Question

7 from unit 1, lesson 9
a. what is the five - number summary for these data?
1
3
3
3
4
8
9
10
10
17
b. when the maximum, 17, is removed from the data set, what is the five - number summary?

Answer:

Part a: Five - number summary for the original data

The data set is: \(1, 3, 3, 3, 4, 8, 9, 10, 10, 17\)

Step 1: Find the minimum value

The minimum value is the smallest number in the data set.
Looking at the data, the minimum value \(= 1\)

Step 2: Find the first quartile (\(Q_1\))

First, we need to find the median of the lower half of the data. The data set has \(n = 10\) values. The median (second quartile, \(Q_2\)) of the entire data set will be the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+ 1)\)-th values.
The lower half of the data (values before the median) is \(1, 3, 3, 3, 4\) (the first 5 values).
The median of the lower half (which is \(Q_1\)): Since there are 5 values (odd number), the median is the \(\frac{5 + 1}{2}=3\)-rd value.
The 3 - rd value in \(1, 3, 3, 3, 4\) is \(3\), so \(Q_1=3\)

Step 3: Find the median (\(Q_2\))

For \(n = 10\) (even), the median is the average of the \(\frac{n}{2}=5\)-th and \((\frac{n}{2}+ 1)=6\)-th values.
The 5 - th value is \(4\) and the 6 - th value is \(8\).
Median \(Q_2=\frac{4 + 8}{2}=\frac{12}{2}=6\)

Step 4: Find the third quartile (\(Q_3\))

The upper half of the data (values after the median) is \(8, 9, 10, 10, 17\) (the last 5 values).
The median of the upper half (which is \(Q_3\)): Since there are 5 values (odd number), the median is the \(\frac{5+ 1}{2}=3\)-rd value.
The 3 - rd value in \(8, 9, 10, 10, 17\) is \(10\), so \(Q_3 = 10\)

Step 5: Find the maximum value

The maximum value is the largest number in the data set.
Looking at the data, the maximum value \(= 17\)

The five - number summary is: Minimum \(= 1\), \(Q_1=3\), Median \(= 6\), \(Q_3 = 10\), Maximum \(= 17\)

Part b: Five - number summary after removing 17

The new data set is: \(1, 3, 3, 3, 4, 8, 9, 10, 10\)

Step 1: Find the minimum value

The minimum value is the smallest number in the new data set.
The minimum value \(= 1\)

Step 2: Find the first quartile (\(Q_1\))

The number of values \(n = 9\) (odd). The median (second quartile, \(Q_2\)) is the \(\frac{9 + 1}{2}=5\)-th value.
The lower half of the data (values before the median) is \(1, 3, 3, 3\) (the first 4 values? Wait, for \(n = 9\), the median is at position 5. The lower half is the first \(\lfloor\frac{n}{2}
floor=4\) values? No, for odd \(n\), the lower half is the first \(\frac{n - 1}{2}\) values and the upper half is the last \(\frac{n - 1}{2}\) values. So for \(n=9\), lower half: first 4 values (\(1, 3, 3, 3\)), upper half: last 4 values (\(9, 10, 10, 8\) (sorted as \(8, 9, 10, 10\))). Wait, first, sort the data: \(1, 3, 3, 3, 4, 8, 9, 10, 10\)
Median (\(Q_2\)) is the 5 - th value, which is \(4\)
Lower half: \(1, 3, 3, 3\) (4 values). The median of the lower half ( \(Q_1\)): since there are 4 values (even), the median is the average of the 2 - nd and 3 - rd values. The 2 - nd value is \(3\), the 3 - rd value is \(3\). So \(Q_1=\frac{3 + 3}{2}=3\)

Step 3: Find the median (\(Q_2\))

For \(n = 9\) (odd), the median is the \(\frac{9+1}{2}=5\)-th value. The 5 - th value in \(1, 3, 3, 3, 4, 8, 9, 10, 10\) is \(4\), so \(Q_2 = 4\)

Step 4: Find the third quartile (\(Q_3\))

Upper half: \(8, 9, 10, 10\) (4 values). The median of the upper half ( \(Q_3\)): since there are 4 values (even), the median is the average of the 2 - nd and 3 - rd values. The 2 - nd value is \(9\), the 3 - rd value is \(10\). So \(Q_3=\frac{9+10}{2}=\frac{19}{2} = 9.5\)

Step 5: Find the maximum value

The maximum value is the largest number in the new data set.
The maximum value \(= 10\)

The five - number summary after removing 17 is: Minimum \(= 1\), \(Q_1 = 3\), Median \(= 4\), \(Q_3=9.5\), Maximum \(= 10\)

Final Answers

Part a

Minimum: \(1\), \(Q_1\): \(3\), Median: \(6\), \(Q_3\): \(10\), Maximum: \(17\)

Part b

Minimum: \(1\), \(Q_1\): \(3\), Median: \(4\), \(Q_3\): \(9.5\), Maximum: \(10\)