Question

Please solve this fast that's a request

Answer 1

Class interval

0 - 5

5 - 10

10 - 15

15 - 20

20 - 25

We will find class marks (xᵢ)

Class marks is average of upper limit and lower limit

(Refer attachment)

We will assume mean "a" from any value of xᵢ

Here we have taken 25/2 as "a"

Then mean is given as

$\sf \bar{x} = a + \frac{\Sigma f_id_i}{\Sigma f_i}$

So mean = $\sf \frac{25}{2} + \frac{20}{40}$

→ mean = $\sf\frac{25}{2} + \frac{1}{2}$

→ mean = 13

Hence the mean of given distribution is 13

Answer 2

$\begin {tabular} {|c|c|c|c|c|} </p><p>\cline {1-5} marks &f_{i}&x_{i} & d_{i} & f_{i}d_{i} \\</p><p>\cline {1-5} 0-5&5&2.5&-10&-50 \\</p><p>\cline {1-5} 5-10&10&7.5&-5&-50 \\</p><p>\cline {1-5} 10-15 &9&12.5(a)&0&0 \\</p><p>\cline {1-5} 15-20&8&17.5&5&40 \\</p><p>\cline {1-5} 20-25&8&22.5&10&80 \\</p><p>\cline {1-5} -&\sum f =40&-&-&\sum f_{i}d_{i} = -100+120=20 \\</p><p>\cline {1-5} </p><p>\end {tabular}$

$From \: above \: table , \sum f_{i} = 40 , \\\sum f_{i}d_{i} = 20 ,$

$Mean \bar {x} = a + \frac{\sum f_{i}d_{i} }{\sum f_{i} }$

$= 12.5 + \frac{20}{40} \\= 12.5 + \frac{1}{2} \\= 12.5 + 0.5 \\= 13$

Therefore.,

$\red { Mean } \green { = 13 }$

••♪