Answer:

Option 2 and Option 3

Solution:

Option 1 is incorrect because splitting on spaces counts words rather than members. Band members can have multi-word names such as "John Smith". Subtracting 1 from the word count does not reliably fix this.

Option 2 is correct because .str.split(", ") produces one list element per member for bands with three or more members. A two-member band has no commas, so its list contains only one element. Applying len counts the elements, and f takes the maximum of that count and 2, which corrects the two-member case.

Option 3 is correct because a band with three or more members has exactly one fewer comma than members, so .str.count(", ") + 1 gives the right count. A two-member band has no commas, so the expression gives 1. Applying f raises this to 2, which is the correct number of members.

Option 4 is incorrect because .str.replace("and", ", ") replaces every occurrence of "and", including inside names such as "Brandon". This inserts extra commas and causes an overcount. We cannot make assumptions about the names themselves.

Answer:

• Input data type: str
• Output data type: int

Solution:

The function g is applied to the Series bands.get("member_names"), whose values are strings of band member names, so each input to g is a string, or str. The purpose of g is to count the members in each band, and a count is a whole number, so the output of g is an integer, or int. Recall that .apply(g) calls g separately on each value in the Series, and each returned integer becomes the corresponding value in the resulting Series.

Answer:

• (a) max(x.count(",") + 1, 2)

An alternative correct answer is max(len(x.split(",")), 2).

Solution:

For bands with three or more members, names are separated by commas, so the member count is x.count(",") + 1. Two-member bands are separated by " and " with no commas, so this expression would incorrectly return 1 for them. Since every band has at least two members, we wrap the expression in max(..., 2) to fix this case.

def g(x):
    return max(x.count(",") + 1, 2)

Alternatively, len(x.split(",")) gives the same count as x.count(",") + 1, so max(len(x.split(",")), 2) also works, assuming names do not contain commas.

Answer:

bands = bands.assign(albums_per_member = bands.get("albums") / bands.get("members"))

Solution:

bands.get("albums") / bands.get("members") performs elementwise division, so each band’s album count is divided by its own member count, such as 12 albums and 4 members giving \frac{12}{4}=3. The .assign(albums_per_member = ...) method adds these values as a new column, and because it returns a new DataFrame rather than modifying bands in place, we assign the result back to bands.

bands = bands.assign(albums_per_member = bands.get("albums") / bands.get("members"))

Answer:

Option 2, Option 4, and Option 5

Solution:

Option 1 is incorrect because index is not in quotation marks, so sort_values(by=index, ...) raises an error, and .iloc[-1] would return a whole row rather than a year.

Option 2 is correct because grouping by "albums_per_member", resetting the index, sorting in descending order, and taking .get("formed").iloc[0] returns a formation year for a band with the highest ratio.

Option 3 is incorrect because groupby("formed") makes the years the index, so .get("formed") fails since "formed" is no longer a column.

Option 4 is correct because after groupby("formed").max() and a descending sort, the year with the highest ratio comes first, and .index[0] retrieves it from the index.

Option 5 is correct because an ascending sort puts the highest ratio in the last row, so .get("formed").iloc[-1] gives its year.

Option 6 is incorrect because grouping by "albums_per_member" makes the ratios the index, so .index[0] returns a ratio rather than a year.

Answer:

Scatter plot

Solution:

A scatter plot is the most appropriate choice because both the formation year and "top_album_sales" are numerical variables, and each point represents one band. It lets us check for an association between the two, such as a downward trend if earlier bands tend to have higher sales.

A line plot is less appropriate because we are comparing individual bands rather than tracking a single quantity over time, and connecting the points would imply a sequence that doesn’t exist.

A bar chart is less appropriate because bar charts compare numerical values across categories, whereas both variables here are numerical.

Answer:

Bar chart

Solution:

A bar chart is the most appropriate choice because "origin" is a categorical variable, and we want to compare a numerical summary, the total number of albums, across countries. We would group the bands by country, sum the albums for each, and draw one bar per country, ideally sorted from highest to lowest so the top countries are easy to spot.

A scatter plot is not appropriate because it displays relationships between two numerical variables, whereas country is categorical.

A line plot is not appropriate because countries have no natural order, and connecting them with lines would suggest a trend that doesn’t exist.

Answer:

Line plot

Solution:

A line plot is the most appropriate choice because we want to see how a numerical quantity changes over time. We would group bands by formation year, compute the average number of members for each year, and plot formation year on the x-axis against that average on the y-axis, with the connecting line making trends in band size easy to spot.

A scatter plot would be better for examining the relationship between formation year and member count across individual bands, rather than a yearly summary.

A bar chart could display the yearly averages, but it is less effective at emphasizing the overall trend across a sequence of years.

Answer:

Bar chart

Solution:

A bar chart is the most appropriate choice because genre is a categorical variable, and we want to compare the median "top_album_sales" across genres. We would group the bands by genre, compute the median top album sales for each, and draw one bar per genre, which lets us directly identify the genre with the highest median.

A scatter plot is less appropriate because it displays relationships between numerical variables, whereas genre is categorical.

A line plot is less appropriate because genres have no natural order, and connecting the medians with lines would suggest a trend that doesn’t exist.

Answer:

Scatter plot

Solution:

A scatter plot is the most appropriate choice because both the number of albums released and "top_album_sales" are numerical variables, and each point represents one band. It lets us examine the direction, strength, and shape of the association, such as whether bands with more albums tend to have higher top sales and whether the relationship looks approximately linear, which is useful for deciding if linear regression is appropriate.

A line plot is less appropriate because the bands are separate observations rather than a continuous sequence, so connecting them would be misleading.

A bar chart is less appropriate because it compares values across categories rather than investigating the relationship between two numerical variables.

Answer:

• (a): sales.mean() - 2 * np.std(sales)
• (b): sales.mean() + 2 * np.std(sales)

Solution:

An interval containing all values within 2 standard deviations of the mean extends from 2 standard deviations below the mean to 2 standard deviations above the mean. We calculate the mean using sales.mean() and the standard deviation using np.std(sales). Therefore, the lower bound is sales.mean() - 2 * np.std(sales), and the upper bound is sales.mean() + 2 * np.std(sales).

Answer:

Option 3: At least 0.75, but we don’t know exactly.

Solution:

By Chebyshev’s inequality, for any distribution with a finite standard deviation, the proportion of values within (k) standard deviations of the mean is at least (1-). Setting (k=2), we obtain (1-=0.75), meaning that at least 75% of the values fall within the interval. Options 1 and 2 refer to the empirical rule for approximately normal distributions, but we are not told that "top_album_sales" is normally distributed. Option 4 is incorrect because Chebyshev’s inequality only guarantees 75% for (k=2), while Option 5 is incorrect because we can determine this minimum proportion.

Answer:

• Mean: 0
• SD: 1

Solution:

Standardization converts each value into standard units by subtracting the mean and dividing by the standard deviation. This transformation produces a distribution with a mean of 0 and a standard deviation of 1, regardless of the original mean and standard deviation (as long as the original SD is nonzero).

Answer:

2 albums

Solution:

To convert a standardized value back into its original units, use x = \text{mean} + z \cdot \text{SD}. Substituting the given values, we get x = 7 + (-2.5)(2) = 2. Therefore, Joy Division released 2 albums.

Answer:

Option 5: We cannot determine anything about this proportion.

Solution:

The interval [5, 9] extends from one standard deviation below the mean to one standard deviation above it, since the mean is 7 and the standard deviation is 2. Chebyshev’s inequality guarantees a proportion of at least 1-\frac{1}{k^2} within k standard deviations, but for k=1, this lower bound is 0 and provides no useful information. Although approximately 68% of values fall within one standard deviation of the mean in a normal distribution, we are not told that the "albums" distribution is normal. Therefore, we cannot determine the proportion from the information given.

Answer:

Option 1 and Option 4

Solution:

To convert a value in standard units back to the original number of albums, we use x = 7 + 2z, where z is the standardized value. For Option 1, z=-4 gives x=7+2(-4)=-1, which is impossible because a band cannot release a negative number of albums. For Option 4, z=1.75 gives x=7+2(1.75)=10.5, which is impossible because the number of albums must be a whole number. The remaining values correspond to possible nonnegative integer album counts: -0.5 \rightarrow 6, 0 \rightarrow 7, and 8.5 \rightarrow 24. Therefore, only Options 1 and 4 could not be valid standardized album counts.

Answer:

0.025

Solution:

To find each bar’s height in a density histogram, we use \text{Height} = \frac{\text{Proportion}}{\text{Bin Width}}, where the proportion is the area of the bar. Since Histogram B combines bins from Histogram A, we can add the areas of the corresponding bars to calculate each new proportion.

The tallest bar is [0, 20), with a height of 0.025.

Answer:

• Minimum h_1: 0
• Maximum h_1: 0.06

Solution:

In Histogram A, the [0, 10) bin has a height of 0.03 and a width of 10, so its area (proportion) is 0.03 \times 10 = 0.30. When this bin is divided into [0, 5) and [5, 10), we do not know how the 0.30 proportion is distributed between the two smaller bins. The minimum possible height occurs when none of the data falls in [0, 5), giving h_1 = \frac{0}{5} = 0. The maximum possible height occurs when all 0.30 of the proportion falls in [0, 5), giving h_1 = \frac{0.30}{5} = 0.06.

Answer:

Option 2

Solution:

The null hypothesis of a permutation test states that the two groups come from the same underlying distribution, so the "active" labels are interchangeable under the null. Option 1 describes the alternative hypothesis that active bands have higher sales, while Option 3 also describes a population-level alternative rather than the null. Option 4 is incorrect because the null hypothesis does not require the two groups to have exactly equal sample means; differences can occur due to random variation even when the null is true.

Answer:

• (a): np.random.permutation(bands.get("active"))
• (b): np.count_nonzero(sim_diffs >= obs_diff)

Solution:

For (a), np.random.permutation(bands.get("active")) randomly rearranges the "active" labels while keeping the number of active and inactive bands unchanged, breaking any association between activity status and album sales. Each iteration then calculates the difference in mean sales between the shuffled active and inactive groups. For (b), since the alternative hypothesis is that active bands have higher average sales, we count simulated differences that are greater than or equal to the observed difference using np.count_nonzero(sim_diffs >= obs_diff). Dividing by 10,000 gives the proportion of simulated test statistics at least as extreme as the observed statistic in the direction of the alternative, which is the p-value.

Answer:

Option 3

Solution:

Since the p-value of 0.03 is less than the significance level of 0.05, we reject the null hypothesis. Because the alternative hypothesis is one-sided, the test provides statistically significant evidence that active bands have higher average top album sales than inactive bands. Option 1 is incorrect because we do not “accept” the null hypothesis, and Option 2 is incorrect because the p-value is small enough to reject it. Option 4 is incorrect because the direction of the alternative allows us to conclude that active bands have higher average sales, rather than merely concluding that the groups differ.

Answer:

Option 2

Solution:

Shuffling the "active" labels and shuffling the "top_album_sales" values are statistically equivalent ways of breaking the association between group membership and album sales while preserving the original group sizes and sales values. Both methods therefore generate the same theoretical null distribution for the difference in group means. However, because the 10,000 simulations are random, the individual simulated differences will not necessarily match between runs, so the resulting histograms should look nearly identical with only slight random variations.

Answer:

• (a): uk.shape[0]
• (b): 7.5
• (c): 92.5

Solution:

Each bootstrap resample must contain the same number of rows as the original DataFrame uk, so we use uk.shape[0]. For an 85% confidence interval, we want the middle 85% of the bootstrap means, leaving 15% outside the interval. Splitting this equally between the two tails gives 7.5% in each tail. Therefore, we use the 7.5th percentile for the lower endpoint and the 92.5th percentile for the upper endpoint.

Answer:

• 1: Option 3
• 2: Option 4

Solution:

1. Adding Swedish bands increases the sample size, which tends to make a confidence interval narrower. Since the combined dataset also has a smaller standard deviation, both factors favor a narrower interval for uk_sweden. Therefore, we expect uk_width > uk_sweden_width.

2. Adding Swedish bands increases the sample size, which tends to narrow the confidence interval, but it also increases the standard deviation, which tends to widen the interval. Since these effects work in opposite directions, we cannot determine which confidence interval will be wider without additional information.

Answer:

Option 3

Solution:

The true population mean is a fixed, unknown value rather than a random variable. Once we calculate the confidence interval [3.8, 5.2], its endpoints are also fixed, so the true mean either lies inside the interval or does not. The 90% confidence level describes the long-run success rate of the confidence interval procedure: if we repeatedly collected new samples and constructed intervals using the same method, approximately 90% of those intervals would contain the true population mean. Therefore, Option 3 correctly identifies the problem with Elise’s interpretation.

Answer:

Option 4

Solution:

The CLT is typically applied to independent observations, whereas sampling without replacement introduces dependence between observations. However, when the sample is much smaller than the population, removing one observation has very little effect on the distribution of the remaining observations. Therefore, sampling without replacement behaves approximately like sampling with replacement, making the CLT a reasonable approximation. The other options do not address the dependence introduced by sampling without replacement.

Answer:

10

Solution:

A 95% CLT-based confidence interval is approximately \text{sample mean} \pm 2 \cdot \frac{\text{sample SD}}{\sqrt{n}}. The interval [20.0, 22.0] has a midpoint of 21 and a margin of error of 1. Since the sample size is n=400, we have 1 = 2 \cdot \frac{\text{SD}}{\sqrt{400}} = \frac{\text{SD}}{10}. Solving gives \text{SD} = 10.

Answer:

0.1

Solution:

For a sample proportion \hat{p}, a 95% CLT-based confidence interval is approximately \hat{p} \pm 2\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, so its total width is 4\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. The expression \hat{p}(1-\hat{p}) is maximized when \hat{p}=0.5, giving a maximum value of 0.25. With n=400, the maximum width is 4\sqrt{\frac{0.25}{400}} = 4(0.025) = 0.1.

Answer:

Options 1, 2, and 4

Solution:

Option 1 is correct because a CLT-based confidence interval for a population mean only requires the sample mean, sample standard deviation, and sample size to calculate the endpoints. Option 2 is correct because the introductory CLT method applies to sample means rather than medians, so we would use bootstrapping to estimate a confidence interval for the population median. Option 3 is incorrect because bootstrapping requires resampling the original observations, and knowing only their mean, standard deviation, and count is insufficient to reconstruct the bootstrap distribution. Option 4 is correct because a proportion is the mean of a binary variable, allowing us to use either a CLT-based interval or a bootstrap interval. Option 5 is incorrect because neither the standard CLT method nor ordinary bootstrapping provides a reliable confidence interval for the population maximum.

Answer:

• Min: 20
• Max: 32

Solution:

When merging on "genre", each genre contributes the product of its row counts in the two DataFrames, so Rock contributes 4 \cdot 2 = 8 rows and Pop contributes 3 \cdot 4 = 12 rows, for a guaranteed total of 20 rows. Michelle has 3 remaining bands and Pranav has 4 remaining bands, all belonging to genres other than Rock or Pop. The minimum occurs when these remaining bands have no genres in common, giving 20 rows. The maximum occurs when all remaining bands share the same third genre, producing an additional 3 \cdot 4 = 12 rows, for a total of 32 rows.

Answer:

• Min: 0
• Max: 8

Solution:

When merging on "band", each matching band appears once in the result, since each tutor’s top ten list contains distinct bands. The minimum is 0 if Michelle and Pranav have no bands in common. For the maximum, they can share at most 2 Rock bands, since Pranav has only 2, and at most 3 Pop bands, since Michelle has only 3. They can also share at most 3 bands from other genres, since Michelle has 3 remaining bands and Pranav has 4. Therefore, the maximum number of matching bands is 2 + 3 + 3 = \mathbf{8}.

Answer:

• Min: 4x + 3y
• Max: 4x + 3y + 3(10 - x - y)

Solution:

Merging on "genre" guarantees 4x matching rows for Rock and 3y matching rows for Pop, giving a minimum of 4x + 3y if Michelle’s and Kate’s remaining bands have no other genres in common. Michelle has 10 - 4 - 3 = 3 bands from other genres, while Kate has 10 - x - y such bands. The maximum occurs when all of these remaining bands share the same genre, creating an additional 3(10 - x - y) matching rows. Therefore, the maximum is 4x + 3y + 3(10 - x - y).

Answer:

400

Solution:

An 80% confidence interval procedure is expected to capture the true population parameter approximately 80% of the time when repeated with new random samples. Since Ray repeats the entire process 500 times, the expected number of intervals containing the true median is 0.80 \times 500 = \boxed{400}.

Answer:

Option 3

Solution:

The biggest challenge is obtaining 500 independent random samples of K-Pop bands from the population, which would require substantial time and resources. Option 1 is incorrect because performing bootstrap resampling computationally is generally much easier than collecting new samples. Option 2 is incorrect because bootstrapping can be used to construct confidence intervals for medians, even though the introductory CLT-based method focuses on means. Option 4 is incorrect because constructing an 80% confidence interval is no more difficult than constructing a 95% confidence interval; we simply use different percentile cutoffs.

Answer:

0.9

Solution:

Coldplay’s album count is \frac{10-7}{2}=1.5 standard deviations above the mean. In standard units, the regression line predicts y_{\text{su}}=r\cdot x_{\text{su}}, where r=0.6. Therefore, Coldplay’s predicted top album sale is 0.6(1.5)=\boxed{0.9} standard deviations above the mean.

Answer:

10

Solution:

The regression line has slope r\frac{\text{SD of }y}{\text{SD of }x}=0.6\frac{S}{2}=0.3S and passes through the point of averages (7,21). Since the line also passes through (0,0), its slope must be \frac{21-0}{7-0}=3. Setting 0.3S=3 gives \boxed{S=10}.

Answer:

Options 2, 5, and 6

Solution:

Option 2 is correct because doubling every x-value doubles its standard deviation, so the regression slope r\frac{\text{SD of }y}{\text{SD of }x} is halved, making Option 1 incorrect. Options 3 and 4 are incorrect because the new regression line has the same intercept as the original line: doubling the mean of x while halving the slope leaves the intercept unchanged. Option 5 is correct because multiplying a variable by a positive constant does not change its correlation with another variable. Option 6 is correct because the fitted predictions and residuals remain unchanged, so the RMSE also stays the same. Option 7 is incorrect because the new prediction at 30 albums equals the original prediction at 15 albums, not twice that prediction, and Option 8 is incorrect because several statements are true.

Answer:

• 1: True
• 2: False

Solution:

1. Assuming the two tutor bands have different album counts, the regression line through their two data points fits both observations perfectly, giving an RMSE of 0. The regression line for bands has correlation r=0.6, so it does not fit all observations perfectly and must have a positive RMSE. Therefore, the two regression lines cannot have the same RMSE.

2. Two different datasets can produce exactly the same regression line without sharing any data points. The regression line depends on the datasets’ means, standard deviations, and correlation, rather than requiring particular observations to appear in both datasets. Therefore, bands does not need to contain either of the tutor bands’ exact data points.

Answer:

0.3

Solution:

To calculate TVD, we first convert the genre counts into proportions separately for active and inactive bands. We then find the absolute difference between the two proportions for each genre, add these differences, and divide the sum by 2. Using the values in the DataFrame gives a TVD of \boxed{0.3}, indicating that the two genre distributions differ by a total variation distance of 0.3.

Answer:

\frac{7}{31}

Solution:

Since we know the selected band is not a rock band, we restrict our attention to the 310 non-rock bands. Of these, 70 are active, so the conditional probability is the number of active non-rock bands divided by the total number of non-rock bands. Therefore, P(\text{Active} \mid \text{Not Rock}) = \frac{70}{310} = \boxed{\frac{7}{31}}.