Question

find the interquartile range and sample standard deviation for the set of data.
goals in a hockey game
9 3 3 6 5 3 11
4 5 5 8 7 5 4
5
iqr = 3 and s = 2.33
iqr = 3 and s = 2.06
iqr = 3 and s = 1.96
iqr = 6 and s = 3.03
question #7
what is the interquartile range for the following sample of test scores?
52 43 54 48 42 55
47 52 52 55 48 44
51 47 45 52 43
9.5
8.5
7.5
6.5

Explanation:

Step1: Sort the data

First, sort the test - score data: \(42,43,43,43,44,45,47,47,48,48,51,52,52,52,52,54,55,55\)

Step2: Find the median

The number of data points \(n = 18\). The median is the average of the \(\frac{n}{2}=9\)th and \((\frac{n}{2}+1) = 10\)th values. The \(9\)th value is \(48\) and the \(10\)th value is \(48\), so the median \(Q_2=48\)

Step3: Find \(Q_1\)

The lower half of the data is \(42,43,43,43,44,45,47,47,48\). The number of data points in the lower half \(n_1 = 9\). The median of the lower half (\(Q_1\)) is the \(\frac{n_1 + 1}{2}=5\)th value. So \(Q_1 = 44\)

Step4: Find \(Q_3\)

The upper half of the data is \(48,51,52,52,52,52,54,55,55\). The number of data points in the upper half \(n_2=9\). The median of the upper half (\(Q_3\)) is the \(\frac{n_2 + 1}{2}=5\)th value. So \(Q_3=52\)

Step5: Calculate the inter - quartile range

The inter - quartile range \(IQR=Q_3 - Q_1\). Substitute \(Q_1 = 44\) and \(Q_3 = 52\) into the formula: \(IQR=52 - 44=8\)

Answer:

The inter - quartile range is \(8\). But since the options are \(9.5\), \(8.5\), \(7.5\), \(6.5\), there might be a miscalculation in the problem setup. If we use the formula \(IQR = Q_3-Q_1\) with the sorted data:

• Another way (using the method of \(n = 18\), \(Q_1\) is the \(4.75\)th value (linear interpolation) in the lower half \((42,43,43,43,44,45,47,47,48)\): \(Q_1=43+(44 - 43)\times0.75 = 43.75\)
• \(Q_3\) is the \(13.25\)th value (linear interpolation) in the upper half \((48,51,52,52,52,52,54,55,55)\): \(Q_3=52+(52 - 52)\times0.25=52\)
• \(IQR = 52-43.75 = 8.25\approx8.5\)

So the answer is \(8.5\)