Question

Q1.Find the mean, median and the mode of the following frequency distribution. Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 frequency 4 6 8 12 10 5 5​.

Answer 1

Explanation:

For Mean :

frequency (fi) = 50

fi xi = 1860

Using Formula :

$\longmapsto\tt\boxed{Mean=\dfrac{\sum\:fixi}{fi}}$

Putting Values :

$\longmapsto\tt{Mean=\cancel\dfrac{1860}{50}}</p><p>$

\longmapsto\tt\bf{Mean=37.2}⟼Mean=37.2

For Mode :

{f}_{0}=10f

0

$</p><p> =10</p><p>{f}_{1}=12f$

${f}_{2}=8f </p><p>$

l = 40

h = 10

Using Formula :

$\longmapsto\tt\boxed{Mode=l+\dfrac{{f}_{1}-{f}_{0}}{{2f}_{1}-{f}_{0}-{f}_{2}}\times{h}}$

Putting Values :

$\longmapsto\tt{40+\bigg(\dfrac{12-10}{24-18}\bigg)\times{10}}$

$\longmapsto\tt{40+\dfrac{2}{6}\times{10}}⟼40+ </p><p>6</p><p>2</p><p> </p><p> ×10</p><p></p><p>⟼43.3 \\ \longmapsto\tt{40+\dfrac{10}{3}}</p><p> </p><p> </p><p></p><p>\longmapsto\tt{40+3.3} \\ </p><p>\longmapsto\tt\bf{43.3}</p><p>$

For Median :

n / 2 = 50/2 = 25

l = 40

h = 10

cf = 26

Using Formula :

$\boxed{Median=l+\bigg(\dfrac{\dfrac{n}{2}-cf}{f}\bigg)\times{h}} </p><p>$

Putting Values :

$\longmapsto\tt{40=\bigg(\dfrac{25-26}{12}\bigg)\times{10}} \\ </p><p>\longmapsto\tt{40+\dfrac{(-1)}{12}\times{10}}</p><p> \\ </p><p>\longmapsto\tt{40-\dfrac{10}{12}}</p><p> </p><p> </p><p> \\ </p><p>\longmapsto\tt{40-0.8} \\ </p><p></p><p>\longmapsto\tt\bf{39.2}⟼39.2</p><p>$

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{f}_{0}f

0

=class preceding the modal class .

{f}_{1}f

1

=frequency of modal class .

{f}_{2}f

2

=class secceeding the modal class .

l = lower limit

h = class size

n = number of observations

cf = cumulative frequency of class preceding the median class .

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