Question

here are the scores of 12 students on a history test.
61, 65, 67, 67, 69, 75, 79, 82, 85, 87, 88, 89
notice that the scores are ordered from least to greatest.
give the five - number summary and the interquartile range for the data set.
five - number summary
minimum:
lower quartile:
median:
upper quartile:
maximum:
interquartile range:

Explanation:

Step1: Find the minimum and maximum

The minimum is the smallest value in the data set, which is \(61\). The maximum is the largest value, which is \(89\).

Step2: Find the median

Since there are \(n = 12\) data points (even number), the median is the average of the \(\frac{n}{2}=6^{th}\) and \((\frac{n}{2}+ 1)=7^{th}\) values. The ordered data set: \(61,65,67,67,69,75,79,82,85,87,88,89\). The \(6^{th}\) value is \(75\) and the \(7^{th}\) value is \(79\). Median \(M=\frac{75 + 79}{2}=77\).

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

The lower half of the data is \(61,65,67,67,69,75\). Since \(n_1 = 6\) (even), \(Q_1=\frac{67+67}{2}=67\).

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

The upper half of the data is \(79,82,85,87,88,89\). Since \(n_2=6\) (even), \(Q_3=\frac{85 + 87}{2}=86\).

Step5: Calculate the inter - quartile range (IQR)

\(IQR=Q_3 - Q_1\). Substitute \(Q_1 = 67\) and \(Q_3=86\), \(IQR=86-67 = 19\).

Answer:

Minimum: \(61\), Median: \(77\), Maximum: \(89\), Lower quartile (\(Q_1\)): \(67\), Upper quartile (\(Q_3\)): \(86\), Inter - quartile range (IQR): \(19\)