Question

How do you find the mean, median and mode of the following frequency distribution table?

Score	Number of students
$ 10 $	$ 6 $
$ 9 $	$ 13 $
$ 8 $	$ 12 $
$ 7 $	$ 11 $
$ 6 $	$ 13 $
$ 5 $	$ 5 $

Answer 1

Hint: First we make the table, the table has a score and the number of students and $ {f_i}{x_i} $ . But we find the value of $ {f_i}{x_i} $ .To find mean, median and mode.
We use the mean, median and mode formula. And then we have $ x $ and $ f $ substitute in the formula, we get mean, median and mode.
Mean:
 $ \overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
Median:
Median $ = \dfrac{{N + 1}}{2} $
Mode:
Mode $ Z = $ Frequency repeated in the set of data, a maximum number of times.

Complete Step by Step Solution:
Consider the data in the given table.
Mean:
First we find the mean in the given distribution table.
The mean $ = \dfrac{{{\text{sum of all data values}}}}{{{\text{the number of data values}}}} $
Therefore, we find
 $ \overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
We make the table, the table has a score and number of students and $ {f_i}{x_i} $ . But we find the value of $ {f_i}{x_i} $ .
Let, $ {x_i} $ is the score and $ {f_i} $ is the number of students
Multiply $ {x_i} $ by $ {f_i} $ for each term. For example, $ {x_i} = 10 $ and $ {f_i} = 6 $ ,we find $ {f_i}{x_i} $ , then $ {f_i}{x_i} = 60 $ .
Let’s make the table

Score $ {x_i} $	Number of students $ {f_i} $	$ {f_i}{x_i} $
$ 10 $	$ 6 $	$ 60 $
$ 9 $	$ 13 $	$ 117 $
$ 8 $	$ 12 $	$ 96 $
$ 7 $	$ 11 $	$ 77 $
$ 6 $	$ 13 $	$ 78 $
$ 5 $	$ 5 $	$ 25 $
Total	$ \sum {{f_i} = } 60 $	$ \sum {{f_i}{x_i} = 453} $

Therefore, we calculate mean,
 $ \overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
Now we substitute $ \sum {{f_i}{x_i}} $ and $ \sum {{f_i}} $ in the mean formula,
 $ \overline x = \dfrac{{453}}{{60}} $
Divide $ 453 $ by $ 60 $
 $ \overline x = 7.55 $
The mean is $ \overline x = 7.55 $
Median:
First we make the table, the table has a score $ (x) $ and number of students $ (f) $ , and we find the cumulative frequency $ (c.f) $

$ x $	$ f $	$ c.f $
$ 10 $	$ 6 $	$ 60 $
$ 9 $	$ 13 $	$ 54 $
$ 8 $	$ 12 $	$ 41 $
$ 7 $	$ 11 $	$ 29 $
$ 6 $	$ 13 $	$ 18 $
$ 5 $	$ 5 $	$ 5 $
Total	$ 60 $	-

Here, $ N = 60 $ .
Hence, The Median $ = \dfrac{{N + 1}}{2} $
Now we substitute $ N $ in the median formula
Median $ = \dfrac{{60 + 1}}{2} = \dfrac{{61}}{2} $
Now divide $ 61 $ by $ 2 $
Median $ = 30.5 $
The value of $ x $ for which the $ c.f $ is just greater than $ 30.5 $ is given by $ x = 8 $
Therefore, $ x = 8 $ is the median of the frequency distribution.
Mode:
Mode $ Z = $ Frequency repeated in the set of data, maximum number of times.
Now see $ {f_i} $ in the table, $ 13 $ has repeated two times. So the mode is $ 13 $
Therefore, Mode $ Z = 13 $

Note: The following are the five measures of central tendencies that are in common use.
1.Arithmetic mean (mean).
2.Median.
3.Mode.
4.Geometric mean.
5.Harmonic mean.
In the case of the discrete frequency distribution we calculate the median as follows.
Calculate $ \dfrac{1}{2}N = \dfrac{1}{2}\sum {{f_i}} $
Find the cumulative frequency just greater than $ \dfrac{1}{2}N $
The corresponding value of the variants is the median.