Answer: 350

Chebyshev’s inequality says something about how much data falls within a given number of standard deviations. The data that doesn’t fall in that range could be entirely below that range, entirely above that range, or split some below and some above. So the idea is that we should figure out the endpoints of the range for which Chebyshev guarantees that at least 375 transactions must fall. Then at most 25 might fall above that range. So we’ll fill in the blank with the upper limit of that range. Now, since there are 400 transactions, 375 as a proportion becomes \frac{375}{400} = \frac{15}{16}. That’s 1 - \frac{1}{16} or 1 - \left(\frac{1}{4}\right)^2, so we should use z=4 in the statement of Chebyshev’s inequality. That is, \frac{15}{16} proportion of the data falls within 4 standard deviations of the mean. The upper endpoint of that range is 70 (the mean) plus 4 \cdot 70 (four standard deviations), or 5 \cdot 70 = 350.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 30%.

Answer: Option B

Here, the main factor which we can use to identify the correct plot is the correlation coefficient. A correlation coefficient of -\frac{1}{4} indicates that the data will have a slight downward trend (values on the y axis will be lower as we go further right). This narrows it down to option A or option B, but option A appears to have too strong of a linear trend. We want the data to look more like a cloud than a line since the correlation is relatively close to zero, which suggests that option B is the more viable choice.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: f = \frac{1}{4}

This problem requires us to make a prediction using the regression line for a given x = 280. We can solve this problem using original units or standard units. Since 280 is a multiple of 70, and the mean and standard deviation are both 70, it’s pretty straightforward to convert 280 to 3 standard units, as \frac{(280-70)}{70} = \frac{210}{70} = 3. To make a prediction in standard units, all we need to do is multiply by r=-\frac{1}{4}, resulting in a predicted lifetime spending of =-\frac{3}{4} in standard units. Since we are told in the previous subpart that both the mean and standard deviation of lifetime spending are c dollars, then converting to original units gives c + -\frac{3}{4} \cdot c = \frac{1}{4} \cdot c, so f = \frac{1}{4}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: c = 32

We start with the formulas for the mean and intercept of the regression line, then set the mean and SD of x both to 70, and the mean and SD of y both to c, as well as the intercept b to 40. Then we can solve for c.

\begin{align*} m &= r \cdot \frac{\text{SD of } y}{\text{SD of }x} \\ b &= \text{mean of } y - m \cdot \text{mean of } x \\ m &= -\frac{1}{4} \cdot \frac{c}{70} \\ 40 &= c - (-\frac{1}{4} \cdot \frac{c}{70}) \cdot 70 \\ 40 &= c + \frac{1}{4} c \\ 40 &= \frac{5}{4} c \\ c &= 40 \cdot \frac{4}{5} \\ c &= 32 \end{align*}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: True

Standard units standardize the data into z scores. When converting to Z scores the scale of both the dependent and independent variables are the same, and consequently, the slope can at most increase by 1. Alternatively, according to the reference sheet, the slope of the regression line, when both variables are measured in standard units, is also equal to the correlation coefficient. And by definition, the correlation coefficient can never be greater than 1 (since you can’t have more than a ‘perfect’ correlation).

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Difficulty: ⭐️

The average score on this problem was 93%.

Answer: Options 1, 2, and 5 are correct.

1. The mean of the new distribution where weight is in pounds would be given by 2.2 \cdot \text{mean} in kilograms. Since the mean in kilograms is positive, this quantity must then increase if we multiply it by 2.2. Intuitively this should make sense as we are basically scaling all the (positive) values by 2.2 (a positive constant), so we expect the values to increase and thus the mean to increase. So, option 1 is correct.
2. The standard deviation of the new distribution where weight is in pounds would be given by 2.2 \cdot \text{standard deviation} in kilograms. Since the standard deviation in kilograms is positive, this quantity must then increase if we multiply it by 2.2. Intuitively this should make sense as we are scaling all the values by 2.2, so the difference between larger and smaller values will be greater once they are scaled. Thus, we expect the spread of our distribution to be larger, and the standard deviation to increase. So, option 2 is correct.
3. We are basically interested in the proportion of x_{i} that satisfy \vert \frac{x_{i} \cdot μ}{σ} \vert<3, with x_{i} being the values of "Weight", μ being the mean in kg and σ being the standard deviation in kg. Once we scale everything by 2.2 to convert from kilograms to pounds, we have that
   \vert \frac{2.2x_{i}-2.2μ}{2.2σ}\vert<3 → \frac{2.2}{2.2}\vert \frac{x~i~*μ}{σ}\vert<3 → \vert \frac{x_{i} \cdot μ}{σ}\vert<3.
   Notice that after we scaled it to pounds, the equation still ends up the exact same as the one in kg. Thus, the proportion of "Weight" within 3 standard deviations of the mean stays the same. Intuitively this should make sense because we are scaling everything by the same amount, so the proportion of points that are a specific number of standard deviations away from the mean should be the same, since the standard deviation and mean get scaled as well. So, option 3 is incorrect.
4. Recall that the correlation coefficient, r, is the average value of the product of two variables, when both variables are measured in standard units. In kilograms, "Weight"(kg) in standard units is \frac{x~i~−μ}{σ}. Similar to option 3, "Weight"(pounds) in standard units is \frac{2.2x_{i}-2.2μ}{2.2σ} = \frac{2.2}{2.2} \frac{x_{i} \cdot μ}{σ} = \frac{x_{i} \cdot μ}{σ}. Again, notice that the equation in pounds ends up the exact same as in kilograms. The same applies for "Height" in standard units. Since none of the variables change when measured in standard units, r doesn’t change. So, option 4 is incorrect.
5. Intuitively, this makes sense when we imagine a scatterplot with the y-axis (value being predicted) representing "Weight" and the x-axis representing "Height". We expect that the taller someone is, the heavier they will be. So we can expect a positive regression line slope between Weight and Height. When we convert Weight from kg to pounds, we are scaling every value in "Weight", making their values increase. When we scale the the weight values (y-values) to become bigger, we are making the regression slope even steeper, because an increase in Height (x) now corresponds to an even larger increase in Weight(y). So, option 5 is correct.
6. Similar to the last problem, we can also imagine a scatterplot except this time with "Weight" on the x-axis and "Height" on the y-axis. Since we are increasing the values of "Height", we can imagine stretching the x-axis’ values without changing the y-values, which makes the line more flat and therefore decreases the slope. So, option 6 is incorrect. Another approach to both 5 and 6 is to utilize the correlation coefficient r, which is equal to the slope times \frac{σ~y~}{σ~x~}. We know that multiplying a set values by a value greater than one increases the spread of the data which increases standard deviation. In option 5, "Weight" is in the y-axis, and increasing the numerator of the fraction \frac{σ~y~}{σ~x~} increases r. In option 6, "Weight" is in the x-axis, and increasing the denominator the fraction \frac{σ~y~}{σ~x~} decreases r.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: \frac{4}{5}

Recall that the formula of the regression line in standard units is y_{su}=r \cdot x_{su}. Since we are predicting # of goals from the # of shots, let x_{su} represent # of shots in standard units and y_{su} represent # of goals in standard units. Using the formula for standard units with information in the problem, we find x_{su}=\frac{20-30}{10}=(-1) and y_{su}=\frac{6-10}{5}=(-\frac{4}{5}). Hence, (-\frac{4}{5})=r \cdot (-1) and r=\frac{4}{5}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: 10

This problem asks about the maximum possible of predicted runs for a team with 27 hits, and the correlation coefficient for the case where the predicted number of runs for a team with 25 hits is maximized. Both of these questions relate to the same concept, so we can answer them both in one fell swoop.

Concept:

The key idea is that the highest correlation coefficient leads to maximum predictions. This is because of a simple fact from lecture: When both x (hits) and y (runs) are converted to standard units,

y_\text{pred} (standard units) = r \cdot x (standard units).

So, the maximum slope in standard units, leading to the highest predictions, is 1, since -1 <= r <= 1. If r were any less than 1, we would be multiplying our x value by a smaller number to get our predicted y value, resulting in lower predictions. In other words,

Max(y_\text{pred}) (standard units) = 1 \cdot x (standard units).

Let’s apply that to answer part 1.

In part 1, we are asked the maximum possible predicted runs for a team with 27 hits. 27 in standard units is:

\dfrac{(27 - 24)}{3} = 0.5,

and the highest value of r (yielding the maximum predictions) is 1. So, we have:

\text{Max}\left(y_\text{pred}\right) \text{(standard units) } = 1 \cdot 0.5

\text{Max}\left(y_\text{pred}\right) \text{(standard units) } = 0.5

Now, we simply need to convert our maximum predicted y in standard units back to its original units, and we have our answer:

y \text{ (standard units)} \cdot \text{SD}_y + \text{Mean}_y = y \text{ (original units)}.

0.5 \cdot 4 + 8 = y \text{ (original units)}

10 = y \text{ (original units)}

So, the maximum number of predicted runs for a team with 27 hits is 10.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.

Answer: ii) 1

Part 2 uses the same information as part 1. The value of r that results in the largest predictions of y is 1 (explained above). So, the correlation coefficient for the case where the predicted number of runs for a team with 25 hits is maximized is 1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.