Answer: The median mass of penguins is less than the average mass of penguins

This is a distribution that is skewed to the right, so mean is greater than median.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: 17.355

Recall, Chebyshev’s inequality states that No matter what the shape of the distribution is, the proportion of values in the range “average ± z SDs” is at least 1 - \frac{1}{z^2}.

To approach the problem, we’ll start by converting 3276 grams and 5124 grams to standard units. Doing so yields \frac{3276 - 4200}{840} = -1.1, similarly, \frac{5124 - 4200}{840} = 1.1. This means that 3276 is 1.1 standard deviations below the mean, and 5124 is 1.1 standard deviations above the mean. Thus, we are calculating the proportion of values in the range “average ± 1.1 SDs”.

When z = 1.1, we have 1 - \frac{1}{z^2} = 1 - \frac{1}{1.1^2} \approx 0.173553719, which as a percentage rounded to three decimal places is 17.355\%.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 75%

Recall: proportion with z SDs of the mean

Percent in Range	All Distributions (via Chebyshev’s Inequality)	Normal Distributions
\text{average} \pm 1 \ \text{SD}	\geq 0\%	\approx 68\%
\text{average} \pm 2\text{SDs}	\geq 75\%	\approx 95\%
\text{average} \pm 3\text{SDs}	\geq 88\%	\approx 99.73\%

To approach the problem, we’ll start by converting 3276 grams and 5124 grams to standard units. Doing so yields \frac{1680 - 4200}{840} = -3, similarly, \frac{5880 - 4200}{840} = 2. This means that 1680 is 3 standard deviations below the mean, and 5880 is 2 standard deviations above the mean.

Proportion of values in [-3 SUs, 2 SUs] >= Proportion of values in [-2 SUs, 2 SUs] >= 75% (Since we cannot assume that the distribution is normal, we look at the All Distributions (via Chebyshev’s Inequality) column for proportion).

Thus, at least 75% of the penguins have a mass between 1680 grams and 5880 grams.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: No

The shape of the distribution does not change since we are scaling the x values for all data.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer: (a) 16 (b) 84

Recall, np.percentile(array, p) computes the pth percentile of the numbers in array. To compute the 68% CI, we need to know the percentile of left tail and right tail.

left percentile = (1-0.68)/2 = (0.32)/2 = 0.16 so we have 16th percentile

right percentile = 1-((1-0.68)/2) = 1-((0.32)/2) = 1-0.16 = 0.84 so we have 84th percentile

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: Option 4 (If we created many confidence intervals using the same method, approximately 68% of them would contain the mean weight of all penguins in Antarctica.)

Recall, what a k% confidence level states is that approximately k% of the time, the intervals you create through this process will contain the true population parameter.

In this question, our population parameter is the mean weight of all penguins in Antarctica. So 86% of the time, the intervals you create through this process will contain the mean weight of all penguins in Antarctica. This is the same as Option 4. However, it will be false if we state it in the reverse order (Option 1) since our population parameter is already fixed.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: The median mass of penguins is less than the average mass of penguins

This is a distribution that is skewed to the right, so mean is greater than median.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: The number of penguins with a mass of at most 3500 grams is greater than the number of penguins with a mass of at least 5500 grams.

Recall, a histogram has intervals on the axis, so we cannot know the frequency of an exact value. Thus, we cannot conclude statements 1, 3, 4. Since the frequency of an exact value is unknown, for statement 3, it is possible that all numbers we have in this distribution are even. Although in the graph, we are only given frequency rather than number, we can justify statement 2 by comparing the area in the left side of 3500, and the area in the right side of 5500. You can either estimate by visually comparing the areas of both parts or compute the area sum of both sides by estimating the bars’ height and windth.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: 17.355

Recall, Chebyshev’s inequality states that No matter what the shape of the distribution is, the proportion of values in the range “average ± z SDs” is at least 1 - \frac{1}{z^2}.

To approach the problem, we’ll start by converting 3276 grams and 5124 grams to standard units. Doing so yields \frac{3276 - 4200}{840} = -1.1, similarly, \frac{5124 - 4200}{840} = 1.1. This means that 3276 is 1.1 standard deviations below the mean, and 5124 is 1.1 standard deviations above the mean. Thus, we are calculating the proportion of values in the range “average ± 1.1 SDs”.

When z = 1.1, we have 1 - \frac{1}{z^2} = 1 - \frac{1}{1.1^2} \approx 0.173553719, which as a percentage rounded to three decimal places is 17.355\%.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 75%

Recall: proportion with z SDs of the mean

Percent in Range	All Distributions (via Chebyshev’s Inequality)	Normal Distributions
\text{average} \pm 1 \ \text{SD}	\geq 0\%	\approx 68\%
\text{average} \pm 2\text{SDs}	\geq 75\%	\approx 95\%
\text{average} \pm 3\text{SDs}	\geq 88\%	\approx 99.73\%

To approach the problem, we’ll start by converting 3276 grams and 5124 grams to standard units. Doing so yields \frac{1680 - 4200}{840} = -3, similarly, \frac{5880 - 4200}{840} = 2. This means that 1680 is 3 standard deviations below the mean, and 5880 is 2 standard deviations above the mean.

Proportion of values in [-3 SUs, 2 SUs] >= Proportion of values in [-2 SUs, 2 SUs] >= 75% (Since we cannot assume that the distribution is normal, we look at the All Distributions (via Chebyshev’s Inequality) column for proportion).

Thus, at least 75% of the penguins have a mass between 1680 grams and 5880 grams.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: No

The shape of the distribution does not change since we are scaling the x values for all data.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer: (a) masses, 2, replace=False (b) abs(two_penguins[0] - two_penguins[1])<=50

1. Recall, np.random.choice( ) can have three parameters array, n, replace=False, and returns n elements from the array at random, without replacement. We are randomly choosing 2 different penguins from the masses array, so we are using np.random.choice( ) without replacement.
2. We want to count the number of pairs of penguins that have body masses difference within 50 grams, so we are using the index to access the two penguins generated from two_penguins and calculating their absolute difference with abs(). And in this if condition, we only want to have penguins with absolute difference less than or equal to 50, so we write a <= condition to justify whether the generated pairs of penguins fulfill this requirement.

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Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: Option 2

Recall, according to the Central Limit Theorem (CLT), the probability distribution of the sum or mean of a large random sample drawn with replacement will be roughly normal, regardless of the distribution of the population from which the sample is drawn.

Thus, our graph should have a normal distribution. We eliminate Option 4.

Recall that the standard normal curve has inflection points at z = +-1, which is 68% proportion of a normal distribution.(inflection point is where a curve goes from “opening down” to “opening up”) Since we have a confidence intervel of 68% in this question, by looking at the inflection point, we can eliminate Option 3

To compute the SD of the sample mean’s distribution, when we don’t know the population’s SD, we can use the sample’s SD (840): \text{SD of Distribution of Possible Sample Means} \approx \frac{\text{Sample SD}}{\sqrt{\text{sample size}}} = \frac{840}{\sqrt{330}} \approx 46.24

Recall: proportion with z SDs of the mean

Percent in Range	All Distributions (via Chebyshev’s Inequality)	Normal Distributions
\text{average} \pm 1 \ \text{SD}	\geq 0\%	\approx 68\%
\text{average} \pm 2\text{SDs}	\geq 75\%	\approx 95\%
\text{average} \pm 3\text{SDs}	\geq 88\%	\approx 99.73\%

In this question, we want 68% confidence interval, given that the distribution of sample mean is roughly normal, our CI should have range \text{sample mean} \pm 1 \ \text{SD}. Thus, the interval is approximately [4200-46.24 = 4153.76, 4200+46.24=4246.24]. We compare the 68% CI in Option 1, 2 and we choose Option 2 since it has a 68% CI with approximately the same interval.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answer: (a) 16 (b) 84

Recall, np.percentile(array, p) computes the pth percentile of the numbers in array. To compute the 68% CI, we need to know the percentile of left tail and right tail.

left percentile = (1-0.68)/2 = (0.32)/2 = 0.16 so we have 16th percentile

right percentile = 1-((1-0.68)/2) = 1-((0.32)/2) = 1-0.16 = 0.84 so we have 84th percentile

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: Option 4 (If we created many confidence intervals using the same method, approximately 68% of them would contain the mean weight of all penguins in Antarctica.)

Recall, what a k% confidence level states is that approximately k% of the time, the intervals you create through this process will contain the true population parameter.

In this question, our population parameter is the mean weight of all penguins in Antarctica. So 86% of the time, the intervals you create through this process will contain the mean weight of all penguins in Antarctica. This is the same as Option 4. However, it will be false if we state it in the reverse order (Option 1) since our population parameter is already fixed.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: \frac{1}{20}

In any density histogram, the total area of all bars is 1. This histogram has five bars, each of which has a width of 1 (e.g. 3 - 2 = 1). Since \text{Area} = \text{Height} \cdot \text{Width}, we have that the area of each bar is equal to its height. So, the total area of the histogram in this case is the sum of the heights of each bar:

\text{Total Area} = x + 3x + 12x + 3x + x = 20x

Since we know that the total area is equal to 1, we have

20x = 1 \implies \boxed{x = \frac{1}{20}}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: 7,500

First, it’s a good idea to understand what the integer we’re trying to find actually means in the context of the information provided. In this case, it’s the number of sample means that are greater than v2 and less than or equal to v4. Here’s how to arrive at that conclusion:

1. First, note that sample_means is an array of length 10,000.
2. sample_means > v2 and sample_means <= v4 are both Boolean arrays of length 10,000.
3. (sample_means > v2) & (sample_means <= v4) is also a Boolean array of length 10,000, which contains True for every sample mean that is greater than v2 and less than or equal to v4 and False for every other sample mean.
4. Finally, np.count_nonzero((sample_means > v2) & (sample_means <= v4)) is a number between 0 and 10,000, corresponding to the number of True elements in the array (sample_means > v2) & (sample_means <= v4).

Remember, the histogram we’re looking at visualizes the distribution of the 10,000 values that result from calling f on every value in sample_means. To proceed, we need to understand how the function f decides what value to return for a given input, v:

• If the input v is greater than v2, then the first two conditions (v <= v1 and v <= v2) are False, and so the only possible values of f(v) are 0, 1, or 2.
• If the input v is less than or equal to v4, the only possible values of f(v) are -2, -1, 0, 1.
• Thus, if the input v is greater than v2 and less than or equal to v4, the only possible values of f(v) are 0 and 1.

Now, our job boils down to finding the number of values in the visualized distribution that are equal to 0 or 1. This is equivalent to finding the number of values that fall in the [0, 1) and [1, 2) bins – since f only returns integer values, the only value in the [0, 1) bin is 0 and the only value in the [1, 2) bin is 1 (remember, histogram bins are left-inclusive and right-exclusive).

To do this, we need to find the proportion of values in those two bins, and multiply that proportion by the total number of values (10,000).

We know that the area of a bar is equal to the proportion of values in that bin. We also know that, in this case, the area of each bar is equal to its height, since the width of each bin is 1. Thus, the proportion of values in a given bin is equal to the height of the corresponding bar. As such, the proportion of values in the [0, 1) bin is 12x, and the proportion of values in the [1, 2) bin is 3x, meaning the proportion of values in the histogram that are equal to either 0 or 1 is 12x + 3x = 15x.

In the previous subpart, we found that x = \frac{1}{20}, so the proportion of values in the histogram that are equal either 0 or 1 is 15 \cdot \frac{1}{20} = \frac{3}{4}*, and since there are 10,000 values being visualized in the histogram total, \frac{3}{4} \cdot 10,000 = 7,500 of them are equal to either 0 or 1.

Thus, 7,500 of the values in sample_means are greater than v2 and less than or equal to v4, so np.count_nonzero((sample_means > v2) & (sample_means <= v4)) evaluates to 7,500.

Note: It’s possible to answer this subpart without knowing the value of x, i.e. without answering the previous subpart. The area of the [0, 1) and [1, 2) bars is 15x, and the total area of the histogram is 20x. So, the proportion of the area in [0, 1) or [1, 2) is \frac{15x}{20x} = \frac{15}{20} = \frac{3}{4}, which is the same value we found by substituting x = \frac{1}{20} into 15x.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.

Answer: 35.84

The table provided above tells us the proportion of values within u standard deviations of the mean in a normal distribution, for various values of u. For instance, it tells us that the proportion of values within 1.28 standard deviations of the mean in a normal distribution is 0.8.

Let’s reflect on what we know at the moment:

• The distribution of the sample mean is roughly normal, by the Central Limit Theorem. Normal distributions are symmetric, and have a “peak” at the center. The histogram above is also symmetric and has its peak at its center.
• The proportion of values in the histogram that are equal to 0 is 12x = 12 \cdot \frac{1}{20} = 0.6.
• The function f returns 0 for all inputs that are greater than v2 and less than or equal to v3. This, combined with the fact above, tells us that the proportion of sample means between v2 (exclusive) and v3 (inclusive) is 0.6.
• From the table provided, we know that in a normal distribution, the proportion of values within 0.84 standard deviations of the mean is 0.6.

Combining the facts above, we have that v2 is 0.84 standard deviations below the mean of the sample mean’s distribution and v3 is 0.84 standard deviations above the mean of the sample mean’s distribution.

The sample mean’s distribution has the following characteristics:

\begin{align*} \text{Mean of Distribution of Possible Sample Means} &= \text{Population Mean} = 35 \\ \text{SD of Distribution of Possible Sample Means} &= \frac{\text{Population SD}}{\sqrt{\text{Sample Size}}} = \frac{10}{\sqrt{100}} = 1 \end{align*}

0.84 standard deviations above the mean of the sample mean’s distribution is:

35 + 0.84 \cdot \frac{10}{\sqrt{100}} = 35 + 0.84 \cdot 1 = \boxed{35.84}

So, the value of v3 is 35.84.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 9%.

Answer: 33.35

The table provided tells us that in a normal distribution, 90% of values are within 1.65 standard deviations of the mean. Since normal distributions are symmetric, it also means that 5% of values are above 1.65 standard deviations of the mean and, more importantly, 5% of values are below 1.65 standard deviations of the mean.

The 5th percentile of a distribution is the smallest value that is greater than or equal to 5% of values, so in this case the 5th percentile is 1.65 SDs below the mean. As in the previous subpart, the mean and SD we are referring to are the mean and SD of the distribution of sample means (sample_means), which we found to be 35 and 1, respectively.

1.65 standard deviations below this mean is

35 - 1.65 \cdot 1 = \boxed{33.35}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: 8

First, we can calculate the standard deviation of the sample using the given formula: \sqrt{0.2\cdot(1-0.2)} = \sqrt{0.16}= 0.4. Additionally, we know that the width of a 95% confidence interval for a population mean (including a proportion) is approximately \frac{4 * \text{sample SD}}{\sqrt{n}}, since 95% of the data of a normal distribution falls within two standard deviations of the mean on either side. Now, plugging the sample standard deviation into this formula, we can set this expression equal to the given formula for the width of the confidence interval: \frac{c}{5 \sqrt{n}} = \frac{4 * 0.4}{\sqrt{n}}. We can multiply both sides by \sqrt{n}, and we’re left with \frac{c}{5} = 4 * 0.4. Now, all we have to do is solve for c by multiplying both sides by 5, which gives c = 8.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: 1600

Here, we have to use the formula for the endpoints of the 95% confidence interval to see what the largest value of n is such that 0.22 will be contained in the interval. The endpoints are given by \text{sample mean} \pm 2 * \frac{\text{sample SD}}{\sqrt{n}}. Since the null hypothesis is that the proportion is 0.22 (which is greater than our sample mean), we only need to work with the right endpoint for this question. Plugging in the values that we have, the right endpoint is given by 0.2 + 2 * \frac{0.4}{\sqrt{n}}. Now we must find a value of n which satisfies the inequality 0.2 + 2 * \frac{0.4}{\sqrt{n}} \geq 0.22, and since we’re looking for the smallest such value of n (i.e, the last n for which this inequality holds), we can simply set the two sides equal to each other, and solve for n. From 0.2 + 2 * \frac{0.4}{\sqrt{n}} = 0.22, we can subtract 0.2 from both sides, then multiply both sides by \sqrt{n}, and divide both sides by 0.02 (from 0.22 - 0.2). This yields \sqrt{n} = \frac{2 * 0.4}{0.02} = \sqrt{n} = 40, which implies that n is 1600.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 21%.

Answer: 1.95

The area we want to find is shown below in two colors. We can find the area in each half of the armchair curve separately and add the results.

For the yellow area, we know that the area within 2 standard deviations of the mean on the standard normal curve is 0.95. The remaining 0.05 is split equally on both sides, so the yellow area is 0.975.

The blue area is the same by symmetry so the total shaded area is 0.975*2 = 1.95.

Equivalently, we can use the fact that the total area under the armchair curve is 2, and the amount of unshaded area on either side is 0.025, so the total shaded area is 2 - (0.025*2) = 1.95.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 1

The area we want to find is shown below in two colors.

As we saw in Problem 12.2, the point on the left half of the armchair curve that corresponds to 8.37 is 0.37. This means that if we move the blue area from the right half of the armchair curve to the left half, it will fit perfectly, as shown below.

Therefore the total of the blue and yellow areas is the same as the area under one standard normal curve, which is 1.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: z>4 or z>=4

The body of the function contains an if statement followed by a return statement, which executes only when the if condition is false. In that case, the function returns scipy.stats.norm.cdf(z), which is the area under the standard normal curve to the left of z. When z is in the left half of the armchair curve, the area under the armchair curve to the left of z is the area under the standard normal curve to the left of z because the left half of the armchair curve is a standard normal curve, centered at 0. So we want to execute the return statement in that case, but not if z is in the right half of the armchair curve, since in that case the area to the left of z under the armchair curve should be more than 1, and scipy.stats.norm.cdf(z) can never exceed 1. This means the if condition needs to correspond to z being in the right half of the armchair curve, which corresponds to z>4 or z>=4, either of which is a correct solution.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: 1+scipy.stats.norm.cdf(z-8)

This blank should contain the value we want to return when z is in the right half of the armchair curve. In this case, the area under the armchair curve to the left of z is the sum of two areas:

1. the area under the entire left half of the armchair curve, which is 1, and
2. the area under the portion of the right half of the armchair curve that falls to the left of z.

Since the right half of the armchair curve is just a standard normal curve that’s been shifted to the right by 8 units, the area under that normal curve to the left of z is the same as the area to the left of z-8 on the standard normal curve that’s centered at 0. Adding the portion from the left half and the right half of the armchair curve gives 1+scipy.stats.norm.cdf(z-8).

For example, if we want to find the area under the armchair curve to the left of 9, we need to total the yellow and blue areas in the image below.

The yellow area is 1 and the blue area is the same as the area under the standard normal curve (or the left half of the armchair curve) to the left of 1 because 1 is the point on the left half of the armchair curve that corresponds to 9 on the right half. In general, we need to subtract 8 from a value on the right half to get the corresponding value on the left half.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 54%.

Answer: area_left_of(y) - area_left_of(x)

In general, we can find the area under any curve between x and y by taking the area under the curve to the left of y and subtracting the area under the curve to the left of x. Since we have a function to find the area to the left of any given point in the armchair curve, we just need to call that function twice with the appropriate inputs and subtract the result.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer: 12

cond is a Series that contains True for each row in usa where "Jul" is greater than or equal to 370 and less than 430. mystery, then, is the percentage of values in usa in which cond is True. This is because np.count_nonzero(cond) is the number of Trues in cond, np.count_nonzero(cond) / usa.shape[0] is the proportion of values in cond that are True, and 100 * np.count_nonzero(cond) / usa.shape[0] is the percentage of values in cond that are True. Our goal here, then, is to use the histogram to find the percentage of values in the histogram between 370 (inclusive) and 430 (exclusive).

We know that in histograms, the area of each bar is equal to the proportion of data points that fall within its bin’s range. Conveniently, there’s only one bar we need to look at – the one corresponding to the bin [370, 430). That bar has a width of 430 - 370 = 60 and a height of 0.002. Then, the area of that bar – i.e. the proportion of values that are between 370 (inclusive) and 430 (exclusive) is:

\text{proportion} = \text{area} = \text{height} \cdot \text{width} = 0.002 \cdot 60 = 0.12

This means that the proportion of values in [370, 430) is 0.12, which means that the percentage of values in [370, 430) is 12%, and that mystery evaluates to 12.

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: 250

In the previous part, we learned that the proportion of cities in the usa DataFrame in the interval [370, 430) (i.e. that have between 370 and 430 sunshine hours in July) is 0.12. To use the fact that there are 5 more cities in the interval [370, 430) than there are in the interval [270, 290), we need to first find the proportion of cities in the interval [270, 290). To do so, we look at the [270, 290) bin, which has a width of 290 - 270 = 20 and a height of 0.005:

\text{proportion} = 0.005 \cdot 20 = 0.10

We are told that there are 5 more cities in the [370, 430) interval than there are in the [270, 290) interval. Given the proportions we’ve computed, we have that:

\text{difference in proportions} = 0.12 - 0.1 = 0.02

If 0.02 \cdot \text{number of cities} is 5, then \text{number of cities} = 5 \cdot \frac{1}{0.02} = 5 \cdot 50 = 250.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 49%.

Answer: m = 0

When we standardize a dataset, the mean of the resulting values is always 0 and the standard deviation of the resulting values is always 1. This tells us right away that the answer is m = 0. Intuitively, we know that a value in standard units represents the number of standard deviations that value is above or below the mean of the column it came from. m is equal to the mean of the column it came from, so m in standard units is 0.

If we’d like to approach this more algebracically, we can remember the formula for converting a value x_i from a column x to standard units:

x_{i \: \text{(su)}} = \frac{x_i - \text{mean of } x}{\text{SD of } x}

Let x be the column (i.e. Series) containing the mean number of sunshine hours in July for all US cities in sun. m, by definition, is the mean of x. Then,

m_{\text{(su)}} = \frac{m - \text{mean of } x}{\text{SD of } x} = \frac{m - m}{\text{SD of }x} = 0

Given that m is the mean of column x, the numerator of m_\text{(su)} is 0, and hence m_\text{(su)} is 0.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: s = 1

As mentioned in the previous solution, when we standardize a dataset, the mean of the resulting values is always 0 and the standard deviation of the resulting values is always 1.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

Answer: -1 < d < 0

In the histogram, we see that the distribution of the number of sunshine hours in July for all US cities in sunis skewed right, or has a right tail. This means that this distribution’s mean is dragged in the direction of its tail and is larger than its median. Since the mean in standard units is 0, and the median is less than the mean, the median in standard units must be negative. There’s no property that states that the median is exactly -1, and the median is only slightly less than the mean, which means that it must be the case that -1 < d < 0.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: False

The original histogram depicting the distribution of the number of sunshine hours in July for all US cities is right-skewed. When data is converted to standard units, the shape of the distribution does not change. Therefore, if the original data is right-skewed, the standardized data will also be right-skewed.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: 306

First, we need to find the position of the 80th percentile using the rule from class:

h = \left(\frac{80}{100}\right) \cdot 15 = \frac{4}{5} \cdot 15 = 12

Since 12 is an integer, we don’t need to round up, so k = 12. Starting from the right-most number, which is the smallest number and hence position 1 here, the 12th number is 306.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: 40

The first step is to find the position of 257 in the collection when we start at position 1, which is 6. Since there are 15 values total, this means that 257 is the smallest value that is greater than or equal to 40% of the values.

If we set p to be any number larger than 40, say, 41, then 257 won’t be larger than p\% of the values in the collection; thus, the largest positive integer value of p that makes 257 the pth percentile is 40.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 30%.

Answer: 34

Let’s look at the next number down from 257, which is 255. 255 is the 5th number out of 15, so it is the smallest number that is greater than or equal to 33.333% of the values. This means the 33rd percentile is also 255, since 33.333 > 33. However, 255 is not greater than or equal to 34% of the values, which makes the 34th percentile 257. Therefore, 34 is the smallest integer value of p such that the pth percentile is 257.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 21%.

Answer: 91%

First, we need to find the number of standard deviations above the mean 300 is, and the number of standard deviations below the mean 200 is.

z = \frac{300 - 250}{15} = \frac{50}{15} = \frac{10}{3}

The above equation tells us that 300 is \frac{10}{3} standard deviations above the mean; you can verify that 200 is the same number of standard deviations below the mean. Chebyshev’s inequality tells us the proportion of values within z SDs of the mean is at least 1 - \frac{1}{z^2}, which here is:

1 - \frac{1}{\left(\frac{10}{3}\right)^2} = 1 - \frac{9}{100} = 0.91

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer:

• (a) 2280
• (b) 2320

scipy.stats.norm.cdf(-0.8) tells us that from the bounds of (-\inf, -0.8], the normal distribution has an area of 0.21. Therefore, if we take it to the other side from [0.8, \inf), it also has an area of 0.21 due to the symmetrical property of the normal distribution. This means that the interval between [-0.8, 0.8] has an area of 1 - 0.21 - 0.21 = 0.58: the confidence interval we are aiming to find.

In the question, we are given the standard deviation of security deposits in a sample, meaning we need to find the standard deviation for the population. To find this, we use the following formula and compute:

\frac{\text{SD of sample}}{\sqrt{\text{sample size}}} = \frac{500}{\sqrt{400}} = \frac{500}{20} = 25.

Now that we have the population standard deviation, we can calculate the endpoints of the confidence interval.

Left endpoint: 2300 - \frac{4}{5} \cdot 25 = 2280

Right endpoint: 2300 + \frac{4}{5} \cdot 25 = 2320

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 29%.

Answer: The mean is approximately equal to the median.

• The histogram appears to be approximately symmetric, with the peak near the center of the distribution.
• For symmetric distributions, the mean and the median are approximately equal because the data is evenly distributed around the central point.
• If the distribution were skewed:
  • A right-skewed distribution would have the mean significantly larger than the median.
  • A left-skewed distribution would have the mean significantly smaller than the median.
• In this case, there is no visible skew, so the correct answer is that the mean is approximately equal to the median.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: 35 = 5\cdot7

• The height of the bin [200, 225) on the density histogram is approximately 7 times the height of the bin [275, 300).
  
• Since the number of stages in a bin is proportional to the bin’s height, the number of stages in [200, 225) is 35 = 5\cdot7.

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Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: \frac{15}{16}

Using Chebyshev’s inequality, we know at least 1 - \frac{1}{z^2} of the data lies within z SDs. Here, z = 4 so we know 1 - \frac{1}{16} = \frac{15}{16} of the data lie in that range.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: 68%

We know about 68% of values lie within 1 standard deviation of the mean of any normal distribution. The distribution of means of samples of size 25 from this dataset is normally distributed with mean 200km and SD \frac{50}{\sqrt{25}} = 10, so 190km to 210km contains 68% of the values.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: Option 3, Option 4, and Option 7

Option 1:
Incorrect. Confidence intervals describe the uncertainty in estimating the population mean, not the proportion of data points. A 95% confidence interval does not imply that 95% of individual stage distances fall between 190 km and 210 km.

Option 2:
Incorrect. Confidence intervals are based on the sampling process, not probability. Once the interval is calculated, the true mean is either inside or outside the interval. We cannot assign a probability to this.

Option 3:
Correct. This is the standard interpretation of confidence intervals: “We are 95% confident that the true mean lies within the interval.”

Option 4:
Correct. Given a sample size of 100 and population standard deviation of 50, the confidence interval ([190, 210]) is consistent with the calculation using the rule of thumb that a 95% confidence interval is approximately 2 standard deviations apart from the mean.

For a 95% confidence interval, the range can be approximated as:

\left[\text{sample mean} - 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}}, \text{sample mean} + 2\cdot \frac{\text{sample SD}}{\sqrt{\text{sample size}}} \right]

Substituting the given values:

• Sample mean = 200
• Sample standard deviation = 50
• Sample size = 100 \text{CI} = \left[ 200 - 2\cdot \frac{50}{\sqrt{100}}, 200 + 2\cdot \frac{50}{\sqrt{100}} \right] Simplify: \text{CI} = \left[ 200 - 2\cdot 5, 200 + 2\cdot 5 \right] \text{CI} = [190, 210]

Option 5:
Incorrect. refer to option 4

Option 6:
Incorrect. The wording “exactly 95%” is overly precise. In practice, confidence intervals are based on the sampling process, and we use “approximately” or “roughly” 95%.

Option 7:
Correct. By definition of a confidence interval, if we repeatedly sampled and constructed 95% confidence intervals, roughly 95% of them would contain the true mean.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer:

• (i): thinner
  • Histogram B would appear thinner because larger sample sizes reduce the variability of the sample means. With a sample size of 1000 (compared to 100 for Histogram A), the standard error decreases, leading to a narrower distribution.
• (ii): not at all
  • Histogram B would not shift left or right because the sample mean does not depend on the sample size. Both histograms have the same mean, as they are based on the same population.
• (iii): unchanged
  • The mean remains unchanged because the mean of the sampling distribution (the population mean) does not depend on sample size.
• (iv): smaller
  • The standard deviation of Histogram B is smaller because \text{SD of Distribution of Possible Sample Means} = \frac{\text{Population SD}}{\sqrt{\text{sample size}}}
  Increasing sample size from 100 to 1000 decreases the Sample standard deviation as the Population standard deviation remains unchanged, leading to a smaller standard deviation for the sampling distribution.

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: Vivek, Hector, Clara

To compare Vivek, Hector, and Clara’s relative performance we want to compare their Z scores to handle standardization. For Vivek, his Z score is (83-75) / 6 = 4/3. For Hector, his score is (77-70) / 5 = 7/5. For Clara, her score is (80-75) / 3 = 5/3. Ranking these, 5/3 > 7/5 > 4/3 which yields the result of Vivek, Hector, Clara.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 92.284\%

Recall, Chebyshev’s inequality states that the proportion of values within z standard deviations of the mean is at least 1 - \frac{1}{z^2}.

To approach the problem, we’ll start by converting 25 points per game to standard units. Doing so yields \frac{25 - 7}{5} = 3.6. This means that 25 is 3.6 standard deviations above the mean. The value 3.6 standard deviations below the mean is 7 - 3.6 \cdot 5 = -11, so when we use Chebyshev’s inequality with z = 3.6, we will get a lower bound on the proportion of values between -11 and 25. However, as the question tells us, points per game must be non-negative, so in this case the proportion of values between -11 and 25 is the same as the proportion of values between 0 and 25 (i.e. the proportion of values less than or equal to 25).

When z = 3.6, we have 1 - \frac{1}{z^2} = 1 - \frac{1}{3.6^2} = 0.922839, which as a percentage rounded to three decimal places is 92.284\%. Thus, at least 92.284\% scored 25 or fewer points per game.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

Answer: 85

For reference, recall that we find the pth percentile of a collection of n numbers as follows:

1. Sort the collection in increasing order.
2. Define h to be p\% of n:

h = \frac{p}{100} \cdot n

3. If h is an integer, define k = h. Otherwise, let k be the smallest integer greater than h.

4. Take the kth element of the sorted collection (start counting from 1, not 0).

To start, it’s worth emphasizing that there are n = 14 numbers in aces total. 14.24 is at position 12 (when the positions are numbered 1 through 14).

Let’s try and find a value of p such that 14.24 is the pth percentile. To do so, we might try and find what “percentage” of the way through the distribution 14.24 is; doing so gives \frac{12}{14} = 85.71\%. If we follow the process outlined above with p = 85, we get that h = \frac{85}{100} \cdot 14 = 11.9 and thus k = 12, meaning that the 85th percentile is the number at position 12, which 14.24.

Let’s see what happens when we try the same process with p = 86. This time, we have h = \frac{86}{100} \cdot 14 = 12.04 and thus k = 13, meaning that the 86th percentile is the number at position 13, which is 14.81.

This means that the value of q is 85 – the 85th percentile is 14.24, while the 86th percentile is 14.81.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: 'Player'

Ideally, the index of a DataFrame is unique, so that we can use it to “identify” the rows. Here, each row is about a player, so 'Player' should be the index. 'Player' is the only column that is likely to be unique; it is possible that two players have the same name, but it’s still a better choice of index than the other three options, which are definitely not unique.

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Difficulty: ⭐️

The average score on this problem was 95%.

Answer: 'Player'

Ideally, the index of a DataFrame is unique, so that we can use it to “identify” the rows. Here, each row is about a player, so 'Player' should be the index. 'Player' is the only column that is likely to be unique; it is possible that two players have the same name, but it’s still a better choice of index than the other three options, which are definitely not unique.

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Difficulty: ⭐️

The average score on this problem was 95%.

Answer: 1

The median of a distribution is the value that is “halfway” through the distribution, i.e. the value such that half of the values in the distribution are larger than it and half the values in the distribution are smaller than it.

Visually, we’re looking for the location on the x-axis where we can draw a vertical line that splits the area of the histogram in half. While it’s impossible to tell the exact median of the distribution, since we don’t know how the values are distributed within the bars, we can get pretty close by using this principle.

Immediately, we can rule out 0.5, 0.75, 1.5, and 1.75, since they are too far from the “center” of the distribution (imagine drawing vertical lines at any of those points on the x-axis; they don’t split the distribution’s area in half). To decide between 1 and 1.25, we can use the fact that the distribution is right-skewed, meaning that its mean is larger than its median (intuitively, the mean is dragged in the direction of the tail, which is to the right). This means that the median should be less than the mean. We are given that the mean of the distribution is 1.25, so the median should be 1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: 0.5

To convert a value x to standard units (denoted by x_{\text{su}}), we use the following formula:

x_{\text{su}} = \frac{x - \text{mean of }x}{\text{SD of }x}

Let’s look at the first line given to us: In “original units”, Sabrina Ionescu had 3.5 turnovers per game. In standard units, her turnovers per game is 3.

Substituting the information we know into the above equation gives us:

3 = \frac{3.5 - 1.25}{\text{SD of }x}

In order to convert future values from original units to standard units, we’ll need to know \text{SD of }x, which we don’t currently but can obtain by rearranging the above equation. Doing so yields

\text{SD of }x = \frac{3.5-1.25}{3} = \frac{2.25}{3} = 0.75

Now, let’s look at the second line we’re given: In standard units, Sami Whitcomb’s turnovers per game is -1. How many turnovers per game did Sami Whitcomb have in original units? Round your answer to 3 decimal places.

We have all the information we need to convert Sami Whitcomb’s turnovers per game from standard units to original units! Plugging in the values we know gives us:

\begin{aligned} x_{\text{su}} &= \frac{x - \text{mean of }x}{\text{SD of }x} \\ -1 &= \frac{x - 1.25}{0.75} \\ -0.75 &= x - 1.25 \\ 1.25 - 0.75 &= x \\ x &= \boxed{0.5} \end{aligned}

Thus, in original units, Sami Whitcomb averaged 0.5 turnovers per game.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: -1.667

The smallest possible number of turnovers per game in original units is 0 (which a player would have if they never had a turnover – that would mean they’re really good!). To find the smallest possible turnovers per game in standard units, all we need to do is convert 0 from original units to standard units. This will involve our work from the previous subpart.

\begin{aligned} x_{\text{su}} &= \frac{x - \text{mean of }x}{\text{SD of }x} \\ &= \frac{0 - 1.25}{0.75} \\ &= -\frac{1.25}{0.75} \\ &= -\frac{5}{3} = \boxed{-1.667} \end{aligned}

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: -2/3

We know that the normal distribution is symmetric about the mean, and that the mean is the “peak” described in the graph. The inflection points occur one standard deviation above and below the mean (the peak), so a point which is one third of the way in between the first inflection point and the peak is -(1-\frac{1}{3}) = -\frac{2}{3} standard deviations from the mean. We can then use stats.norm.cdf(-2/3) to calculate the area under the curve to the left of this point.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

Answer: 68

Since we know that both slopes are normal distributions (just scaled and shifted), we can derive this answer by writing Keenan’s crouch point in terms of standard deviations from the mean. He typically crouches at 92 feet, whose distance from the mean (in standard deviations) is given by \frac{92 - 65}{12} = 2.25. So, all we need to do is find what number is 2.25 standard deviations from the mean in the Olympic mountain. This is given by 50 + (2.25 * 8) = 68

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: 0.3

We know that when Aaron reaches the mean (50), exactly 0.5 of the mountain’s area is behind him, since the mean and median are equal for normal distributions like this one. We also see that 54 is one half of a standard deviation away from the mean. So, all we have to do is find out what proportion of the area is within half a standard deviation of the mean. Using the 68-95-99.7 rule, we know that 68% of the values lie within one standard deviation of the mean to both the right and left side. So, this means 34% of the values are within one standard deviation on one side and at least 17% are within half a standard deviation on one side. Since the area is 1, the area would be 0.17. So, by the time Aaron reaches an x-coordinate of 54, 0.5 + 0.17 = 0.67 of the mountain is behind him. From here, we simply calculate the area in front by 1 - 0.67 = 0.33, so we conclude that approximately 0.3 of the area is in front of Aaron.

Note: As a clafrification, the 0.17 is an estimate, specifically, an underestimate, due to the shape of the normal distribution. Thse area under a normal distribution is not proportional to how many standard deviations far away from the mean you are.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 50%.

Answer: 1 - (scipy.stats.norm.cdf(1.5) - scipy.stats.norm.cdf(-2)) & scipy.stats.norm.cdf(-2) + scipy.stats.norm.cdf(-1.5)

• Option 1: This code calculates the probability that a value lies outside the range between z = -2 and z = 1.5, which corresponds to x \leq 10 or x \geq 80. This is done by subtracting the area under the normal curve between -2 and 1.5 from 1. This is correct because it accurately captures the combined probability in the left and right tails of the distribution.

• Option 2: This code multiplies the cumulative distribution function (CDF) at z = 1.75 by 2. This assumes symmetry around the mean and is used for intervals like |z| \geq 1.75, but that’s not what we want. The correct z-values for this problem are -2 and 1.5, so this option is incorrect.

• Option 3: This code adds the probability of z \leq -2 and z \geq 1.5, using the fact that P(z \geq 1.5) = P(z \leq -1.5) by symmetry. So, while the code appears to show both as left-tail calculations, it actually produces the correct total tail probability. This option is correct.

• Option 4: This is a static value with no basis in the z-scores of -2 and 1.5. It’s likely meant as a distractor and does not represent the correct probability for the specified conditions. This option is incorrect.

Answer: x = 30 and x = 70

Recall that the inflection points of a normal distribution are located one standard deviation away from the mean. In this problem, the mean is x = 50 and the standard deviation is 20, so the inflection points occur at x = 30 and x = 70. These are the points where the curve changes concavity and where the height is changing the fastest. Therefore, the correct answer is x = 30 and x = 70.

Answer: 0.00

We are told that the distribution is not necessarily normal. The mean is 130 and the standard deviation is 30. We’re asked for the minimum proportion of area between x = 100 and x = 190.

Since the distribution isn’t normal and we don’t know its shape, we can’t use the empirical rule (68-95-99.7) or z-scores. We might try using Chebyshev’s Inequality, but that only works for intervals that are equally far below the mean as above the mean. This interval is not like that (it’s 1 standard deviation below the mean and 2 above), so Chebyshev’s Inequality doesn’t apply. The most we can say using Chebyshev’s Inequality is that in the interval from 1 standard deviation below the mean to 1 standard deviation above the mean, we can get at least 1 - \frac{1}{0^2} = 0 percent of the data. We can’t make any additional guarantees. So, the minimum possible proportion of area is 0.00.