Question

Find the mean , median & mode age in years for the following frequency distribution

$\sf{\fbox{Age in years : (10 - 19) (20 - 29) (30 - 39) (40 - 49) (50 - 59) (60 - 69)}}$$\fbox{Number of persons : 8 , 8 , 10 , 14 , 28 , 32 }$

Answer 1

Answer:

Step-by-step explanation:

Refer to attachment for values of Median and mode.

To find mean, we will not use the formula for mean, that will make the answer too long.

For mean, we can use :-

3 Median = Mode + 2 Mean

Where we already got the values of Median and Mode.

So, Median = 54.92 years (~55)

Mode = 59.5 years (~60)

3(54.92) = 59.5 + 2 Mean

164.76 = 59.5 + 2 Mean

164.76 - 59.5 = 2 Mean

105.26 = 2 Mean

Mean = $\dfrac{105.36}{2}$

Answer :- Mean = 52.63 years

Mean = 53 years (Rounded off, approximate value)

Answer 2

Given :-

Table of years and number of person

To find :-

Median mode and mean

Solution :-

We know that

$\sf Median = l + \bigg(\dfrac{\dfrac{n}{2} - cf}{f}\bigg)\times h$

$\sf Median = 39.5+\bigg( \dfrac{\dfrac{100}{2} - 26}{14}\bigg)\times 9$

$\sf Median =39.5 + \bigg( \dfrac{50-26}{14}\bigg)\times 9$

$\sf Median = 39.5 \times \dfrac{24}{14} \times 9$

$\sf Median = 39.5 \times \dfrac{12}{7}\times 9$

$\sf Median = 54.92 = 55 year$

Finding mode

$\sf Mode =l \bigg( \dfrac{f_1 - f_0}{2f_1-f_0-f_2}\bigg)\times h$

$\sf Mode = 59.5 \times\bigg(\dfrac{32-28}{2(32) - 28}\bigg)\times9$

$\sf Mode = 59.5 + \bigg(\dfrac{32 - 28}{64 - 28}\bigg)\times 9$

$\sf Mode = 59.5 \times \bigg(\dfrac{4}{36}\bigg)\times9$

$\sf Mode = 535.5 \times \dfrac{1}{9}$

$\sf Mode = 59.5 = 60 \; years$

Finding mean

$\bf \dfrac{ \sum f_i x_i}{\sum x_i}$

Number of times happened = fi = 52.63

And

Number of frequency = xi = 8 + 8 + 10 + 14 + 28 + 32 = 100

By multiplying

$\sf \dfrac{52.63 \times 100}{100}$

$\dfrac{52.63}{1}$

$\sf 52.63 = 53 years$