Question

what is the interquartile range of the data set shown below? {1, 1, 1, 4, 5, 6, 7, 15} a. 5 b. 6.5 c. 5.5 d. 14

Explanation:

Step1: Find the median (Q2)

The data set is \(\{1, 1, 1, 4, 5, 6, 7, 15\}\) with \(n = 8\) (even). The median is the average of the 4th and 5th values.
4th value: \(4\), 5th value: \(5\).
Median \(Q2=\frac{4 + 5}{2}=4.5\).

Step2: Find Q1 (median of lower half)

Lower half: \(\{1, 1, 1, 4\}\) (first 4 values). \(n = 4\) (even), median of lower half is average of 2nd and 3rd values.
2nd value: \(1\), 3rd value: \(1\).
\(Q1=\frac{1 + 1}{2}=1\)? Wait, no, wait: Wait, lower half for \(n = 8\) is first 4 elements: positions 1 - 4: \(1,1,1,4\). The median of lower half (Q1) is the median of these 4. For even \(n\) in subset, average of middle two. Middle two are 2nd and 3rd: \(1\) and \(1\), so Q1 = 1? Wait, no, maybe I made a mistake. Wait, another method: for \(n = 8\), the quartiles can be calculated as:

Position of \(Q1=\frac{n + 1}{4}=\frac{8 + 1}{4}=2.25\). So Q1 is \(1 + 0.25\times(1 - 1)=1\)? No, maybe better to use the method where we split the data into lower half (first 4) and upper half (last 4).

Lower half: \(\{1, 1, 1, 4\}\), median (Q1) is average of 2nd and 3rd: \(1\) and \(1\), so Q1 = 1.

Upper half: \(\{5, 6, 7, 15\}\), median (Q3) is average of 2nd and 3rd: \(6\) and \(7\), so \(Q3=\frac{6 + 7}{2}=6.5\).

Wait, no, wait, maybe I messed up the lower and upper halves. Wait, when \(n\) is even, the lower half is first \(n/2\) elements, upper half is last \(n/2\) elements. So for \(n = 8\), lower half: first 4: \(1,1,1,4\); upper half: last 4: \(5,6,7,15\).

Then Q1 is median of lower half: for lower half (4 elements), median is average of 2nd and 3rd: \(1\) and \(1\), so Q1 = 1. Q3 is median of upper half: average of 2nd and 3rd of upper half: \(6\) and \(7\), so Q3 = 6.5.

Then interquartile range (IQR) = Q3 - Q1 = \(6.5 - 1 = 5.5\)? Wait, but let's check again.

Wait, maybe another approach: Let's list the data in order (already ordered): \(1,1,1,4,5,6,7,15\).

Number of observations \(n = 8\).

The formula for quartiles when \(n\) is even:

\(Q1\) is the median of the first \(n/2\) observations.

\(Q3\) is the median of the last \(n/2\) observations.

First \(n/2 = 4\) observations: \(1,1,1,4\). Median of these 4: average of 2nd and 3rd: \((1 + 1)/2 = 1\). Wait, no, 2nd is \(1\), 3rd is \(1\), so median is \(1\).

Last \(n/2 = 4\) observations: \(5,6,7,15\). Median of these 4: average of 2nd and 3rd: \((6 + 7)/2 = 6.5\).

Then IQR = Q3 - Q1 = \(6.5 - 1 = 5.5\). Wait, but let's check the answer options. Option C is 5.5. So that's the answer.

Wait, but let's verify again. Maybe I made a mistake in Q1. Wait, another way: using the percentile formula.

\(Q1\) is the 25th percentile, \(Q3\) is the 75th percentile.

For 25th percentile: \(i = \frac{25}{100}\times n=\frac{25}{100}\times8 = 2\). So the 2nd and 3rd values? Wait, no, when \(i\) is integer, it's average of \(i\)th and \((i + 1)\)th? Wait, no, different methods.

Wait, the data set is \(1,1,1,4,5,6,7,15\).

Let's list the positions: 1:1, 2:1, 3:1, 4:4, 5:5, 6:6, 7:7, 8:15.

\(Q1\) (25th percentile): The value at 25% of 8 data points. \(0.25\times8 = 2\), so between 2nd and 3rd value. 2nd value is 1, 3rd value is 1. So \(Q1 = 1 + 0.25\times(1 - 1)=1\).

\(Q3\) (75th percentile): \(0.75\times8 = 6\), so between 6th and 7th value. 6th value is 6, 7th value is 7. So \(Q3 = 6 + 0.75\times(7 - 6)=6.75\)? Wait, no, that's a different method. Wait, now I'm confused.

Wait, maybe the correct method for quartiles in even \(n\) is:

For \(n = 8\), the quartiles are calculated as:

- \(Q1\): median of the first 4 observations (positions 1 - 4: \(1,1,1,4\)) → median is \((1 + 1)/2 = 1\) (since 2nd and 3rd are 1 and 1).

- \(Q2\): median of all 8 → \((4 + 5)/2 = 4.5\).

- \(Q3\): median of the last 4 observations (positions 5 - 8: \(5,6,7,15\)) → median is \((6 + 7)/2 = 6.5\).

Then IQR = Q3 - Q1 = \(6.5 - 1 = 5.5\). Which matches option C.

Yes, that makes sense. So the interquartile range is 5.5.

Answer:

C. 5.5