Question

5 from unit 1, lesson 9
a. what is the five - number summary for the data?
0 2 2 4 5 5 5 5 7 11
b. when the minimum, 0, is removed from the data set, what is the five - number summary?

Explanation:

Step1: Arrange data in ascending order (already done: 0, 2, 2, 4, 5, 5, 5, 5, 7, 11)

Step2: Find the minimum

The minimum is 0.

Step3: Find the 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, $Q_1=2+(0.75)\times(2 - 2)=2$.

Step4: Find the median

The position of the median for $n = 10$ is $\frac{n+1}{2}=5.5$. So, the median is $\frac{5 + 5}{2}=5$.

Step5: Find the third - quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(10 + 1)}{4}=8.25$. So, $Q_3=5+(0.25)\times(7 - 5)=5.5$.

Step6: Find the maximum

The maximum is 11.

Step1: Remove 0 from the data set. New data set: 2, 2, 4, 5, 5, 5, 5, 7, 11

Step2: Find the new minimum

The new minimum is 2.

Step3: Find the new 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, $Q_1=2+(0.5)\times(4 - 2)=3$.

Step4: Find the new median

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

Step5: Find the new third - quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(9+1)}{4}=7.5$. So, $Q_3=5+(0.5)\times(7 - 5)=6$.

Step6: Find the new maximum

The new maximum is 11.

Answer:

a. Minimum: 0, $Q_1$: 2, Median: 5, $Q_3$: 5.5, Maximum: 11