side-by-side box-and-whisker plots can be used to compare two or more different data sets. each box-and-whisker plot is drawn on the same number line to compare the data sets more easily. the lengths (in seconds) of songs played at two different concerts are shown. complete parts a through d below.

(a) describe the shape of each distribution. which concert has less variation in song lengths?
the distribution of concert 1 is skewed left.
the distribution of concert 2 is symmetric.
which concert has less variation in song lengths?
concert 2 has less variation in song lengths.

(b) which distribution is more likely to have outliers? explain your reasoning.
concert 1 is more likely to have outliers because it has more variation.

(c) which concert do you think has a standard deviation of 15.3? explain your reasoning.
a. concert 1, because the interquartile range of concert 2 is greater than twice 15.3. for each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $q_1$ and $q_3$. therefore, the difference between $q_1$ and $q_3$ should be smaller than twice 15.3.
b. concert 2 because it has a smaller median.
c. concert 1 because it has a larger median.
d. concert 2, because the interquartile range of concert 1 is greater than twice 15.3. for each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $q_1$ and $q_3$. therefore, the difference between $q_1$ and $q_3$ should be smaller than twice 15.3.

Question

side-by-side box-and-whisker plots can be used to compare two or more different data sets. each box-and-whisker plot is drawn on the same number line to compare the data sets more easily. the lengths (in seconds) of songs played at two different concerts are shown. complete parts a through d below.

(a) describe the shape of each distribution. which concert has less variation in song lengths?
the distribution of concert 1 is skewed left.
the distribution of concert 2 is symmetric.
which concert has less variation in song lengths?
concert 2 has less variation in song lengths.

(b) which distribution is more likely to have outliers? explain your reasoning.
concert 1 is more likely to have outliers because it has more variation.

(c) which concert do you think has a standard deviation of 15.3? explain your reasoning.
a. concert 1, because the interquartile range of concert 2 is greater than twice 15.3. for each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $q_1$ and $q_3$. therefore, the difference between $q_1$ and $q_3$ should be smaller than twice 15.3.
b. concert 2 because it has a smaller median.
c. concert 1 because it has a larger median.
d. concert 2, because the interquartile range of concert 1 is greater than twice 15.3. for each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $q_1$ and $q_3$. therefore, the difference between $q_1$ and $q_3$ should be smaller than twice 15.3.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Box-and-Whisker Plot", "Interquartile Range", "Measures of Variation" ], "new_concepts": [], "current_concepts": [ "Box-and-Whisker Plot", "Interquartile Range", "Measures of Variation", "Standard Deviation Estimation" ] } ## Analyze the box-and-whisker plot data Using the *Box-and-Whisker Plot* and *Interquartile Range* knowledge points $$ \begin{aligned} &\text{Concert 1:} \quad Q_1 = 250, \quad Q_2 = 274, \quad Q_3 = 310 \ &\text{IQR}_1 = Q_3 - Q_1 = 310 - 250 = 60 \ &\text{Concert 2:} \quad Q_1 = 202, \quad Q_2 = 215, \quad Q_3 = 228 \ &\text{IQR}_2 = Q_3 - Q_1 = 228 - 202 = 26 \end{aligned} $$ ## Evaluate standard deviation of 15.3 Using the *Measures of Variation* knowledge point $$ \begin{aligned} &\sigma \approx 15.3 \implies 2\sigma \approx 30.6 \ &\text{For Concert 2, } \text{IQR}_2 = 26 < 2(15.3) = 30.6 \ &\text{For Concert 1, } \text{IQR}_1 = 60 > 2(15.3) = 30.6 \end{aligned} $$ Since $50\%$ of the data lies within the IQR (between $Q_1$ and $Q_3$), and for a distribution with standard deviation $\sigma = 15.3$, at least $75\%$ of the data lies within $2\sigma = 30.6$ of the mean (by Chebyshev's Inequality), the IQR (which contains only $50\%$ of the data) should be smaller than $2\sigma = 30.6$. This holds true for Concert 2 ($\text{IQR} = 26 < 30.6$), but not for Concert 1 ($\text{IQR} = 60 > 30.6$). Thus, Concert 2 is the distribution with a standard deviation of $15.3$.

A. Concert 1, because the interquartile range of Concert 2 is greater than twice 15.3. For each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $Q_1$ and $Q_3$. Therefore, the difference between $Q_1$ and $Q_3$ should be smaller than twice 15.3.

B. Concert 2 because it has a smaller median.

C. Concert 1 because it has a larger median.

D. Concert 2, because the interquartile range of Concert 1 is greater than twice 15.3. For each distribution, at least 75% of the data lies within two standard deviation of the mean. 50% of the data lies between $Q_1$ and $Q_3$. Therefore, the difference between $Q_1$ and $Q_3$ should be smaller than twice 15.3.

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Box-and-Whisker Plot" ] }

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QUESTION IMAGE

Question

Explanation:

Analyze the box-and-whisker plot data

Evaluate standard deviation of 15.3

Answer: