Question

5 from unit 1, lesson 4
a. what is the five - number summary for the data?
2 2 4 4 5 5 6 7 9 15
b. when the maximum, 15, is removed from the data set, what is the five - number summary?

Explanation:

Step1: Arrange data in ascending order

The data set is already in ascending order: 2, 2, 4, 4, 5, 5, 6, 7, 9, 15.

Step2: Find the minimum

The minimum value is 2.

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, $Q_1=4$.

Step4: Find the median ($Q_2$)

The position of the median for $n = 10$ (even number of data points) is $\frac{n}{2}=5$ and $\frac{n}{2}+1 = 6$. So, $Q_2=\frac{5 + 5}{2}=5$.

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, $Q_3=7$.

Step6: Find the maximum

The maximum value is 15.

Answer:

a. Minimum = 2, $Q_1$ = 4, Median = 5, $Q_3$ = 7, Maximum = 15

Step1: Remove the maximum value

The new data set is 2, 2, 4, 4, 5, 5, 6, 7, 9.

Step2: Find the new minimum

The new minimum value 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, $Q_1=\frac{4+4}{2}=4$.

Step4: Find the new median ($Q_2$)

The position of the median for $n = 9$ (odd number of data points) is $\frac{n + 1}{2}=5$. So, $Q_2=5$.

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, $Q_3=\frac{6+7}{2}=6.5$.

Step6: Find the new maximum

The new maximum value is 9.