Animated Solution for Mathematics - Statistics: If the variance of the frequency distribution $$\begin{array}{cccccccc}{\mathbf{x}}_{\mathbf{i}}& 2& 3& 4& 5& 6& 7& 8\\ \text{Frequency }{\mathbf{f}}_{\mathbf{i}}& 3& 6& 16& \alpha & 9& 5& 6\end{array}$$ is 3, then $\alpha$ is ______.

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering the hidden symmetry within a frequency distribution.

We have a distribution of $x_i$ values ranging from 2 to 8, with frequencies $f_i$ that include an unknown $\alpha$. Our goal is to find $\alpha$ given that the variance ${\sigma }^2$ is 3.

Many students would immediately reach for the standard variance formula:

While correct, this path is fraught with the danger of large numbers and complex algebraic manipulation. Instead, we adopt the mindset of a seasoned problem solver and use the assumed mean method.

By setting our assumed mean $A=5$, we define a new deviation variable $d_i=x_i-5$. We choose 5 because it is the value associated with our unknown $\alpha$.

When we calculate the deviation $d_i$ for $x_i=5$, we get $5-5=0$. This means the term involving $\alpha$ in our sum $\sum f_i{d}_i$ will vanish entirely.

Let us construct our table. The total frequency $N$ is the sum of all $f_i$:

Next, we calculate the product $f_i{d}_i$: - For $x_i=2,d_i=-3,f_i{d}_i=-9$ - For $x_i=3,d_i=-2,f_i{d}_i=-12$ - For $x_i=4,d_i=-1,f_i{d}_i=-16$ - For $x_i=5,d_i=0,f_i{d}_i=0$ - For $x_i=6,d_i=1,f_i{d}_i=9$ - For $x_i=7,d_i=2,f_i{d}_i=10$ - For $x_i=8,d_i=3,f_i{d}_i=18$

Summing these, we get $\sum f_i{d}_i=-9-12-16+0+9+10+18=0$. The $\alpha$ term has vanished, just as we planned.

Now, for the final column, $f_i{d}_i^2$, we multiply $d_i$ by $f_i{d}_i$:

With our sums ready, we apply the variance formula:

Substituting our values, we get:

The second term is zero, leaving us with the simple equation:

Cross-multiplying, we find $45+\alpha =50$, which leads us directly to $\alpha =5$. It is a beautiful result, isn't it? By choosing the right tool, we turned a potentially messy algebraic problem into a clean, logical victory.