Question

answer the statistical measures and create a box and whiskers plot for the following set of data. e decimals. 5,6,8,9,9,9,11,11,12,13,13,15,16,16,17 min: □ q1: □ med: □ q3: □ max: □ create the box plot by dragging the lines: box plot image with x - axis from 0 to 20 answer attempt 1 out of 3 you must answer all questions above in order to submit.

Explanation:

Step1: Find the minimum value

The minimum value is the smallest number in the data set. Looking at the data \(5, 6, 8, 9, 9, 9, 11, 11, 12, 13, 13, 15, 16, 16, 17\), the smallest number is \(5\).

Step2: Find the median (Med)

First, the data set has \(n = 15\) values (odd number of observations). The median is the middle value, which is the \(\frac{n + 1}{2}=\frac{15+ 1}{2}=8\)-th value when the data is ordered. The ordered data is as given, and the 8 - th value is \(11\).

Step3: Find the first quartile (Q1)

The first quartile is the median of the lower half of the data. The lower half of the data (excluding the median) has \(n_1=\frac{n - 1}{2}=7\) values? Wait, no. For \(n = 15\), the lower half is the first \(7\) values? Wait, actually, when \(n\) is odd, for the lower half, we take the first \(\frac{n-1}{2}=\frac{15 - 1}{2}=7\) values? Wait, no, the data is \(5,6,8,9,9,9,11,11,12,13,13,15,16,16,17\). The median is at position \(8\) (value \(11\)). The lower half is the data before the median: \(5,6,8,9,9,9,11\) (wait, no, the median is the 8 - th value, so the lower half is the first \(7\) values? Wait, no, when \(n = 15\), the positions are \(1\) to \(15\). The median is at position \(8\). The lower half is positions \(1\) to \(7\), and the upper half is positions \(9\) to \(15\). So the lower half data is \(5,6,8,9,9,9,11\)? Wait, no, the 7 - th value is \(11\)? Wait, no, the data is: position \(1:5\), \(2:6\), \(3:8\), \(4:9\), \(5:9\), \(6:9\), \(7:11\), \(8:11\) (median), \(9:12\), \(10:13\), \(11:13\), \(12:15\), \(13:16\), \(14:16\), \(15:17\). Wait, I made a mistake. The 7 - th value is \(11\)? No, the 7 - th value: let's count again. \(1:5\), \(2:6\), \(3:8\), \(4:9\), \(5:9\), \(6:9\), \(7:11\)? Wait, no, the data is \(5,6,8,9,9,9,11,11,12,13,13,15,16,16,17\). So the first seven values are \(5,6,8,9,9,9,11\)? Wait, no, the seventh value is \(11\)? Wait, no, the sixth value is \(9\), seventh is \(11\). Then the median of the lower half (first seven values) is the 4 - th value (since \(7\) is odd, \(\frac{7 + 1}{2}=4\)-th value). The first seven values: \(5,6,8,9,9,9,11\). The 4 - th value is \(9\). So Q1 is \(9\).

Step4: Find the third quartile (Q3)

The third quartile is the median of the upper half of the data. The upper half of the data (excluding the median) is the values from position \(9\) to \(15\): \(12,13,13,15,16,16,17\). There are \(7\) values. The median of these \(7\) values is the 4 - th value (since \(\frac{7+1}{2}=4\)-th value). The data \(12,13,13,15,16,16,17\), the 4 - th value is \(15\). So Q3 is \(15\).

Step5: Find the maximum value

The maximum value is the largest number in the data set. Looking at the data, the largest number is \(17\).

Answer:

Min: \(5\)
Q1: \(9\)
Med: \(11\)
Q3: \(15\)
Max: \(17\)

For the box - and - whisker plot:

• The left whisker starts at the minimum value (\(5\)) and ends at Q1 (\(9\)).
• The box starts at Q1 (\(9\)), has a line at the median (\(11\)) and ends at Q3 (\(15\)).
• The right whisker starts at Q3 (\(15\)) and ends at the maximum value (\(17\)).