Find the Upper or Third Quartile 7 , 12 , 3 , 14 , 17 , 20 , 5 , 3 , 17 , 4 , 13 , 2 , 15 , 9 , 15 , 18 , 16 , 9 , 1 , 6

$7$ , $12$ , $3$ , $14$ , $17$ , $20$ , $5$ , $3$ , $17$ , $4$ , $13$ , $2$ , $15$ , $9$ , $15$ , $18$ , $16$ , $9$ , $1$ , $6$

There are $20$ observations, so the median is the mean of the two middle numbers 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

Arrange the terms in ascending order.

$1, 2, 3, 3, 4, 5, 6, 7, 9, 9, 12, 13, 14, 15, 15, 16, 17, 17, 18, 20$

Find the median of $1, 2, 3, 3, 4, 5, 6, 7, 9, 9, 12, 13, 14, 15, 15, 16, 17, 17, 18, 20$ .

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.

$\frac{9 + 12}{2}$

Remove parentheses.

$\frac{9 + 12}{2}$

Add $9$ and $12$ .

$\frac{21}{2}$

Convert the median $\frac{21}{2}$ to decimal.

$10.5$

The upper half of data is the set above the median.

$12, 13, 14, 15, 15, 16, 17, 17, 18, 20$

The median for the upper half of data $12, 13, 14, 15, 15, 16, 17, 17, 18, 20$ is the upper or third quartile. In this case, the third quartile is $15.5$ .

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.

$\frac{15 + 16}{2}$

Remove parentheses.

$\frac{15 + 16}{2}$

Add $15$ and $16$ .

$\frac{31}{2}$

Convert the median $\frac{31}{2}$ to decimal.

$15.5$

Find the Upper or Third Quartile 7 , 12 , 3 , 14 , 17 , 20 , 5 , 3 , 17 , 4 , 13 , 2 , 15 , 9 , 15 , 18 , 16 , 9 , 1 , 6

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