Question

a modified box - and - whisker plot is a box - and - whisker plot that uses symbols to identify outliers. the horizontal line of a modified box - and - whisker plot extends as far as the minimum data entry that is not an outlier and the maximum data entry that is not an outlier. (a) identify any outliers and (b) draw a modified box - and - whisker plot that represents the data set.
10 13 7 10 7 15 13 11 5
12 26 10 15 13 12 2
(a) identify any outliers. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the outlier(s) is(are)
(type an integer or decimal. use a comma to separate answers as needed.)
b. there are no outliers.
(b) choose the correct modified box - and - whisker plot below.
a.
b.
c.
d.

Explanation:

Step1: Organize the data

First, we list the data set: \(2, 5, 7, 7, 10, 10, 10, 11, 12, 12, 13, 13, 13, 15, 15, 26\) (note: we assume the "5" was a typo and the data is as above, sorted).

Step2: Find Q1, Q2 (median), Q3

- The number of data points \(n = 16\) (even). The median (Q2) is the average of the 8th and 9th values: \(\frac{11 + 12}{2} = 11.5\)? Wait, no, wait, let's re - sort correctly. Wait, original data: 10, 13, 7, 10, 7, 15, 13, 11, 5, 12, 26, 10, 15, 13, 12, 2. Let's sort them: \(2, 5, 7, 7, 10, 10, 10, 11, 12, 12, 13, 13, 13, 15, 15, 26\). So \(n = 16\).
- Q1 (first quartile) is the median of the first 8 data points: \(2, 5, 7, 7, 10, 10, 10, 11\). The median of these 8 (even number) is \(\frac{7 + 10}{2}=8.5\)
- Q3 (third quartile) is the median of the last 8 data points: \(12, 12, 13, 13, 13, 15, 15, 26\). The median of these 8 is \(\frac{13 + 15}{2}=14\)

Step3: Calculate IQR

IQR (Inter - Quartile Range) \(= Q3 - Q1=14 - 8.5 = 5.5\)

Step4: Find lower and upper fences

- Lower fence \(= Q1-1.5\times IQR=8.5 - 1.5\times5.5=8.5 - 8.25 = 0.25\)
- Upper fence \(= Q3 + 1.5\times IQR=14+1.5\times5.5=14 + 8.25 = 22.25\)

Step5: Identify outliers

We check each data point. The data point 26 is greater than the upper fence (22.25). Also, 2 is greater than 0.25? Wait, no, 2>0.25, 5>0.25, etc. Wait, 26 is the only one greater than 22.25. Wait, wait, our calculation of Q1 and Q3 was wrong. Wait, when \(n = 16\), the first quartile is the median of the first 8, which are positions 1 - 8: \(2, 5, 7, 7, 10, 10, 10, 11\). The median of these 8 (since \(n = 8\), the median is the average of the 4th and 5th? Wait, no, for a set with \(n\) elements, the median of a subset of size \(m\): if \(m\) is even, median is \(\frac{x_{\frac{m}{2}}+x_{\frac{m}{2}+1}}{2}\). For \(m = 8\), \(\frac{x_{4}+x_{5}}{2}=\frac{7 + 10}{2}=8.5\) (Q1). For the upper half (positions 9 - 16): \(12, 12, 13, 13, 13, 15, 15, 26\). Median is \(\frac{x_{12}+x_{13}}{2}=\frac{13 + 15}{2}=14\) (Q3). Then IQR \(=14 - 8.5 = 5.5\). Upper fence \(=14+1.5\times5.5 = 14 + 8.25=22.25\). Lower fence \(=8.5-1.5\times5.5 = 8.5 - 8.25 = 0.25\). Now, check data points: 2 is above 0.25, 5 above, 7 above, etc. The only data point above upper fence is 26. Wait, but also, is 2 an outlier? 2>0.25, so no. So the outlier is 26? Wait, no, wait, maybe I made a mistake in sorting. Wait, original data: 10, 13, 7, 10, 7, 15, 13, 11, 5, 12, 26, 10, 15, 13, 12, 2. Let's count the number of data points: 16. So sorted: 2, 5, 7, 7, 10, 10, 10, 11, 12, 12, 13, 13, 13, 15, 15, 26. So Q1: the median of the first 8 (indices 0 - 7: 2,5,7,7,10,10,10,11). The median of these 8 is the average of the 4th and 5th (0 - based: 4th is 10, 5th is 10? Wait, no, 0 - based index: for 8 elements (indices 0 - 7), the median is between index 3 and 4. Wait, no, for a set of \(m\) numbers, the median position is \(\frac{m + 1}{2}\) for odd, and for even \(m\), it's the average of \(\frac{m}{2}\) and \(\frac{m}{2}+1\) - th terms (1 - based). So for the first 8 numbers (1 - based: 1 - 8), the 4th and 5th terms. 1 - based: 1:2, 2:5, 3:7, 4:7, 5:10, 6:10, 7:10, 8:11. So 4th term is 7, 5th term is 10. So Q1 \(=\frac{7 + 10}{2}=8.5\) (correct). For the last 8 numbers (1 - based: 9 - 16): 9:12, 10:12, 11:13, 12:13, 13:13, 14:15, 15:15, 16:26. 4th term (1 - based in this subset) is 13, 5th term is 13? Wait, no, 1 - based in the subset: 9:12 (index 8 in original), 10:12 (index 9), 11:13 (index 10), 12:13 (index 11), 13:13 (index 12), 14:15 (index 13), 15:15 (index 14), 16:26 (index 15). So 4th term is 13 (index 11), 5th term is 13 (index 12). Wait, no, I think I messed up. Wait, the correct way to find quartiles for even \(n\):

For \(n = 16\), the median (Q2) is the average of the 8th and 9th terms (1 - based): 8th term is 11, 9th term is 12, so Q2 \(=\frac{11 + 12}{2}=11.5\). Then Q1 is the median of the first 8 terms (1 - based: 1 - 8): terms are 2,5,7,7,10,10,10,11. Median of these 8 (1 - based) is average of 4th and 5th terms: 4th term is 7, 5th term is 10, so Q1 \(=\frac{7+10}{2}=8.5\). Q3 is the median of the last 8 terms (1 - based: 9 - 16): terms are 12,12,13,13,13,15,15,26. Median of these 8 is average of 4th and 5th terms: 4th term is 13, 5th term is 13, so Q3 \(=\frac{13 + 13}{2}=13\). Wait, I see my mistake earlier! Oh no, I took the wrong subset for Q3. The last 8 terms after the median: since the median is between 8th and 9th terms, the first half is terms 1 - 8, the second half is terms 9 - 16. So Q3 is the median of terms 9 - 16. So terms 9 - 16: 12,12,13,13,13,15,15,26. So median of these 8 (1 - based in the subset: 1 - 8) is average of 4th and 5th terms (1 - based in subset: 4th term is 13, 5th term is 13), so Q3 \( = 13\). Then IQR \(=Q3 - Q1=13 - 8.5 = 4.5\). Then upper fence \(=Q3+1.5\times IQR=13 + 1.5\times4.5=13 + 6.75 = 19.75\). Lower fence \(=Q1-1.5\times IQR=8.5-1.5\times4.5=8.5 - 6.75 = 1.75\). Now, check data points: 26 is greater than 19.75, so 26 is an outlier. Also, 2 is greater than 1.75, 5>1.75, etc. So the outlier is 26.

Answer:

(a) The outlier(s) is(are) 26. So the correct choice is A. The outlier(s) is(are) \(\boldsymbol{26}\)