consider the following set of data: 2.0, 12.8, 3.2, 4.0, 2.8, 6.6, 3.8, 6.6, 2.8, 2.8
notice that we have 10 values. this means the median will be the 5th value, the lower quartile will be halfway between the 2nd and 3rd values, and the upper quartile will be halfway between the 7th and 8th values.
which statement is true regarding how the five critical points changed after removing the outlier?
options (partially visible):
a: the median and the upper extreme decreased.
b: all of the five critical points increased.
c: the lower extreme remained unchanged, but the median and upper quartile increased.
d: the median remained the same, while the lower extreme decreased and the upper extreme increased.

Question

Accepted Answer

To solve this, we first identify the outlier. The data set is $2.0, 12.8, 3.2, 4.0, 2.8, 6.6, 3.8, 6.6, 2.8, 2.8$. The outlier is $12.8$ (much larger than others).

### Step 1: Original Five - Number Summary (with outlier)
- **Sort the data**: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6, 12.8$
- **Minimum (lower extreme)**: $2.0$
- **Lower Quartile ($Q_1$)**: The lower half is $2.0, 2.8, 2.8, 2.8, 3.2$. The median of the lower half (2nd and 3rd values average) $\frac{2.8 + 2.8}{2}=2.8$
- **Median ($Q_2$)**: The 5th and 6th values (since $n = 10$, median is average of 5th and 6th). $\frac{3.2+3.8}{2}=3.5$
- **Upper Quartile ($Q_3$)**: The upper half is $3.8, 4.0, 6.6, 6.6, 12.8$. The median of the upper half (7th and 8th values average) $\frac{4.0 + 6.6}{2}=5.3$
- **Maximum (upper extreme)**: $12.8$

### Step 2: Five - Number Summary after removing outlier ($12.8$)
- **Sort the new data**: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6$ ($n=9$)
- **Minimum (lower extreme)**: $2.0$ (remains same)
- **Lower Quartile ($Q_1$)**: The lower half is $2.0, 2.8, 2.8, 2.8$. The median of the lower half (2nd and 3rd values average) $\frac{2.8 + 2.8}{2}=2.8$ (remains same)
- **Median ($Q_2$)**: The 5th value (since $n = 9$, median is 5th value) $3.2$ (was $3.5$ before, decreased? Wait, no, wait let's re - check. Wait original $n = 10$, median was $\frac{3.2 + 3.8}{2}=3.5$. After removing $12.8$, $n=9$, median is the 5th value. Sorted data: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6$. 5th value is $3.2$, which is less than $3.5$? Wait, no, maybe I made a mistake. Wait, no, let's re - evaluate the options. Wait the option C says "The lower extreme remained unchanged, but the median and upper quartile increased." Wait no, let's re - calculate the upper quartile after removal. After removal, upper half is $3.8, 4.0, 6.6, 6.6$ (wait $n = 9$, so the upper half is from the 5th value? No, for $n=9$, the positions are: $Q_1$ is the $\frac{n + 1}{4}=2.5$th term, $Q_2$ is the $\frac{n + 1}{2}=5$th term, $Q_3$ is the $\frac{3(n + 1)}{4}=7.5$th term. So $Q_1$: average of 2nd and 3rd term: $\frac{2.8+2.8}{2}=2.8$ (same as before). $Q_2$: 5th term $3.2$ (original $Q_2$ was $3.5$, so it decreased? No, maybe the option is mis - read. Wait the option C: "The lower extreme remained unchanged, but the median and upper quartile increased." Wait no, maybe my initial calculation is wrong. Wait let's re - sort the original data: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6, 12.8$. After removing $12.8$, the data is $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6$. Now, $Q_3$ for $n = 9$ is the 7.5th term: average of 7th and 8th term. 7th term is $4.0$, 8th term is $6.6$, so $Q_3=\frac{4.0 + 6.6}{2}=5.3$ (same as before? No, original $Q_3$ was average of 7th ($4.0$) and 8th ($6.6$) term (since $n = 10$, upper half is 5th to 10th terms: $3.2, 3.8, 4.0, 6.6, 6.6, 12.8$? Wait no, I made a mistake in dividing the data for quartiles. For $n = 10$ (even), the lower half is first 5 terms: $2.0, 2.8, 2.8, 2.8, 3.2$, upper half is last 5 terms: $3.8, 4.0, 6.6, 6.6, 12.8$. So $Q_1$ is median of lower half (3rd term? Wait no, for even $n$, the lower quartile is the median of the first $\frac{n}{2}$ terms. So $\frac{n}{2}=5$ terms, median of 5 terms is the 3rd term. So $Q_1 = 2.8$ (3rd term of lower half: $2.0, 2.8, 2.8, 2.8, 3.2$ is $2.8$). $Q_3$ is median of upper half (3rd term of upper half: $3.8, 4.0, 6.6, 6.6, 12.8$ is $6.6$? Wait I see my mistake earlier. Oh no! I messed up the quartile calculation. Let's use the correct method:

For a data set with $n$ values:

- If $n$ is even:
    - Lower half: first $\frac{n}{2}$ values, upper half: last $\frac{n}{2}$ values.
    - $Q_1$: median of lower half, $Q_3$: median of upper half.
- If $n$ is odd:
    - Lower half: first $\frac{n - 1}{2}$ values, upper half: last $\frac{n - 1}{2}$ values, median is the middle value.

### Correct Original Calculation ($n = 10$, even)
- Sorted data: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6, 12.8$
- Lower half ($n_1=5$): $2.0, 2.8, 2.8, 2.8, 3.2$, $Q_1$ (median of lower half) is the 3rd term: $2.8$
- Median ($Q_2$): average of 5th and 6th terms: $\frac{3.2 + 3.8}{2}=3.5$
- Upper half ($n_2 = 5$): $3.8, 4.0, 6.6, 6.6, 12.8$, $Q_3$ (median of upper half) is the 3rd term: $6.6$
- Minimum: $2.0$, Maximum: $12.8$

### Correct After Removal ($n = 9$, odd)
- Sorted data: $2.0, 2.8, 2.8, 2.8, 3.2, 3.8, 4.0, 6.6, 6.6$
- Lower half ($n_1=\frac{9 - 1}{2}=4$): $2.0, 2.8, 2.8, 2.8$, $Q_1$ (median of lower half) is average of 2nd and 3rd terms: $\frac{2.8+2.8}{2}=2.8$ (same as before)
- Median ($Q_2$): 5th term: $3.2$ (was $3.5$ before, so it decreased? No, wait the option C says "The lower extreme remained unchanged, but the median and upper quartile increased." Wait no, let's calculate $Q_3$ after removal. Upper half ($n_2=\frac{9 - 1}{2}=4$): $3.8, 4.0, 6.6, 6.6$, $Q_3$ (median of upper half) is average of 2nd and 3rd terms: $\frac{4.0+6.6}{2}=5.3$ (original $Q_3$ was $6.6$, so it decreased? This is confusing. Wait maybe the outlier is $12.8$, and when we remove it, let's re - check the options. Wait the option C: "The lower extreme remained unchanged, but the median and upper quartile increased." Wait no, maybe I made a mistake in the outlier. Wait $12.8$ is much larger than the other values. Let's check the other options:

- Option A: "The median and the upper extreme decreased." Upper extreme (maximum) was $12.8$, after removal it's $6.6$ (decreased). Median was $3.5$, after removal it's $3.2$ (decreased). Wait but $Q_1$ remained same. But let's check the other options.

- Option B: "All of the five critical points increased." No, minimum is same, median decreased, so B is wrong.

- Option C: "The lower extreme remained unchanged, but the median and upper quartile increased." Lower extreme (minimum) is $2.0$ (unchanged). Median before: $3.5$, after: $3.2$ (decreased), so C is wrong.

Wait, maybe my initial quartile calculation was wrong. Let's use the method of finding quartiles as the values at positions $\frac{n + 1}{4}$, $\frac{n + 1}{2}$, $\frac{3(n + 1)}{4}$

### Original ($n = 10$)
- Position of $Q_1$: $\frac{10+1}{4}=2.75$th term. So $Q_1=2.8+0.75\times(2.8 - 2.8)=2.8$
- Position of $Q_2$: $\frac{10 + 1}{2}=5.5$th term. $Q_2=3.2+0.5\times(3.8 - 3.2)=3.5$
- Position of $Q_3$: $\frac{3(10 + 1)}{4}=8.25$th term. $Q_3=6.6+0.25\times(6.6 - 6.6)=6.6$

### After Removal ($n = 9$)
- Position of $Q_1$: $\frac{9+1}{4}=2.5$th term. $Q_1=\frac{2.8 + 2.8}{2}=2.8$ (same as before)
- Position of $Q_2$: $\frac{9+1}{2}=5$th term. $Q_2 = 3.2$ (decreased from $3.5$)
- Position of $Q_3$: $\frac{3(9 + 1)}{4}=7.5$th term. $Q_3=\frac{4.0+6.6}{2}=5.3$ (decreased from $6.6$)
- Minimum: $2.0$ (same), Maximum: $6.6$ (decreased from $12.8$)

Wait, this is conflicting with the options. But maybe the intended outlier is $12.8$, and the correct option is A: "The median and the upper extreme decreased." Because median went from $3.5$ to $3.2$ (decreased), upper extreme (maximum) went from $12.8$ to $6.6$ (decreased). $Q_1$ remained same, but the option A says "The median and the upper extreme decreased" which is true.

# Answer:
A. The median and the upper extreme decreased.

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QUESTION IMAGE

Question

Explanation:

Step 1: Original Five - Number Summary (with outlier)

Step 2: Five - Number Summary after removing outlier (\(12.8\))

Answer: