If the mean and variance of five observations are $$\frac{24}{5}$$ and $$\frac{194}{25}$$ respectively and the mean of first four observations is $$\frac{7}{2}$$, then the variance of the first four observations is equal to

We are given five observations with mean $$\frac{24}{5}$$ and variance $$\frac{194}{25}$$, and the mean of the first four observations is $$\frac{7}{2}$$. We need to find the variance of the first four observations.

Since the mean of all five observations is $$\frac{24}{5}$$, their sum is $$\sum_{i=1}^{5}x_i = 5 \times \frac{24}{5} = 24$$, and because the mean of the first four observations is $$\frac{7}{2}$$, we have $$\sum_{i=1}^{4}x_i = 4 \times \frac{7}{2} = 14$$. It follows that the fifth observation is $$x_5 = 24 - 14 = 10$$.

Using the variance formula $$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$$ for the five observations, we write $$\frac{194}{25} = \frac{\sum_{i=1}^{5}x_i^2}{5} - \left(\frac{24}{5}\right)^2$$. Since $$\left(\frac{24}{5}\right)^2 = \frac{576}{25}$$, this gives $$\frac{\sum_{i=1}^{5}x_i^2}{5} = \frac{194 + 576}{25} = \frac{770}{25}$$ and hence $$\sum_{i=1}^{5}x_i^2 = 5 \times \frac{770}{25} = 154$$.

For the first four observations we then have $$\sum_{i=1}^{4}x_i^2 = 154 - x_5^2 = 154 - 100 = 54$$. Their variance is therefore $$\sigma_4^2 = \frac{\sum_{i=1}^{4}x_i^2}{4} - \left(\frac{\sum_{i=1}^{4}x_i}{4}\right)^2 = \frac{54}{4} - \left(\frac{14}{4}\right)^2 = \frac{216}{16} - \frac{196}{16} = \frac{20}{16} = \frac{5}{4}$$.

Answer: Option 3 — $$\frac{5}{4}$$