Question

the following are the ages of 12 music teachers in a school district.
26, 28, 32, 34, 35, 36, 38, 38, 45, 45, 57, 60
notice that the ages are ordered from least to greatest.
give the five - number summary and the interquartile range for the data set.
five - number summary
minimum: 26
lower quartile: 33
median: 37
upper quartile: 51
maximum: 60
interquartile range: 18

Explanation:

Step1: Find the minimum and maximum

The minimum is the smallest value in the data set, which is \(26\). The maximum is the largest value, which is \(60\).

Step2: Find the median

Since there are \(n = 12\) data points. The median is the average of the \(\frac{n}{2}\) - th and \((\frac{n}{2}+ 1)\) - th values. \(\frac{12}{2}=6\), \(\frac{12}{2}+1 = 7\). The \(6\) - th value is \(36\) and the \(7\) - th value is \(38\). Median \(M=\frac{36 + 38}{2}=37\).

Step3: Find the lower quartile (\(Q_1\))

The lower half of the data is \(26,28,32,34,35,36\). There are \(n_1=6\) data points. The lower quartile is the average of the \(\frac{n_1}{2}\) - th and \((\frac{n_1}{2}+ 1)\) - th values. \(\frac{6}{2}=3\), \(\frac{6}{2}+1 = 4\). \(Q_1=\frac{32+34}{2}=33\).

Step4: Find the upper quartile (\(Q_3\))

The upper half of the data is \(38,45,45,57,60\). There are \(n_2 = 6\) data points. The upper quartile is the average of the \(\frac{n_2}{2}\) - th and \((\frac{n_2}{2}+ 1)\) - th values. \(\frac{6}{2}=3\), \(\frac{6}{2}+1 = 4\). \(Q_3=\frac{45 + 57}{2}=51\).

Step5: Calculate the inter - quartile range (IQR)

\(IQR=Q_3 - Q_1\). Substitute \(Q_1 = 33\) and \(Q_3=51\). \(IQR=51-33 = 18\).

Answer:

Minimum: \(26\), Lower quartile: \(33\), Median: \(37\), Upper quartile: \(51\), Maximum: \(60\), Interquartile range: \(18\)