Question

The heights (in inches) of 13 plants are 6, 9, 10, 10, 10, 11, 11, 12, 12, 13, 14, 16, and 17. What is the interquartile range of this data set? A) 3.5 B) 6 C) 10.5 D) 11

Answer 1

In order to find the interquartile range we need to find the first, second and third quartiles. These three numbers are called quartiles (think quarters = 4) because they divide the data into 4 groups. In order to find these quartiles we need the data to be in order (least to greatest) which your data already is.

Let's first find the second quartile. This is also known as the Median. It is the middle value in your list of numbers. The middle number in your data is 11.

6, 9, 10, 10, 10, 11, (11 = Median), 12, 12, 13, 14, 16, 17

There are six numbers to the left of the middle number and six to the right. We have divided the data into two groups.

The first quartile is the middle number of the first group. Since there are 6 numbers there technically isn't a middle number, so we take the average of the two numbers closest to the middle. These are in parenthesis below:
6, 9, (10), (10), 11
Since the two number are the same (10) their average is also 10. The first quartile (denoted q1 = 10).

Now let's look at the numbers to the right of the median. Again there are 6 numbers here so none is technically in the middle but I have put in parenthesis the two closest to the middle.
12, 12, (13), (14), 16, 17.
The third quartile q3 is the average of 13 & 14. This is found by adding 13 + 14 and dividing the sum by 2. The third quartile is 13.5

Going back to our original list of data we have now split the data into four groups of 3 numbers each as follows:
6, 9, 10, (q1= 10), 10, 10, 11, (Median = 11), 12, 12, 13, (q3=13.5), 14, 16, 17

The interquartile range is the third quartile minus the first quartile. Here it is 13.5 - 10 = 3.5. It gives the range (the distance) between the first and third quartiles.

Answer 2

The given heights:

6, 9, 10, 10, 10, 11, 11, 12, 12, 13, 14, 16, 17.

The first step is to find Median in order to divide the above series into lower and upper quartiles.

To find the median, just cross out the first and the last value until you left with the single value, as the number of elements is odd.

Like first 6 would cancel the last 17,
second number 9 would cancel the second last number 16,
and so on.

By doing above exercise, we are left with 11.

So the median = 11.

Now any number below the median 11 would be in lower quartile series, while other half would be in upper one.

To find the lower quartile, find the median of 6, 9, 10, 10, 10, 11:
Now the numbers are even; therefore, we would do the above-mentioned exercise of cancelling first and the last element until we are left with just 2 elements:

The two elements are: 10, 10.
Take average = (10 + 10) / 2 = 10

So lower quartile = 10

To find the upper quartile, find the median of 12, 12, 13, 14, 16, 17:
Now the numbers are even; therefore, we would do the above-mentioned exercise of cancelling first and the last element until we are left with just 2 elements:

The two elements are: 13, 14.
Take average = (13 + 14) / 2 = 13.5

So Upper quartile = 13.5

The interquartile range = Upper Quartile - Lower Quartile = 13.5 - 10.0 = 3.5

So the correct option is (A) 3.5


-i