from unit 1, lesson 9\na. what is the five - number summary for these data?\n1 3 3 3 4 8 9 10 10 17\nb. when…

Explanation:

Step1: Arrange data in ascending order (already done: 1, 3, 3, 3, 4, 8, 9, 10, 10, 17)

Step2: Find minimum

The minimum is 1.

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, $Q_1=3+(0.75)\times(3 - 3)=3$.

Step4: Find median

The position of the median for $n = 10$ is $\frac{n}{2}=5$ and $\frac{n}{2}+1 = 6$. The median is $\frac{4 + 8}{2}=6$.

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, $Q_3=10+(0.25)\times(10 - 10)=10$.

Step6: Find maximum

The maximum is 17.

Answer:

Minimum: 1, $Q_1$: 3, Median: 6, $Q_3$: 10, Maximum: 17

Explanation for part b:

Step1: Remove 17 from the data set. New data set: 1, 3, 3, 3, 4, 8, 9, 10, 10

Step2: Find new minimum

The new minimum is 1.

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

Step4: Find new median

The position of the median for $n = 9$ is $\frac{n + 1}{2}=5$. The median is 4.

Step5: Find 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, $Q_3=9+(0.5)\times(10 - 9)=9.5$.

Step6: Find new maximum

The new maximum is 10.

Answer for part b:

Minimum: 1, $Q_1$: 3, Median: 4, $Q_3$: 9.5, Maximum: 10