Question

the data set represents the number of rings each person in a room is wearing.
0, 2, 4, 0, 2, 3, 2, 8, 6
what is the interquartile range of the data?
3
6
2
4

Explanation:

Step1: Order the data set

First, we need to order the data set from least to greatest. The given data set is \(0, 2, 4, 0, 2, 3, 2, 8, 6\). After ordering, we get: \(0, 0, 2, 2, 2, 3, 4, 6, 8\).

Step2: Find the median (Q2)

The median is the middle value of the ordered data set. Since there are 9 data points (an odd number), the median is the 5th value. The ordered data set is \(0, 0, 2, 2, \boldsymbol{2}, 3, 4, 6, 8\), so the median (Q2) is \(2\).

Step3: Find the lower quartile (Q1)

The lower quartile (Q1) is the median of the lower half of the data. The lower half of the data (excluding the median) is \(0, 0, 2, 2\). Since there are 4 data points (an even number), the median of this subset is the average of the 2nd and 3rd values. The 2nd value is \(0\) and the 3rd value is \(2\), so \(Q1=\frac{0 + 2}{2}=1\). Wait, no, wait. Wait, the lower half is the data before the median. The median is the 5th value, so the lower half is the first 4 values: \(0, 0, 2, 2\). The median of these 4 values: the middle two are the 2nd and 3rd values. So \(\frac{0 + 2}{2}=1\)? Wait, no, maybe I made a mistake. Wait, let's check again. The ordered data is \(0, 0, 2, 2, 2, 3, 4, 6, 8\). The median is the 5th term, which is \(2\). The lower half is the first 4 terms: \(0, 0, 2, 2\). The median of the lower half (Q1) is the average of the 2nd and 3rd terms: \(\frac{0 + 2}{2}=1\)? Wait, no, that doesn't seem right. Wait, maybe the method is different. Alternatively, for a data set with \(n = 9\) (odd), the positions are:

- Position of Q1: \(\frac{n + 1}{4}=\frac{9 + 1}{4}=2.5\)th term.
- Position of Q2: \(\frac{2(n + 1)}{4}=\frac{2(9 + 1)}{4}=5\)th term (which we already found as \(2\)).
- Position of Q3: \(\frac{3(n + 1)}{4}=\frac{3(9 + 1)}{4}=7.5\)th term.

So, Q1 is the value at the 2.5th position. To find this, we take the average of the 2nd and 3rd terms. The 2nd term is \(0\) and the 3rd term is \(2\), so \(Q1=\frac{0 + 2}{2}=1\).

Step4: Find the upper quartile (Q3)

The upper quartile (Q3) is the median of the upper half of the data. The upper half of the data (excluding the median) is \(3, 4, 6, 8\). Since there are 4 data points (an even number), the median of this subset is the average of the 2nd and 3rd values. The 2nd value is \(4\) and the 3rd value is \(6\), so \(Q3=\frac{4 + 6}{2}=5\). Wait, using the position method: the 7.5th term. The 7th term is \(4\) and the 8th term is \(6\), so \(Q3=\frac{4 + 6}{2}=5\).

Step5: Calculate the interquartile range (IQR)

The interquartile range is \(Q3 - Q1\). We found \(Q1 = 1\) and \(Q3 = 5\)? Wait, that can't be right. Wait, maybe I made a mistake in finding Q1. Let's re-examine the ordered data: \(0, 0, 2, 2, 2, 3, 4, 6, 8\). Let's list the positions:

1: 0

2: 0

3: 2

4: 2

5: 2 (median, Q2)

6: 3

7: 4

8: 6

9: 8

Now, the lower quartile (Q1) is the median of the first 4 terms (positions 1 - 4): \(0, 0, 2, 2\). The median of these is the average of the 2nd and 3rd terms: \(\frac{0 + 2}{2}=1\). The upper quartile (Q3) is the median of the last 4 terms (positions 6 - 9): \(3, 4, 6, 8\). The median of these is the average of the 7th and 8th terms: \(\frac{4 + 6}{2}=5\). Then IQR = Q3 - Q1 = \(5 - 1 = 4\)? Wait, but let's check again. Wait, maybe the lower half is considered as including the median? No, the median is excluded from both halves when the number of data points is odd. Wait, another method: for a data set with \(n\) elements, the median is at position \(\frac{n + 1}{2}\). For \(n = 9\), that's position 5. Then the lower quartile is at position \(\frac{n + 1}{4}=\frac{10}{4}=2.5\), and the upper quartile is at position \(\frac{3(n + 1)}{4}=\frac{30}{4}=7.5\). So for position 2.5, we take the average of the 2nd and 3rd terms: \(\frac{0 + 2}{2}=1\) (Q1). For position 7.5, we take the average of the 7th and 8th terms: \(\frac{4 + 6}{2}=5\) (Q3). Then IQR = \(5 - 1 = 4\). Wait, but let's check with another approach. Let's use the formula for IQR: IQR = Q3 - Q1, where Q1 is the 25th percentile and Q3 is the 75th percentile.

Alternatively, let's list the data again: \(0, 0, 2, 2, 2, 3, 4, 6, 8\).

Number of data points \(n = 9\).

Q1 is the value at the \(\frac{n + 1}{4}\)th position: \(\frac{9 + 1}{4}=2.5\)th position. So we take the average of the 2nd and 3rd values: \(\frac{0 + 2}{2}=1\).

Q3 is the value at the \(\frac{3(n + 1)}{4}\)th position: \(\frac{3(9 + 1)}{4}=7.5\)th position. So we take the average of the 7th and 8th values: \(\frac{4 + 6}{2}=5\).

Then IQR = Q3 - Q1 = \(5 - 1 = 4\). Wait, but let's check with another method. Let's use the median to split the data into lower and upper halves.

Lower half (below median): \(0, 0, 2, 2\) (4 values)

Upper half (above median): \(3, 4, 6, 8\) (4 values)

Median of lower half (Q1): \(\frac{0 + 2}{2}=1\)

Median of upper half (Q3): \(\frac{4 + 6}{2}=5\)

IQR = \(5 - 1 = 4\).

Wait, but let's check the answer options. The options are 3, 6, 2, 4. So 4 is an option. So that must be the answer.

Answer:

\(4\)