Chapter 12.4, Problem 6P

Find the minimum, first quartile, median, third quartile, and maximum of each dataset.

90 47 88 53 59 72 68 62 79

To determine

Find minimum , first quartile , median , third quartile , and maximum of the given data set.

Answer to Problem 6P

Minimum = 47

First quartile = 56

Median = 68

Third quartile = 83.5

Maximum = 90

Explanation of Solution

Given:

90 , 47 , 88 , 53 , 59 , 72 , 68 , 62 , 79

Calculation:

Arrange the data in ascending order:

47 , 53 , 59 , 62 , 68 , 72 , 79 , 88 , 90

The minimum is 47.

The first quartile is the median of the lower half of the data set.

Since the total entries are 9, which is odd , So, the lower half is the first $\frac{9-1}{2}=4$ entries.

So , the first quartile is $\frac{53+59}{2}=\frac{112}{2}=56$

Median = The middle entry since there are 9 entries , which is odd = $\frac{9+1}{2}=5^{th}$ entry = 68

The third quartile is the median of the upper half of the data set.

Since the total entries are 9, which is odd , So, the upper half is the last $\frac{9-1}{2}=4$ entries.

So , the third quartile is $\frac{79+88}{2}=\frac{167}{2}=83.5$

The maximum is 90