Find mean, median and mode of the given data
$7, 3, 4, 5, 2, 3, 4, 5, 7, 3, 4, 5, 7, 7, 5, 7, 3, 6, 7, 7$

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Solution

Step 1: Finding the mean
We know that,
$M e a n = \frac{S u m o f a l l t h e o b s e r v a t i o n s}{T o t a l n u m b e r o f o b s e r v a t i o n}$
$= \frac{7 + 3 + 4 + 5 + 2 + 3 + 4 + 5 + 7 + 3 + 4 + 5 + 7 + 7 + 5 + 7 + 3 + 6 + 7 + 7}{20}$
$= \frac{101}{20} = 5.05$
Step 2: Finding the median
Arranging the given data in ascending order
$2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 7, 7, 7, 7, 7, 7, 7$
We can observe that total number of observation are even i.e.; $20$ .
So, median $=$ Mean of $(\frac{N}{2}) t h$ and $(\frac{N}{2} + 1) t h$ observation.
$=$ Mean of $(\frac{20}{2}) t h$ and $(\frac{20}{2} + 1) t h$ observation.
$=$ Mean of $10 t h$ and $11 t h$ observation.
$= \frac{5 + 5}{2} = \frac{10}{2} = 5$
Step 3: Finding the mode
By observing the data, we can see, $7$ is having the heighest frequency i.e.; $7$
So, the mode of given data is $7$ .
Hence, the mean, median and mode of given data are $5.05, 5$ and $7$ respectively.

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Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

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Key

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Find mean, median and mode of the given data7,3,4,5,2,3,4,5,7,3,4,5,7,7,5,7,3,6,7,7

Find mean, median and mode of the given data
$7, 3, 4, 5, 2, 3, 4, 5, 7, 3, 4, 5, 7, 7, 5, 7, 3, 6, 7, 7$