Answer: 6

Observe that len(df.get("high_temp")) evaluates to the number of rows in the DataFrame df, and the for-loop is iterating through the numerical row indices in df. During each iteration, we check whether the value of "low_temp" is less than -10. If it is, we terminate the loop and return i, the row index at which the first low temperature is below -10. From here, we can refer to the preview of the snow DataFrame. Notice that in the 7th row, there is a low temperature below -10. Our function would hence return 6, corresponding to that row’s index.

Answer: "snow"

Let’s first understand what DataFrame we are passing into f2: snow.sample(snow.shape[0], replace=False).

Since we are sampling "snow" without replacement, and snow.shape[0] evaluates to the number of rows in snow, the above expression is actually going to produce a DataFrame with the same rows as the original snow DataFrame. The only difference is that rows are now potentially in a different order. Let us refer to this as the “shuffled” snow DataFrame.

Note that the index of snow consists of regular integers. By using iloc[-k], we are essentially iterating backwards through all entries of the "snowfall" column in the original snow DataFrame (with the exception of the first index 0, which we check first because 0 = -0). (Note that .sample() “preserves” the original index values, but reorders them.)

Finally note the two “if”-statements. If the current entry in the "snowfall" column is less than 1, we append it to an array called out. However, if there are any "snowfall" entries greater than 10, we return "snow", terminating the for-loop entirely.

Thus: it is sufficient to check the preview of snow in the data description, and see if any entries in the "snowfall" column are greater than 10. At index 7, there is one with 10.5, so we know that the function must return "snow".

Answer: True

Outside of the for-loop, we initialize an empty array results. Then, observe that the line results = np.append(results, output) within the for-loop is a typical representation of an accumulator pattern.

Answer: False

Note that .apply() is a Series method, meaning that it can only use values contained within a single Series. However, the function f has two arguments, reliant on two distinct Series: the "prev_snow" and "snowfall" columns. Since .apply() cannot access values from multiple Series at once given how we’ve written f, the apply() method is not appropriate.

Answer: Bar chart

Since output is the output of the function f, we can look at what can be returned by f. The possibilities are "blizzard buildup", "fresh snowfall", "light accumulation", and "no snow". We can view these values as part of a categorical variable to describe (qualitatively) how much snow is on the ground.

Thus, a scatter plot, line plot, and histogram are inappropriate: we need quantitative/numerical variables for such plots.

This leaves the bar chart as the only acceptable choice.

Answer: "light accumulation"

Note that results[2] corresponds to loc[2] entries in our for-loop, by nature of the accumulator pattern. Since the index of snow is already numerical, loc[2] and iloc[2] are equivalent.

We refer to the preview of snow in the data description. The value snow.get("prev_snow").loc[2] is 3.0, and snow.get("snowfall").loc[2] is 2.5. We are thus finding the output of f(3.0, 2.5), where a = 3.0 and b = 2.5.

Since a = 3 \leq 20, the body of the first if-statement is not executed. We check the second. Since a = 3 \neq 0, the body of the second if-statement is also not executed. We check the third. Observe that a = 3 > 0, b = 2.5 > 0, and b = 2.5 \leq 5. This condition is satisfied, so we execute the body of the third if-statement. This returns "light accumulation".

Answer (a): snow.get("coastal")

Answer (b): ["date", "city"] (or any list that includes these two variables plus additional column names)

Answer (c): median(), min(), max(), sum()

Explanation (a): The trip DataFrame must contain date and coastal city pairs, which means that we must filter to only include coastal cities. Thus, we need to query snow according to the "coastal" column.

Explanation (b): Again, trip must contain all date and coastal city pairs. If we group by only one of date or coastal status, then we will lose data for pairs after the grouping. Thus, we must group by at least both "date" and "city".

Explanation (c): Every choice except count() works to produce the snowfall for each date-coastal city pair. Because there is only one snowfall value per date-coastal city pair in snow, the median, min, max, and sum will all return the same numerical value.

For the same reason, note that count() doesn’t work because it will output 1. This is not the snowfall for a date in a coastal city, so we can’t use it.

Answer (d): sort_values(by="elevation")

Answer (e): get("city")

Answer (f): 0

It’s also fine to sort the other way in blank (d) and use -1 in blank (f).

Explanation (d): Recall that there is one entry in snow per date-city pair. When we create trip and aggregate using one of the correct functions in blank (c), this means that "elevation" data is preserved for that city. Since we want the cities with lowest elevation, we can thus sort by the "elevation" column in trip. (Either ascending or descending order works.)

Explanation (e): We need the name of the coastal city with lowest elevation, so after sorting, we must extract the city name from the "city" column.

Explanation (f): Depending on how you sorted in blank (d), we must now extract the entry associated with lowest elevation. If ascending order is used, this is the first entry in the sorted DataFrame: we use iloc[0]. If descending order is used, this is the last entry in the sorted DataFrame: we use iloc[-1]. (It may also be possible to use iloc[trip.shape[0] - 1].)

Answer: \frac{1}{5}

The TVD is the sum of absolute differences in proportions, all divided by 2. We are essentially comparing categorical distributions of "continent" in the first five rows of snow (snow1) against the next five rows of snow (snow2).

We refer to the preview of snow in the data description. The counts and corresponding proportions of continents in the first five rows are as follows:

1. North America: 2 \Rightarrow 2/5
2. Europe: 1 \Rightarrow 1/5
3. Asia: 1 \Rightarrow 1/5
4. Africa: 1 \Rightarrow 1/5

In the next five rows, the counts and corresponding proportions of continents are:

1. North America: 2 \Rightarrow 2/5
2. Europe: 2 \Rightarrow 2/5
3. Asia: 1 \Rightarrow 1/5
4. Africa: 0 \Rightarrow 0

Let p_k^{(1)} be the proportion associated with continent k in snow1. Similarly let p_k^{(2)} be the proportion associated with continent k in snow2. Computing the TVD by its definition, we get:

\frac{1}{2} \sum_{k=1}^{4} \left|p_k^{(1)} - p_k^{(2)}\right| = \frac{1}{2}\left(\left|\frac{2}{5} - \frac{2}{5}\right| + \left|\frac{1}{5} - \frac{2}{5}\right| + \left|\frac{1}{5} - \frac{1}{5}\right| + \left|\frac{1}{5} - 0\right|\right) = \frac{1}{2}\left(0 + \frac{1}{5} + 0 + \frac{1}{5}\right) = \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{5}

Answer: \frac{1}{6}

If we add a new row to snow1 and an identical new row to snow2, there will be 6 total rows instead of 5. Hence, our proportions will be out of 6, instead of 5. However, the numerator from the TVD computation will not change. Suppose c_1 and c_2 are counts for the original snow1 and snow2 respectively, corresponding to the new continent we’ve added. In the new snow1 and snow2, these counts become:

c_1 + 1 \quad \text{and} \quad c_2 + 1

and the corresponding proportions are

\frac{c_1 + 1}{6} \quad \text{and} \quad \frac{c_2 + 1}{6}.

When computing TVD, we must then compute:

\left|\left(\frac{c_1 + 1}{6}\right) - \left(\frac{c_2 + 1}{6}\right)\right| = \left|\frac{c_1}{6} - \frac{c_2}{6}\right| = \frac{5}{6}\left|\frac{c_1}{5} - \frac{c_2}{5}\right|

Since |c_1/5 - c_2/5| is the absolute difference between original proportions in part (a), the above would suggest that the only difference in TVD is the denominator changing from 5 to 6.

Answer: 7

Recall our continent counts for snow1 and snow2 from part (a).

snow1:

1. North America: 2
2. Europe: 1
3. Asia: 1
4. Africa: 1

snow2:

1. North America: 2
2. Europe: 2
3. Asia: 1
4. Africa: 0

For each entry in snow1, merging adds an amount of rows that describes the number of “matches” that entry has with entries in snow2. For example, the first North America entry in snow1 will have 2 matches with North America entries in snow2. Since there are 2 North America entries in snow1 and 2 North America entries in snow2, there are a total of 2 \cdot 2 = 4 row matches for North America across snow1 and snow2. Following this logic, the resulting number of rows is:

(2 \cdot 2) + (1 \cdot 2) + (1 \cdot 1) + (1 \cdot 0) = 4 + 2 + 1 = 7 \text{ rows}

Answer: 12

Instead of continents, we are now matching values in the "coastal" column. The counts are as follows:

snow1:

1. True: 2
2. False: 3

snow2:

1. True: 3
2. False: 2

The logic is the same as in part (c), and computing accordingly, we get:

(2 \cdot 3) + (3 \cdot 2) = 6 + 6 = 12 \text{ rows}

Answer: t^2 + (n-t)^2 = n^2 - 2nt + 2t^2

If there are t rows that contain True and there are n total rows, we know that there must be n - t rows that contain False.

If we merge snow with itself on the "coastal" column, each True entry in snow will have t matches with True entries in the “other” snow DataFrame (since we are working with the same DataFrame). Similarly, each False entry will have n - t matches with other False entries. Since there are t True entries and n - t False entries, this results in

(t \cdot t) + \Big((n-t) \cdot (n-t)\Big) = t^2 + (n-t)^2

rows in the resulting DataFrame.

Answer: The mean of the "snowfall" column of snow.

Note that the Central Limit Theorem works only for means and sums.

Answer: The median, the mean, and the standard deviation of the "snowfall" column of snow.

Note that only the largest value — the maximum — doesn’t make sense here. The true maximum must be greater than or equal to the maximum of our sample. (Otherwise, it is not the true maximum!) Instead of bootstrapping for the maximum, it makes more sense to immediately take the largest value of our given sample (since we can’t bootstrap a larger maximum out of a fixed sample).

Answer: A = 7, B = 2, C = 21

Thinking of the proportion as a mean of 1s and 0s, we can use the Central Limit Theorem. This gives us that the distribution of sample proportions is approximately normal. It will be centered around the sample mean, which in this case is 0.7. The standard deviation will be the sample standard deviation divided by the square root of the sample size. The standard deviation in this case is \sqrt{0.7(1 - 0.7)}, and the square root of the sample size is \sqrt{100}. For a 95% confidence interval, we need to go out two standard deviations in either direction, so our confidence interval is:

\left[0.7 - 2 \cdot \frac{\sqrt{0.7(1-0.7)}}{\sqrt{100}},\ 0.7 + 2 \cdot \frac{\sqrt{0.7(1-0.7)}}{\sqrt{100}}\right]

Answer: 400

The width of a 95% confidence interval is 4 \cdot \frac{\text{Sample SD}}{\sqrt{\text{Sample Size}}}. For 0s and 1s, the biggest possible standard deviation would be \sqrt{0.5 \cdot 0.5}, so we can assume that’s our sample standard deviation to get a conservative estimate. That gives:

0.1 \geq 4 \cdot \frac{\sqrt{0.5 \cdot 0.5}}{\sqrt{N}}

Solving for N, we get:

\sqrt{N} \geq 4 \cdot \frac{0.5}{0.1}

\sqrt{N} \geq 4 \cdot 5

N \geq 400

So N = 400 is the smallest sample size we could take to guarantee a 95% confidence interval width of 0.1 or less.

Answer (a): 100, replace=True or flurry.shape[0], replace=True

Answer (b): resample.get("snowfall").mean()

Explanation (a): When bootstrapping, each resample must have the same size as the original sample and be taken with replacement.

Explanation (b): Since we are estimating mean "snowfall", we must compute the corresponding quantity within each resample.

Answer: np.std(snow.get("snowfall")) / 10 and np.std(means)

The means from (a) and the sample means of these 1000 new samples are approximating the true distribution of sample means for "snowfall". From lecture, the sample mean and SD are likely to be close to the population mean and SD. Thus, the SD of means and the SD of the new collection of sample means will be similar to each other (because they are both approximating the same population parameter). The choice np.std(means) is correct.

We look for other options that approximate the SD of sample means distribution. From the original sample, note that this can be approximated as follows:

\frac{\text{Sample SD}}{\sqrt{\text{sample size}}}

Since our original sample is given by snow.get("snowfall"), the sample SD is given by np.std(snow.get("snowfall")). We now need to divide by the square root of the sample size, which is given to be 100. Since \sqrt{100} = 10, it follows that np.std(snow.get("snowfall")) / 10 is correct.

The other options are incorrect because they do not approximate the SD of the distribution of sample means.

Answer (i.): 4

Note that the values X and 24 * X must be between 0 and 100. Furthermore, a bootstrapped C% confidence interval places equal probability mass in the left and right tails of the bootstrapped distribution. Here, this property can be expressed as follows:

|X - 0| = |100 - 24X|

Solving for X yields:

X - 0 = 100 - 24X

25X = 100

X = 4

Indeed if X = 4, then 24X = 24(4) = 96, such that 100 - 96 = 4. This is a valid bootstrapped confidence interval.

Answer (ii.): 92

We saw that X = 4 in part (a). Our left endpoint corresponds to the 4th percentile. Our right endpoint corresponds to the 96th percentile. This means that C = 96 - 4 = 92.

Answer: \frac{8}{9}

Since we have made no assumptions about the data, we use Chebyshev’s inequality. We want to be within 3 SDs of the mean. Therefore, the proportion is:

1 - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{8}{9}

Answer: (chi_low >= -25) & (chi_low <= 11) or (chi_low >= -7 - 6 * 3) & (chi_low <= -7 + 6 * 3) or -25 <= chi_low <= 11 or equivalent

The mean is -7 and the SD is 6. Thus the range that is within 3 SDs of the mean is given by

[-7 - 3 \cdot 6,\ -7 + 3 \cdot 6] = [-7 - 18,\ -7 + 18] = [-25, 11]

If x is a Boolean series, then taking the mean of x is equivalent to finding the proportion of True values in x. (Recall that True = 1 and False = 0, so taking the mean is equivalent to counting all instances of True then dividing by the total number of entries in the Series.)

All answer options above create a Boolean series x that checks whether each entry in chi_low belongs to the range [-25, 11]. As a result, x.mean() will tell us the proportion of values within 3 SDs of the mean.

Answer: 0.98

We use our answer to part (b) and check the proportion of values that belong to the interval [-25, 11]. Note that all values are at least as large as -25. However, there are three “bars” that lie above 11. These are the values that lie outside 3 SDs of the mean. If we find the proportion of values represented in these three bars, we can subtract this proportion from 1 to find the proportion of values that lie within 3 SDs of the mean.

To compute proportion, we find the area of each bar. The width of each bar is 1, and the heights of the three bars are 0.01, 0.005, 0.005 (approximately). Their total area is therefore:

1(0.01 + 0.005 + 0.005) = 0.02

Hence 1 - 0.02 = 0.98 is the proportion of values within 3 SDs of the mean.

Answer: 4.5

Recall the formula for x in standard units:

\frac{x - \text{mean}}{\text{SD}}

When x = 20, mean is -7, and SD is 6, the corresponding value in standard units is:

\frac{20 - (-7)}{6} = \frac{27}{6} = \frac{9}{2} = 4.5

Answer: No substantial change.

Note that this is not a distribution of sample statistics, but a distribution of a sample itself. Hypothetically, if we collected many more individual low temperatures, we would expect them to be distributed similarly to our current sample. Thus, the SD will stay generally the same.

(It would also help to recall that the sample standard deviation can be used to approximate the population standard deviation. Thus, if the initial sample size is “large enough,” increasing the sample size should not substantially change the approximation of the population standard deviation.)

Answer: Empirical distribution.

The probability distribution is a theoretical model and is generally not known. Since we only have a sample of low temperatures, there are many more not included in the distribution. Thus, the histogram cannot represent the true probability distribution.

This is not a categorical distribution because "low_temp" is a quantitative variable.

This is an empirical distribution, because it depicts the distribution of our observed data.

Answer (a): "city"

Answer (b): max()

Answer (c): "continent"

Answer (d): mean()

Answer (e): na - asia

We are told that to compute average maximum snowfall, we must first compute the maximum snowfall per city, then take the average within each continent. The first step is completed in blanks (a) and (b), where we must first group by "city" then aggregate by the max() function. Blanks (c) and (d) correspond to the second step, where we must now group maximum snowfall by "continent" and take the average.

Blank (e) requires you to realize that the alternative hypothesis is phrased differently than how it is expressed in the code.

Let AMS stand for “average maximum snowfall.” The alternative hypothesis is that Asian AMS is greater than North American AMS. However, the extreme variable in the code is checking for values that are smaller than the observed statistic test_stat(cities). Thus, when Asian AMS is much larger than North American AMS, the quantity na - asia will be more negative and thus be considered more extreme under the alternative hypothesis. Blank (e) should be na - asia.

Note that asia - na is wrong, because this test statistic corresponds to a different alternative hypothesis: North American AMS is greater than Asian AMS.

Answer: Line number to change: 18 or No lines should be changed.

The code in this line should be changed to the following:

extreme = (np.abs(stats) >= abs(test_stat(cities))) or extreme = (np.abs(stats) >= abs(test_stat(asia_na)))

Another solution is to change line 9 to return -abs(na-asia) or -abs(asia-na).

The alternative is now two-sided. To reflect this, we can either edit our test statistic in Line 9, or edit which values are considered extreme in Line 18.

The absolute difference is a good candidate for the new test statistic. However, since we are only changing the test statistic, we must be careful to remember that the extreme variable counts values that are smaller than the observed statistic. Thus, we must add a negative sign outside the absolute difference. This allows us to find evidence for the correct alternative hypothesis.

We can also change the extreme region. By taking absolute value of the difference test statistics, we are effectively covering both “tails” of our distribution and satisfying the two-sided nature of the alternative.

Answer: Yes. Divide p_b in half to approximate p_a, because the difference of means is symmetric around 0 under the null hypothesis.

The test in (a) is one-sided, while the test in (b) is two-sided. Thus, the area corresponding to the “extreme” region will be twice as large on a two-sided test compared to a one-sided test. We must divide p_b in half to approximate p_a.

The first choice (“No”) has an irrelevant explanation, because the distribution of sample average maximum values is approximately normal, according to CLT. Hence, the distribution is symmetric. Because the distribution is symmetric, we expect approximately equal probability mass in the left and right tails, so dividing by 2 is a valid way to approximate p_a from p_b.

Answer (a): np.random.multinomial(snow_nyc40.shape[0], [0.4, 0.6])[0]

Answer (b): <=

If it snows on a day, the corresponding entry in "snowfall" will be greater than 0. We have a proposed distribution under the null hypothesis, which can be described as 40% days with more than zero snowfall, and 60% days with zero snowfall. To simulate counts, according to this null hypothesis, we can use np.random.multinomial(). Note that since we are comparing counts, we must use the same total as our original sample, snow_nyc40 which is given by snow_nyc40.shape[0].

Our alternative states that less than 40% of days are snow days. Note that obs_stat is computing the number of days in snow_nyc40 that experienced more than zero snowfall. So, test statistics that are smaller than or equal to the observed will be in favor of the alternative hypothesis.

Answer: np.count_nonzero(snow_nyc40.get("snowfall") == 0), (snow_nyc40.get("snowfall") > 0).sum() / snow_nyc40.shape[0], and (snow_nyc40.get("snowfall") > 0).mean() - 0.40

The first choice can be used to test the following alternative: “There is no snowfall on more than 60% of the days for which the news predicts a 40% chance of snow.” This is equivalent to our original alternative hypothesis.

The second choice simply converts the original test statistic (a count) into its corresponding proportion, which isn’t going to affect our hypotheses.

The third choice also converts the test statistic into a proportion (using mean()). It additionally subtracts a constant. Since 0.40 is subtracted from all test statistics, however, the conclusion we make will not change and thus the corresponding hypotheses remain the same.

Note that snow_nyc40.get("snowfall") is not a Boolean series, therefore snow_nyc40.get("snowfall").sum() will not help us determine whether (fewer than) 40% of days are snow days.

Answer: Because this confidence interval does not contain the value of 0.40, it indicates that the p-value from the hypothesis test above should be below 0.05.

We can use CLT since we are working with proportions. Hence, the 95% CLT-based confidence interval for the population parameter (the true proportion) can be used to reject/fail to reject the null hypothesis. Since 0.4 (the proposed proportion under the null) does not belong in our CLT-based confidence interval for the true proportion, it lies in the (100 - 95)\% = 5\% region not contained by the confidence interval. The p-value of the hypothesis test must therefore be smaller than 0.05.

Answer: -0.6

Recall that the slope of a linear regression line can be computed as follows:

\text{slope} = r \cdot \frac{\text{SD}_y}{\text{SD}_x}

where r is the correlation, \text{SD}_x is the standard deviation of the x variable (independent/explanatory variable), and \text{SD}_y is the standard deviation of the y variable (dependent/response variable).

From the plot, we can see that x is the Low Temperature, and y is the Snowfall. Plugging in the corresponding values:

\text{slope} = -0.4 \cdot \frac{9}{6} = -\frac{2}{5} \cdot \frac{9}{6} = -\frac{2}{5} \cdot \frac{3}{2} = -\frac{3}{5} = -0.6

Answer: When the temperature is 0°C, the predicted snowfall is 3cm.

We are predicting snowfall using temperature. Our regression line is given by \hat{y} = mx + b, where m is the slope, b = 3 is the intercept, x represents the lowest temperature, and \hat{y} represents the predicted snowfall. If we let x = 0, then \hat{y} = m(0) + 3 = 3. Indeed, the predicted snowfall (\hat{y}) is 3cm when temperature (x) is 0°C.

Answer: 25

For a temperature-snowfall pair (x_i, y_i), let y_i be the actual snowfall and \hat{y}_i be the predicted snowfall when you input x_i into the regression line \hat{y} = mx + b. The residual is given by the following:

y_i - \hat{y}_i

Our primary goal is to find a temperature-snowfall pair where the prediction \hat{y}_i is the “farthest off” from the actual value y_i. If you observe low temperatures within the range [-20, -15], there is a snowfall value that is considerably “farther away” from the regression line compared to all other points. The point is approximately located at (-17.5, 40), where 40 is the actual snowfall. We compare this to the value of the regression line at approximately x = -17.5. Since we are rounding to the nearest multiple of 5, we visually approximate the value on the regression line as (-17.5, 15). The residual is therefore 40 - 15 = 25.

Answer: Residuals have more vertical spread on the left than on the right.

A trick you can use to visualize residuals is to angle your vision such that the linear regression line becomes “horizontal.” Then, imagine that the “horizontal” regression line is the x-axis and the points around it are the residuals.

We can see that the residuals become more “vertically squeezed together” in the plot as you travel left to right along the regression line. This is equivalent to saying that residuals have more vertical spread (variation) on the left than on the right.

Answer: The variability of the residuals changes with temperature. The linear model is appropriate, but prediction uncertainty is not constant across temperatures.

The first choice follows from our answer to part (d). If residuals have more vertical spread on the left than on the right, this implies that lower temperatures indicate higher variability in residuals than higher temperatures. In other words, we can expect variability of residuals to decrease as temperature increases.

A quadratic/curved model would require a quadratic/curved shape in the original data. However, from our plot, it is unclear whether such a model will fit better than linear regression. As a result, we cannot select this choice.

The pattern seen in the errors is heteroskedasticity, with bigger spread at lower values of Low Temperature and smaller spread at higher values of Low Temperature. This is due to the fact that the variability in snowfall is greater at lower values of Low Temperature than it is at higher values of Low Temperature. Therefore, any other prediction line would not be able to change this pattern.

The linear model is appropriate because there does appear to be a linear relationship between Low Temperature and Snowfall. Due to the greater variability at lower values of Low Temperature, however, the prediction uncertainty is greater at those values than it is at higher values of Low Temperature.

Answer: The slope of the regression line. The intercept of the regression line.

The slope depends on the standard deviation of low_temp. Note that, by properties of variance:

\text{Var}(1.8 \cdot \texttt{low\_temp} + 32) = 1.8^2 \text{Var}(\texttt{low\_temp})

Hence the new standard deviation of low_temp_f is

\text{SD}_{(°F)} = \sqrt{1.8^2 \text{Var}(\texttt{low\_temp})} = 1.8 \cdot \text{SD}_{(°C)}

which is different.

The intercept is also affected by the transformation because it uses the mean of low_temp_f. The mean using Fahrenheit is clearly different from the mean when using Celsius:

\text{mean}_{(°F)} = \text{mean}(1.8 \cdot \texttt{low\_temp} + 32) = 1.8 \cdot \text{mean}(\texttt{low\_temp}) + 32 = 1.8 \cdot \text{mean}_{(°C)} + 32

By definition, the correlation is the average value of the product of x and y when both are measured in standard units. Let x_{(°F)} = 1.8 \cdot x_{(°C)} + 32 be the temperature in Fahrenheit and x_{(°C)} be the temperature in Celsius. Let z_{(°F)} and z_{(°C)} be the standard units of x_{(°F)} and x_{(°C)}, respectively. Observe that:

z_{(°F)} = \frac{x_{(°F)} - \text{mean}_{(°F)}}{\text{SD}_{(°F)}} = \frac{x_{(°F)} - 1.8 \cdot \text{mean}_{(°C)} - 32}{1.8 \cdot \text{SD}_{(°C)}} = \frac{1.8 \cdot x_{(°C)} + 32 - 1.8 \cdot \text{mean}_{(°C)} - 32}{1.8 \cdot \text{SD}_{(°C)}} = \frac{x_{(°C)} - \text{mean}_{(°C)}}{\text{SD}_{(°C)}} = z_{(°C)}

Therefore, the standard units are equal after the transformation, and thus the correlation will also stay the same.

Note that errors are computed using the predicted values. Since we’re predicting snowfall and only transforming the low_temp values, our actual snowfall values (predicted and actual) will not change. Therefore, RMSE will stay the same if we only transform the temperature variable.

As before, we are not transforming our snowfall values. We are only transforming temperature. Therefore, our predictions will stay the same.

Answer: A predicted snowfall value at 1°C produced by one bootstrap regression line.

In essence, the question is asking us to interpret what the sum of the slope and intercept is. In our regression line,

\hat{y} = mx + b

where \hat{y} is the predicted snowfall, x is the Lowest Temperature, m is the slope, and b is the intercept. Notice that if we set x = 1, then we get m + b, which is precisely the sum of the slope and intercept. Since x = 1 refers to the temperature, this means that the sum of the slope and intercept is the predicted snowfall when temperature is 1°C. Since we’re bootstrapping, this is one such prediction made by producing a bootstrapped regression line.

Answer: 1 - \left(\dfrac{4}{5}\right)^5

It is easier to find the probability that it doesn’t snow on all 5 days. The probability that it doesn’t snow on any given day is described by two disjoint cases.

1. There is no precipitation at all. Since the probability that there is precipitation is 2/5, by complement rule we get:

\mathbb{P}(\text{no precipitation}) = 1 - \mathbb{P}(\text{precipitation}) = 1 - \frac{2}{5} = \frac{3}{5}

2. There is precipitation, but it rains instead of snows. This is given by the event that there is precipitation and, given that there is precipitation, it rains. In other words:

\mathbb{P}(\text{rains}) = \mathbb{P}(\text{precipitation}) \cdot \mathbb{P}(\text{rain} \mid \text{precipitation}) = \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5}

Since the two cases are disjoint (rain requires precipitation), we can sum the probabilities of the two cases and get

\mathbb{P}(\text{no snow}) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}.

(Note that, by complement rule, \mathbb{P}(\text{snow}) = 1/5.)

Thus, the probability that it doesn’t snow on 5 days in a row is

\mathbb{P}(\text{no snow for 5 days}) = \left(\frac{4}{5}\right)^5

and by complement rule, the probability that it snows on at least one of those 5 days is:

\mathbb{P}(\text{snow on at least 1 of 5 days}) = 1 - \left(\frac{4}{5}\right)^5

Answer: \left(\dfrac{1}{5}\right)\left(\dfrac{4}{5}\right)^4

It helps to use our computations from part (a). We can interpret the event as follows:

• NO snow on Monday \Rightarrow \mathbb{P}(\text{no snow}) = \dfrac{4}{5}
• and NO snow on Tuesday \Rightarrow \mathbb{P}(\text{no snow}) = \dfrac{4}{5}
• and NO snow on Wednesday \Rightarrow \mathbb{P}(\text{no snow}) = \dfrac{4}{5}
• and NO snow on Thursday \Rightarrow \mathbb{P}(\text{no snow}) = \dfrac{4}{5}
• and snow on Friday \Rightarrow \mathbb{P}(\text{snow}) = \dfrac{1}{5}

This results in

\mathbb{P}(\text{snow only on Friday}) = \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{1}{5} = \left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)

Answer: 5\left(\dfrac{1}{5}\right)\left(\dfrac{4}{5}\right)^4 = \left(\dfrac{4}{5}\right)^4

In part (b), we computed the probability that it snows only on Friday. In part (c), the event is more general in that any of the 5 days can be considered. Observe that the probability in part (b) is not unique to Friday, but can be used to describe the probability of it snowing on exactly one day of the week, for any specific day of the week. Thus the probability that it snows on exactly one day, without choosing which day in advance, is

5\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^4 = \left(\frac{4}{5}\right)^4