Question

find each measure for the given set of data: 11, 13, 17, 20, 22, 25, 27, 31, 31, 33 mean = median = range = interquartile range =

Explanation:

Step1: Calculate the mean

The mean is calculated by summing all the data points and dividing by the number of data points.
The data set is \(11, 13, 17, 20, 22, 25, 27, 31, 31, 33\).
The sum of the data points is \(11 + 13+17 + 20+22 + 25+27 + 31+31 + 33=\sum_{i = 1}^{10}x_{i}=240\).
The number of data points \(n = 10\).
The mean \(\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}=\frac{240}{10}=24\).

Step2: Calculate the median

Since \(n = 10\) (an even - numbered data set), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th ordered data points.
\(\frac{n}{2}=5\) and \(\frac{n}{2}+1 = 6\).
The 5th data point is \(22\) and the 6th data point is \(25\).
The median \(M=\frac{22 + 25}{2}=\frac{47}{2}=23.5\).

Step3: Calculate the range

The range is the difference between the maximum and minimum values in the data set.
The maximum value \(x_{\text{max}}=33\) and the minimum value \(x_{\text{min}}=11\).
The range \(R=x_{\text{max}}-x_{\text{min}}=33 - 11=22\).

Step4: Calculate the inter - quartile range

First, find the first quartile \(Q_{1}\) and the third quartile \(Q_{3}\).
For \(n = 10\), the lower half of the data is \(11, 13, 17, 20, 22\). The median of the lower half (first quartile) \(Q_{1}\) is the 3rd value of the lower - half data set. So \(Q_{1}=17\).
The upper half of the data is \(27, 31, 31, 33\). The median of the upper half (third quartile) \(Q_{3}\) is the 3rd value of the upper - half data set. So \(Q_{3}=31\).
The inter - quartile range \(IQR = Q_{3}-Q_{1}=31 - 17=14\).

Answer:

Mean \(=24\), Median \(=23.5\), Range \(=22\), Interquartile range \(=14\)