Animated Solution for Mathematics - Statistics: If the mean of the frequency distribution $$\begin{array}{llllll}\text{Class :}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \text{Frequency :}& 2& 3& x& 5& 4\end{array}$$ is 28, then its variance is ________ .

To begin, we must identify the missing frequency $x$ using the given mean $\bar{x}=28$. We first determine the class marks $x_i$ for each interval: $5,15,25,35,$ and $45$.

Next, we calculate the product $f_i{x}_i$ for each row: $2(5)=10$ $3(15)=45$ $x(25)=25x$ $5(35)=175$ $4(45)=180$

The sum of frequencies is $N=2+3+x+5+4=14+x$. The sum of products is $\sum f_i{x}_i=410+25x$.

We set up the mean formula as follows:

Cross-multiplying yields $28(14+x)=410+25x$, which simplifies to $392+28x=410+25x$. Solving for $x$, we find $3x=18$, which results in $x=6$.

With $x$ determined, our total frequency is $N=14+6=20$.

Variance, denoted by ${\sigma }^2$, measures the average of the squared deviations from the mean. It quantifies how far each data point is from the center of the distribution.

The formula for variance is:

We square the deviations because simply summing $(x_i-\bar{x})$ would result in zero due to the cancellation of positive and negative differences. Squaring ensures that every deviation contributes positively to the total spread.

We calculate the squared deviations for each class mark relative to the mean $\bar{x}=28$:

For $x_1=5$: $2(5-28{)}^2=2(-23{)}^2=2(529)=1058$ For $x_2=15$: $3(15-28{)}^2=3(-13{)}^2=3(169)=507$ For $x_3=25$: $6(25-28{)}^2=6(-3{)}^2=6(9)=54$ For $x_4=35$: $5(35-28{)}^2=5(7{)}^2=5(49)=245$ For $x_5=45$: $4(45-28{)}^2=4(17{)}^2=4(289)=1156$

Summing these values, we obtain:

Finally, dividing by the total frequency $N=20$:

The final variance of the distribution is $151$.