Name: Mean of Ungrouped data: Example
Uploaded: 2019-07-19
Duration: 2 min 7 s
Description: Familiarise yourself with the calculation of the arithmetic means of the values collected in the data. This example involves marks scored by students.

Question

Find the mean and median of all the positive factors of $72$ .

Accepted Answer

The positive factors of $72$ are $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$

We know that, if $x_{1}, x_{2}, x_{3}, x_{4}, \dots \dots, x_{n}$ are $n$ observations in an ungrouped data, then their mean is denoted by $\bar{x}$ , is given by $\bar{x} = \frac{x_{1} + x_{2} + x_{3} + \dots \dots . . + x_{n}}{n}$ .

Now, $\bar{x} = \frac{1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72}{12} = \frac{195}{12} = 16.25$

Thus, the mean of the given data is $\bar{x} = 16.25$ .

The given numbers in ascending order is $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$ .

Here, number of observations $(n)$ $= 12$

We know that, if $n$ is odd, median $= \frac{1}{2} [value of {(\frac{n}{2})}^{th} term + value of {(\frac{n}{2} + 1)}^{th} term]$ .

Now, median $= \frac{1}{2} [value of {(\frac{12}{2})}^{th} term + value of {(\frac{12}{2} + 1)}^{th} term]$

$= \frac{1}{2} [value of {(6)}^{th} term + value of {(7)}^{th} term]$

$= \frac{1}{2} [8 + 9]$

Thus, median of the given data $= 8.5$ .

Find the mean and median of all the positive factors of $72$ .

Important Questions on Mean and Median [For Ungrouped Data Only]

Learn and Prepare for any exam you want !

Find the mean and median of all the positive factors of 72.

Important Questions on Mean and Median [For Ungrouped Data Only]

Learn and Prepare for any exam you want !

Find the mean and median of all the positive factors of $72$ .