Question

the mean of the following frequency distribution is 25 find the value of k class interval is equal to 0-10 10-20 20-30 30-40 40-50 frequency is equal to 4 6 10 6 K​

Answer 1

The value of k will be 4

Step-by-step explanation:

Given:

Class intervals- 0-10  10-20  20-30  30-40  40-50  

Frequency- 4  6  10  6  k

mean of the distribution ($\bar x$)= 25

To find:

value of k

Solution:

First, we take out the classmarks of the classes

Classmark($x_{i}$)= Upper limit of class + Lower limit of class

                                                    2

Classmark for the first interval 0-10= 10+0/2= 5

Classmark for the second interval 10-20= 20+10/2= 15

Classmark for the third interval 20-30= 30+20/2 = 25

Classmark for the fourth interval 30-40= 40+30/2 = 35

Classmark for the fifth interval 40-50= 40+50/2 = 45

Formula for mean ($\bar x$)= $\frac{\sum x_{i} f_{i}}{\sum f_{i} }$

Let's multiply the frequencies($f_{i}$) & classmarks($x_{i}$) to get ($x_{i} f_{i}$)

For the first interval: 5(4)=20

For the second interval: 15(6)= 90

For the third interval: 25(10)= 250

For the fourth interval: 35(6)= 210

For the fifth interval: 45(k)= 45k

mean ($\bar x$)= $\frac{\sum x_{i} f_{i}}{\sum f_{i} }$

25= 20+90+250+210+45k

             4+6+10+6+k

25=  570+45k

         26+k

25 (26+k)= 570+45k

650+25k= 570+45k

650-570= 45k-25k

80 = 20k

∴ 20k = 80

∴ k = $\frac{80}{20}$

∴ k = 4

The value of k will be 4

#SPJ3

Answer 2

Answer:

The value of k is 4.

Step-by-step explanation:

Given - Class intervals : 0-10, 10-20, 20-30, 30-40, 40-50
            Frequency - 4, 6, 10, 6, k
            Mean of distribution ($\bar x$) = 25

To find - K

Solution -

We first need to find the classmarks of every class.

$x_i = \frac{upper\;limit + lower\;limit}{2}$

$x_1 = \frac{10+0}{2} = 5\\\\ x_2 = \frac{20+10}{2} = 15\\\\x_3 = \frac{30+20}{2} = 25\\\\x_4 = \frac{40+30}{2} = 35\\\\x_5 = \frac{50+40}{2} = 45$

Next we find the value of $\sum x_if_i$

$x_1f_1 = 5\times 4 = 20\\x_2f_2 = 15\times 6 = 90\\x_3f_3 = 25\times 10 = 250\\x_4f_4 = 35\times 6 = 210\\x_5f_5 = 45\times k = 45k$

Now, we find the mean using the formula $\bar x = \frac{\sum x_if_i}{\sum f_i}$

$25 = \frac{20+90+250+210+45k}{4+6+10+6+k} \\\\25 = \frac{570+45k}{26+k} \\ 25 (26+k) = 570 + 45k\\650 + 25k = 570 + 45k\\80 = 20k\\k = 4$

Thus, the value of k is 4.

#SPJ2