Question

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

For Mean :

• frequency (fi) = 50
• fi xi = 1860

Using Formula :

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

Putting Values :

$\longmapsto\tt{Mean=\cancel\dfrac{1860}{50}}$

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

For Mode :

• ${f}_{0}=10$
• ${f}_{1}=12$
• ${f}_{2}=8$
• 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 :

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

$\longmapsto\tt{40+\dfrac{2}{6}\times{10}}$

$\longmapsto\tt{40+\dfrac{10}{3}}$

$\longmapsto\tt{40+3.3}$

$\longmapsto\tt\bf{43.3}$

For Median :

• n / 2 = 50/2 = 25
• l = 40
• h = 10
• cf = 26

Using Formula :

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

Putting Values :

$\longmapsto\tt{40=\bigg(\dfrac{25-26}{12}\bigg)\times{10}}$

$\longmapsto\tt{40+\dfrac{(-1)}{12}\times{10}}$

$\longmapsto\tt{40-\dfrac{10}{12}}$

$\longmapsto\tt{40-0.8}$

$\longmapsto\tt\bf{39.2}$

___________________

• ${f}_{0}$=class preceding the modal class .
• ${f}_{1}$=frequency of modal class .
• ${f}_{2}$=class secceeding the modal class .
• l = lower limit
• h = class size
• n = number of observations
• cf = cumulative frequency of class preceding the median class .

___________________

Answer 2

Answer:

• Mean =7.14

• Median =7

• Mode = 4

Step-by-step explanation:

Mean = sum of observation/no. of observation

=(4+4+5+7+8+10+12)/7

=50/7

= 7.14

Median = (n+1)/2

n= no. of observations

{(7+1)/2}n^th term

4 th term

arrange in ascending order

4,4,5,7,8,10,12

Median =7

Mode is the number occurring most of the times of all observations

Mode =4

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