Find the Lower or First Quartile 45 , 8 , 51 , 47 , 31 , 45 , 12 , 55 , 48

$45$ , $8$ , $51$ , $47$ , $31$ , $45$ , $12$ , $55$ , $48$

Step 1

There are $9$ observations, so the median is the middle number of the arranged set of data. Splitting the observations either side of the median gives two groups of observations. The median of the lower half of data is the lower or first quartile. The median of the upper half of data is the upper or third quartile.

The median of the lower half of data is the lower or first quartile

The median of the upper half of data is the upper or third quartile

Step 2

Arrange the terms in ascending order.

$8, 12, 31, 45, 45, 47, 48, 51, 55$

Step 3

The median is the middle term in the arranged data set.

$45$

Step 4

The lower half of data is the set below the median.

$8, 12, 31, 45$

Step 5

The median for the lower half of data $8, 12, 31, 45$ is the lower or first quartile. In this case, the first quartile is $21.5$ .

The median is the middle term in the arranged data set. In the case of an even number of terms, the median is the average of the two middle terms.

$\frac{12 + 31}{2}$

Remove parentheses.

$\frac{12 + 31}{2}$

Add $12$ and $31$ .

$\frac{43}{2}$

Convert the median $\frac{43}{2}$ to decimal.

$21.5$

Find the Lower or First Quartile 45 , 8 , 51 , 47 , 31 , 45 , 12 , 55 , 48

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