Question

the following data shows the points the average points scored by the top 20 nfl teams last season 37, 34.3, 34.3, 34, 29.3, 27, 26.7, 25.7, 25.3, 24.7, 24.3, 24, 24, 23.3, 23.3, 23, 22.7, 21.3, 20.7, 20.7. generate a boxplot given the data. find the 5 - number summary as well and label all parts onto your boxplots. refer to the data. select true if the statement accurately describes the data. select false if the statement does not accurately represent the data. statement true or false statement a: the boxplot and data both show a distribution that is not symmetric. statement b: the data shows a skewed right distribution. statement c: removing the number 37 from the data set will decrease the mean, but not change the median.

Explanation:

Step1: Sort the data

$20.7, 20.7, 21.3, 22.7, 23, 23.3, 23.3, 24, 24, 24.3, 24.7, 25.3, 25.7, 26.7, 27, 29.3, 34, 34.3, 34.3, 37$

Step2: Find the minimum

The minimum value is $20.7$.

Step3: Find the first - quartile ($Q_1$)

The first half of the data is $20.7, 20.7, 21.3, 22.7, 23, 23.3, 23.3, 24, 24, 24.3$. The median of this half is $\frac{23 + 23.3}{2}=23.15$. So $Q_1 = 23.15$.

Step4: Find the median

Since $n = 20$ (even), the median is $\frac{24.3+24.7}{2}=24.5$.

Step5: Find the third - quartile ($Q_3$)

The second half of the data is $24.7, 25.3, 25.7, 26.7, 27, 29.3, 34, 34.3, 34.3, 37$. The median of this half is $\frac{27 + 29.3}{2}=28.15$. So $Q_3 = 28.15$.

Step6: Find the maximum

The maximum value is $37$.

Step7: Analyze Statement A

The box - plot and data show that the distance from the median to $Q_1$ is $24.5−23.15 = 1.35$ and the distance from the median to $Q_3$ is $28.15−24.5 = 3.65$. Since these distances are not equal, the distribution is not symmetric. So Statement A is True.

Step8: Analyze Statement B

Since $Q_3 -$ median $> $ median $- Q_1$, the data is skewed right. So Statement B is True.

Step9: Analyze Statement C

The mean of the original data $\bar{x}=\frac{20.7\times2 + 21.3+22.7+23+23.3\times2+24\times2+24.3+24.7+25.3+25.7+26.7+27+29.3+34+34.3\times2+37}{20}=\frac{517}{20}=25.85$.
The mean of the data without $37$ is $\bar{y}=\frac{517 - 37}{19}=\frac{480}{19}\approx25.26$. The mean decreases. The median of the original data is $24.5$ and the median of the data without $37$ (with $n = 19$) is still the 10th - ordered value which is $24.5$. So Statement C is False.

Answer:

Minimum: $20.7$
Quartile 1: $23.15$
Median: $24.5$
Quartile 3: $28.15$
Maximum: $37$
Statement A: True
Statement B: True
Statement C: False