Question

practice 5 (from unit 1, lesson 9)
a. what is the five - number summary for the data?
2 2 4 4 5 5 6 7 9 15
type the answers in the boxes.
2 4 6 7 18
b. when the maximum, 15, is removed from the data set, what is the five - number summary?
type the answers in the boxes.

Explanation:

Step1: Arrange data in ascending order (already done: 2, 2, 4, 4, 5, 5, 6, 7, 9, 15)

Step2: Find minimum

The minimum value of the data - set is 2.

Step3: Find first - quartile ($Q_1$)

There are $n = 10$ data points. The position of $Q_1$ is $\frac{n + 1}{4}=\frac{10+1}{4}=2.75$. So, $Q_1=4$.

Step4: Find median

The position of the median for $n = 10$ data points is $\frac{n}{2}=5$ and $\frac{n}{2}+1 = 6$. The median is $\frac{5 + 5}{2}=5$.

Step5: Find third - quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(10 + 1)}{4}=8.25$. So, $Q_3=7$.

Step6: Find maximum

The maximum value is 15.

for part b:

Step1: Remove 15 from the data - set. New data - set is 2, 2, 4, 4, 5, 5, 6, 7, 9

Step2: Find minimum

The minimum value of the new data - set is 2.

Step3: Find first - quartile ($Q_1$)

There are $n = 9$ data points. The position of $Q_1$ is $\frac{n + 1}{4}=\frac{9+1}{4}=2.5$. So, $Q_1=\frac{2 + 4}{2}=3$.

Step4: Find median

The position of the median for $n = 9$ data points is $\frac{n + 1}{2}=5$. The median is 5.

Step5: Find third - quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(9 + 1)}{4}=7.5$. So, $Q_3=\frac{6+7}{2}=6.5$.

Step6: Find maximum

The maximum value is 9.

Answer:

Minimum: 2, $Q_1$: 4, Median: 5, $Q_3$: 7, Maximum: 15