← return to practice.dsc10.com
Below are practice problems tagged for Lecture 15 (rendered directly from the original exam/quiz sources).
According to Chebyshev’s inequality, at least 80% of San Diego apartments have a monthly parking fee that falls between $30 and $70.
What is the average monthly parking fee?
Answer: \$50
We are given that the left and right bounds of Chebyshev’s inequality are $30 and $70 respectively. Thus, to find the middle of the two, we compute the following equation (the midpoint equation):
\frac{\text{right} + \text{left}}{2}
\frac{70 + 30}{2} = 50
Therefore, 50 is the average monthly parking fee.
The average score on this problem was 92%.
What is the standard deviation of monthly parking fees?
\frac{20}{\sqrt{5}}
\frac{40}{\sqrt{5}}
20\sqrt{5}
20\sqrt{5}
Answer: \frac{20}{\sqrt{5}}
Chebyshev’s inequality states that at least 1 - \frac{1}{z^2} of values are within z standard deviations of the mean. In addition, z can be represented as \frac{\text{bound} - \text{mean of x}}{\text{SD of x}}.
Therefore, we can set up the equation like so: \frac{4}{5} = 1 - \frac{1}{(\frac{\text{bound} - \text{mean of x}}{\text{SD of x}})^2}
Then, we can solve: \frac{1}{5} = \frac{1}{(\frac{\text{bound} - \text{mean of x}}{\text{SD of x}})^2}
Now since we know both bounds, we can plug one of them in. Since the mean was computed in the earlier step, we also plug this in.
\frac{1}{5} = \frac{1}{(\frac{70 - 50}{\text{SD of x}})^2} 5 = (\frac{20}{\text{SD of x}})^2 \sqrt{5} = \frac{20}{\text{SD of x}} \text{SD of x} = \frac{20}{\sqrt{5}}
The average score on this problem was 70%.
The data visualization below shows all Olympic gold medals for women’s gymnastics, broken down by the age of the gymnast.
Based on this data, rank the following three quantities in ascending order: the median age at which gold medals are earned, the mean age at which gold medals are earned, the standard deviation of the age at which gold medals are earned.
mean, median, SD
median, mean, SD
SD, mean, median
SD, median, mean
Answer: SD, median, mean
The standard deviation will clearly be the smallest of the three
values as most of the data is encompassed between the range of
[14-26]. Intuitively, the standard deviation will have to
be about a third of this range which is around 4 (though this is not the
exact standard deviation, but is clearly much less than the mean and
median with values closer to 19-25). Comparing the median and mean, it
is important to visualize that this distribution is skewed right. When
the data is skewed right it pulls the mean towards a higher value (as
the higher values naturally make the average higher). Therefore, we know
that the mean will be greater than the median and the ranking is SD,
median, mean.
The average score on this problem was 72%.
Which of the following is larger for this dataset?
the difference between the 50th percentile of ages and the 25th percentile of ages
the difference between the 75th percentile of ages and the 50th percentile of ages
both are the same
Answer: the difference between the 75th percentile of ages and the 50th percentile of ages
Since the distribution is right skewed, the 75th percentile will have a larger difference from the 50th percentile than the 25th percentile. With right skewness, values above the 50th percentile will be more different than those smaller than the 50th percentile (and thus more spread out according to the graph).
The average score on this problem was 78%.
The data visualization below shows all Olympic gold medals for women’s gymnastics, broken down by the age of the gymnast.
Based on this data, rank the following three quantities in ascending order: the median age at which gold medals are earned, the mean age at which gold medals are earned, the standard deviation of the age at which gold medals are earned.
mean, median, SD
median, mean, SD
SD, mean, median
SD, median, mean
Answer: SD, median, mean
The standard deviation will clearly be the smallest of the three
values as most of the data is encompassed between the range of
[14-26]. Intuitively, the standard deviation will have to
be about a third of this range which is around 4 (though this is not the
exact standard deviation, but is clearly much less than the mean and
median with values closer to 19-25). Comparing the median and mean, it
is important to visualize that this distribution is skewed right. When
the data is skewed right it pulls the mean towards a higher value (as
the higher values naturally make the average higher). Therefore, we know
that the mean will be greater than the median and the ranking is SD,
median, mean.
The average score on this problem was 72%.