If the variance of the frequency distribution

is 3, then $$\alpha$$ is equal to ______.

To find the value of $$\alpha$$, we first determine the mean $$\bar{x}$$ of the given frequency distribution.

The total frequency $$N$$ is:

$$N = \sum f_i = 3 + 6 + 16 + \alpha + 9 + 5 + 6 = 45 + \alpha$$

The sum of the products of observations and their frequencies $$\sum f_i x_i$$ is:

$$\sum f_i x_i = (2 \times 3) + (3 \times 6) + (4 \times 16) + (5 \times \alpha) + (6 \times 9) + (7 \times 5) + (8 \times 6)$$

$$\sum f_i x_i = 6 + 18 + 64 + 5\alpha + 54 + 35 + 48 = 225 + 5\alpha$$

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Step 1: Calculate the mean $$\bar{x}$$

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{225 + 5\alpha}{45 + \alpha} = \frac{5(45 + \alpha)}{45 + \alpha} = 5$$

Since the mean is independent of $$\alpha$$, we find $$\bar{x} = 5$$.

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Step 2: Set up the variance formula

The variance $$\sigma^2$$ is given by:

$$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = 3$$

Now, we calculate the term $$\sum f_i (x_i - 5)^2$$ for all observations:

For $$x_i = 2$$: $$3 \times (2 - 5)^2 = 3 \times 9 = 27$$

For $$x_i = 3$$: $$6 \times (3 - 5)^2 = 6 \times 4 = 24$$

For $$x_i = 4$$: $$16 \times (4 - 5)^2 = 16 \times 1 = 16$$

For $$x_i = 5$$: $$\alpha \times (5 - 5)^2 = \alpha \times 0 = 0$$

For $$x_i = 6$$: $$9 \times (6 - 5)^2 = 9 \times 1 = 9$$

For $$x_i = 7$$: $$5 \times (7 - 5)^2 = 5 \times 4 = 20$$

For $$x_i = 8$$: $$6 \times (8 - 5)^2 = 6 \times 9 = 54$$

Summing these deviations multiplied by their frequencies:

$$\sum f_i (x_i - 5)^2 = 27 + 24 + 16 + 0 + 9 + 20 + 54 = 150$$

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Step 3: Solve for $$\alpha$$

Substituting this value into the variance equation:

$$\frac{150}{45 + \alpha} = 3$$

$$150 = 3(45 + \alpha)$$

$$50 = 45 + \alpha \implies \alpha = 5$$

Therefore, the value of $$\alpha$$ is equal to 5.