Question

pls solve this question ​

Answer 1

Given:

↬ Incomplete Distribution Table

$\begin{tabular}{c | l}Variable & Frequencies & \\\cline{1-2}0-10 & 10 \\10-20 & 20 \\20-30 & ? \\30-40 & 40 \\40-50 & ? \\50-60 & 25 \\60-70 & 15 \\ \end{tabular}$

↬ Median= 35

↬ Sum of all frequencies= 170

To Find:

✏ Value of missing frequencies

Solution:

We know that,

➺ Class or interval which contains the median is called Median Class

➺ Median of grouped frequency distribution

$\bf{Median=l+\Bigg(\dfrac{\frac{N}{2}-C}{f}\Bigg)\times h}$

where,

l is Lower limit of median class

N is Total frequency

C is Cumulative frequency of the class preceding the median class

f is Frequency of the median class

h is Width(size) of median class

Let the first and second missing values be x and y respectively

$\begin{tabular}{c | c l c l}Variable & Frequencies & Cumulative Frequencies \\\cline{1-3}0-10 & 10 & 10 \\10-20 & 20 & 30\\20-30 & x & 30+x \\30-40 & 40 & 70+x \\40-50 & y & 70+x+y \\50-60 & 25 & 95+x+y \\60-70 & 15 & 110+x+y\\Total & N=110+x+y \end{tabular}$

Now,

Class 30-40 is the median class because it contains the Median value.

So here,

• l= 30
• N= 170
• C is 30+x
• f is 40
• h is 10

Now,

$\longrightarrow\sf{Median=l+\Bigg(\dfrac{\frac{N}{2}-C}{f}\Bigg)\times h}$

$\longrightarrow\sf{35=30+\Bigg(\dfrac{\frac{170}{2}-(30+x)}{40}\Bigg)\times 10}$

$\longrightarrow\sf{35-30=\Bigg(\dfrac{85-30-x}{40}\Bigg)\times 10}$

$\longrightarrow\sf{35-30=\Bigg(\dfrac{55-x}{40}\Bigg)\times 10}$

$\longrightarrow\sf{5=\Bigg(\dfrac{55-x}{\cancel{40}}\Bigg)\times \cancel{10}}$

$\longrightarrow\sf{5=\Bigg(\dfrac{55-x}{4}\Bigg)}$

$\longrightarrow\sf{55-x=5\times4}$

$\longrightarrow\sf{55-x=20}$

$\longrightarrow\sf{55-20=x}$

$\longrightarrow\sf{x=55-20}$

$\longrightarrow\sf{x=35}--------(1)$

Also, it is given that

Sum of all frequencies= 170

110+x+y= 170

x+y= 170-110

x+y= 60

From (1)

35+y= 60

y= 60-35

y= 25

Hence, the values of missing frequencies are 35 and 25.