# Statistics - Median & Mean

• Jun 2nd 2008, 04:56 AM
Risrocks
Statistics - Median & Mean
The table gives marks out of 50 by 40 candidates:
Class Frequency
1-10 7
11-20 8
21-30 15
31-40 7
41-50 3
(i) In which class deos the median mark lie?
(ii) calculate the mean mark.

I know the median is the value of the middle term. But the two 7's has got me confused. Thanks
• Jun 2nd 2008, 05:38 AM
janvdl
Quote:

Originally Posted by Risrocks
The table gives marks out of 50 by 40 candidates:
Class Frequency
1-10 7
11-20 8
21-30 15
31-40 7
41-50 3
(i) In which class deos the median mark lie?
(ii) calculate the mean mark.

I know the median is the value of the middle term. But the two 7's has got me confused. Thanks

You have to use Quantiles for this.
Calculate the 50th percentile.

We have 40 students, and we want the 50th percentile.
$
\frac{(i)(n)}{100} = \frac{(50)(40)}{100} = 20$

Now we have to look for the interval where the cumulative frequency is greater than or equal to 20.

That interval is the [21;30) interval.

The percentile formula for grouped data is:

$p_{i} = k_{l} + \frac{(k_{u} - k_{l})(\frac{in}{100} - F_{kl})}{f_{pi}}$

$p_{50} = 21 + \frac{(30-21)(20 - 15)}{15} = 24$

And that is your median :D

============

Now for the mean:
$
(\bar{x}) = \sum_{i = 1}^{k} \frac{f_{i} x_{i}}{n}$

$
= \sum_{i = 1}^{40} \frac{f_{i} x_{i}}{40}$
( $x_{i}$ are the class midpoints.)

$= \frac{930}{40} = 23.25$
• Jun 2nd 2008, 10:35 PM
Risrocks
Great stuff ... thank you (Rofl)