}{3}. This says P(D) = \frac{1}{3}.

In other words, knowledge of C is irrelevant to the probability of D occurring, which means C and D are independent.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: p^5

The probability of selecting Family W in the first round is p, which is the same for the second round, the third round, and so on. Each of the chosen letters is drawn independently from the others because the result of one draw does not affect the result of the next. We can apply the multiplication rule here and multiply the probabilities of choosing Family W in each round. This comes out to be p\cdot p\cdot p\cdot p\cdot p, which is p^5.

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Difficulty: ⭐️

The average score on this problem was 91%.

Answer: 1 - (1 - p)^5

Since there are too many ways that Family W can be selected to meet the condition that it is selected at least once, it is easier if we calculate the probability that Family W is never selected and subtract that from 1. The probability that Family W is not selected in the first round is 1-p, which is the same for the second round, the third round, and so on. We want this to happen for all five rounds, and since the events are independent, we can multiply their probabilities all together. This comes out to be (1-p)^5, which represents the probability that Family W is never selected. Finally, we subtract (1-p)^5 from 1 to find the probability that Family W is selected at least once, giving the answer 1 - (1-p)^5.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: p \cdot (1 - p)^4

We want to find the probability of Family W being selected only as the last draw, and not in the first four draws. The probability that Family W is not selected in the first draw is (1-p), which is the same for the second, third, and fourth draws. For the fifth draw, the probability of choosing Family W is p. Since the draws are independent, we can multiply these probabilities together, which comes out to be (1-p)^4 \cdot p = p\cdot (1-p)^4.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: replace=True

A family can be selected more than once, as indicated by “placing the number back into the hat” in the problem statement. Therefore we use replace=True to allow for the same family to get picked more than once.

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Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: np.count_nonzero(fams == 4) >= 2 or equivalent

Notice that inside the body of the if statement, the first line defines a variable children which selects two children from among ages. We are told in the problem statement that if a family’s number is written down on the paper at least twice, then two of the three children in that family are randomly selected to win a prize. Therefore, the condition that we want to check in the if statement should correspond to Chiachan’s family number (4) being written down on the paper at least twice.

When we compare the entire fams array to the value 4 using fams == 4, the result is an array of True or False values, where each True represents an instance of Chiachan’s family being chosen. Then np.count_nonzero(fams == 4) evaluates to the number of Trues, because in Python, True is 1 and False is 0. That is, np.count_nonzero(fams == 4) represents the number of times Chichan’s family is selected, and so our condition is np.count_nonzero(fams == 4) >= 2.

There are many equivalent ways to write this same condition, including np.count_nonzero(fams == 4) > 1 and (fams == 4).sum() >= 2.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 17%.

Answer: replace=False

A child cannot win a prize twice, so we remove them from the pool after being selected.

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: "Outcome R"

Chiachan is the middle child in the family, and recall that each outcome corresponds to either Chiachan winning ("Outcome Q"), Chiachan not winning but his siblings winning ("Outcome R"), or nobody in his family winning ("Outcome S").

This condition checks the negation of the middle child being selected, which evaluates to True when Chiachan’s siblings win but he doesn’t, so we append "Outcome R" to the outcomes array in this case.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: "Outcome Q"

Chiachan is the middle child in the family, and recall that each outcome corresponds to either Chiachan winning ("Outcome Q"), Chiachan not winning but his siblings winning ("Outcome R"), or nobody in his family winning ("Outcome S").

This condition corresponds to the middle child being selected, so we append "Outcome Q" to the outcomes array in this case.

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Difficulty: ⭐️⭐️

The average score on this problem was 75%.

Answer: "Outcome S"

Chiachan is the middle child in the family, and recall that each outcome corresponds to either Chiachan winning ("Outcome Q"), Chiachan not winning but his siblings winning ("Outcome R"), or nobody in his family winning ("Outcome S").

This condition is that Chichan’s family was not selected two or more times, which means nobody in his family will win a prize, so we append "Outcome S" to the outcomes array in this case.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.