Question

Complete the following table.

Answer 1

Hi my dear friend

1. 6

2. 24

3. 16

4. 10


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Answer 2

Given:

Table of calculation.

To find:

The missing values of the table.

Solution:

We know that, cumulative frequency (CF) is the sum of frequency of that class and preceding frequencies.

$CF_n=\sum f_{n}$

Using this, we get

$CF_1=\sum f_{1}$

$6=\sum f_{1}$

Therefore, frequency of class 100-109 is 6.

$CF_2=\sum f_{2}$

$CF_2=f_1+f_2$

$CF_2=6+18$

$CF_2=24$

Therefore, CF of 100-119 is 24.

$CF_3=\sum f_3$

$CF_3=f_1+f_2+f_3$

$CF_3=CF_2+f_3$

$40=24+f_3$

$40-24=f_3$

$16=f_3$

Therefore, the frequency of class 120-129 is 16.

$CF_4=\sum f_4$

$CF_3=f_1+f_2+f_3+f_4$

$CF_3=CF_3+f_4$

$50=40+f_4$

$50-40=f_4$

$10=f_4$

Therefore, the frequency of class 130-139 is 10.