Question

find the interquartile range of the data set below.
33, 36, 40, 43, 43, 45, 46, 50, 50, 51, 53, 53

question 10

find the interquartile range of the data set below.
9, 9, 11, 12, 15, 17, 17, 20, 20, 21, 22

question 11
2 pt
find the variance of the data set below. if necessary, round to the nearest hundredth.
79, 82, 88, 88, 92, 92, 95, 96, 98

Explanation:

Step1: Find the median of the data set

For the first data set \(33,36,40,43,43,45,46,50,50,51,53,53\), since \(n = 12\) (even), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+ 1)\)th values. \(\frac{n}{2}=6\), \(\frac{n}{2}+1 = 7\). The 6th value is \(45\) and the 7th value is \(46\), so the median \(Q_2=\frac{45 + 46}{2}=45.5\).
The lower half of the data is \(33,36,40,43,43,45\). For \(n = 6\) (even) in the lower - half, \(\frac{n}{2}=3\), \(\frac{n}{2}+1 = 4\). The 3rd value is \(40\) and the 4th value is \(43\), so \(Q_1=\frac{40+43}{2}=41.5\).
The upper half of the data is \(50,50,51,53,53,53\). For \(n = 6\) (even) in the upper - half, \(\frac{n}{2}=3\), \(\frac{n}{2}+1 = 4\). The 3rd value is \(51\) and the 4th value is \(53\), so \(Q_3=\frac{51 + 53}{2}=52\).
The inter - quartile range \(IQR=Q_3 - Q_1=52-41.5 = 10.5\).

For the second data set \(9,9,11,12,15,17,17,20,20,21,22\), since \(n = 11\) (odd), the median \(Q_2\) is the \(\frac{n + 1}{2}\)th value. \(\frac{n+1}{2}=6\)th value, so \(Q_2 = 17\).
The lower half of the data is \(9,9,11,12,15\). Since \(n = 5\) (odd) in the lower - half, the median of the lower - half \(Q_1\) is the \(\frac{n + 1}{2}\)th value (\(3\)rd value), so \(Q_1=11\).
The upper half of the data is \(20,20,21,22\). Since \(n = 4\) (even) in the upper - half, \(\frac{n}{2}=2\), \(\frac{n}{2}+1 = 3\). The 2nd value is \(20\) and the 3rd value is \(21\), so \(Q_3=\frac{20 + 21}{2}=20.5\).
The inter - quartile range \(IQR=Q_3 - Q_1=20.5-11=9.5\).

Step2: Calculate the variance for the third data set

First, find the mean \(\bar{x}=\frac{79+82+88+88+92+92+95+96+98}{9}=\frac{810}{9}=90\).
Then, use the formula for variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^{2}}{n - 1}\).
\((79 - 90)^{2}=(-11)^{2}=121\), \((82-90)^{2}=(-8)^{2}=64\), \((88 - 90)^{2}=(-2)^{2}=4\), \((88 - 90)^{2}=(-2)^{2}=4\), \((92 - 90)^{2}=2^{2}=4\), \((92 - 90)^{2}=2^{2}=4\), \((95 - 90)^{2}=5^{2}=25\), \((96 - 90)^{2}=6^{2}=36\), \((98 - 90)^{2}=8^{2}=64\).
\(\sum_{i = 1}^{n}(x_i-\bar{x})^{2}=121+64 + 4+4+4+4+25+36+64=326\).
\(s^{2}=\frac{326}{9 - 1}=\frac{326}{8}=40.75\).

Answer:

10.5
9.5
40.75