Let the mean and the variance of 5 observations $$x_1, x_2, x_3, x_4, x_5$$ be $$\frac{24}{5}$$ and $$\frac{194}{25}$$ respectively. If the mean and variance of the first 4 observations are $$\frac{7}{2}$$ and $$a$$ respectively, then $$(4a + x_5)$$ is equal to

Step 1: Find $$x_5$$ using the means.

Mean of 5 observations = $$\frac{24}{5}$$, so $$\sum_{i=1}^{5} x_i = 24$$.

Mean of first 4 observations = $$\frac{7}{2}$$, so $$\sum_{i=1}^{4} x_i = 14$$.

Therefore $$x_5 = 24 - 14 = 10$$.

Step 2: Find $$\sum x_i^2$$ for all 5 observations.

Variance of 5 observations = $$\frac{194}{25}$$:

$$\frac{\sum_{i=1}^{5} x_i^2}{5} - \left(\frac{24}{5}\right)^2 = \frac{194}{25}$$

$$\frac{\sum x_i^2}{5} = \frac{194}{25} + \frac{576}{25} = \frac{770}{25} = \frac{154}{5}$$

$$\sum_{i=1}^{5} x_i^2 = 154$$

Step 3: Find the variance $$a$$ of the first 4 observations.

$$\sum_{i=1}^{4} x_i^2 = 154 - x_5^2 = 154 - 100 = 54$$

$$a = \frac{\sum_{i=1}^{4} x_i^2}{4} - \left(\frac{7}{2}\right)^2 = \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$$

Step 4: Compute $$4a + x_5$$.

$$4a + x_5 = 4 \times \frac{5}{4} + 10 = 5 + 10 = 15$$

The answer is Option B: 15.