Question

1. fill in the values below and use them to create a box and whiskers plot for the following set of data. 1, 3, 3, 4, 6, 6, 7, 9, 9, 10, 14 min: 1 q1: 3 med: 6 q3: 9 max: 14 6. fill in the values below and use them to create a box and whiskers plot for the following set of data. 3, 4, 5, 8, 9, 13, 13, 13, 14, 14, 14, 17, 18, 18 min: 3 q1: _ med: _ q3: ___ max: 18

Explanation:

Step1: Find the minimum and maximum values

The minimum value (\(Min\)) is the smallest number in the data set. The maximum value (\(Max\)) is the largest number in the data set.
For the data set \(3,4,5,8,9,13,13,13,14,14,14,17,18,18\), \(Min = 3\) and \(Max=18\).

Step2: Find the median (\(Med\))

The median is the middle value of an ordered data set. Since there are \(n = 14\) data points (an even number of data points), the median is the average of the \(\frac{n}{2}\) - th and \((\frac{n}{2}+ 1)\) - th values.
\(\frac{n}{2}=\frac{14}{2}=7\) and \(\frac{n}{2}+1 = 8\). The 7 - th value is \(13\) and the 8 - th value is \(13\). So, \(Med=\frac{13 + 13}{2}=13\).

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 is \(3,4,5,8,9,13,13\). Since there are \(n_1=7\) (an odd number of data points) in the lower half, the median of the lower half (first quartile) is the \(\frac{n_1 + 1}{2}\) - th value. \(\frac{7+1}{2}=4\) - th value. The 4 - th value of \(3,4,5,8,9,13,13\) is \(8\). So, \(Q1 = 8\).

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 is \(14,14,14,17,18,18\). Since there are \(n_2 = 6\) (an even number of data points) in the upper half, the median of the upper half (third quartile) is the average of the \(\frac{n_2}{2}\) - th and \((\frac{n_2}{2}+1)\) - th values. \(\frac{n_2}{2}=3\) and \(\frac{n_2}{2}+1 = 4\). The 3 - rd value is \(14\) and the 4 - th value is \(17\). So, \(Q3=\frac{14 + 17}{2}=15.5\).

Answer:

\(Min:3\), \(Q1:8\), \(Med:13\), \(Q3:15.5\), \(Max:18\)