1. Use the mean of all 5 observations

Given mean of 5 observations is $\bar{x} = \frac{24}{5}.$

So their sum is $x_1+x_2+x_3+x_4+x_5 = 5\cdot \frac{24}{5} = 24.$

2. Use the mean of the first 4 observations

Given mean of first 4 observations is $\frac{x_1+x_2+x_3+x_4}{4} = \frac{7}{2}.$

Hence, $x_1+x_2+x_3+x_4 = 4\cdot \frac{7}{2} = 14.$

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

3. Use variance formula for first 4 observations

For the first 4 observations, mean is $\frac{7}{2}$ and variance is $a$.

Using $a = \frac{1}{4}\sum_{i=1}^4 (x_i-\tfrac{7}{2})^2,$ we get $\sum_{i=1}^4 (x_i-\tfrac{7}{2})^2 = 4a.$

Also, $\sum_{i=1}^4 (x_i-\tfrac{7}{2})^2 = \sum_{i=1}^4 x_i^2 - 4\left(\frac{7}{2}\right)^2.$ Since $\sum_{i=1}^4 x_i = 14$, the standard identity gives $\sum_{i=1}^4 (x_i-\bar{x})^2 = \sum_{i=1}^4 x_i^2 - 4\bar{x}^2.$ So, $4a = \sum_{i=1}^4 x_i^2 - 4\cdot \frac{49}{4} = \sum_{i=1}^4 x_i^2 - 49.$ Thus, $\sum_{i=1}^4 x_i^2 = 4a+49.$

4. Use variance formula for all 5 observations

Mean of all 5 observations is $\frac{24}{5}$ and variance is $\frac{194}{25}$.

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

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

Using the identity $\sum_{i=1}^5 (x_i-\bar{x})^2 = \sum_{i=1}^5 x_i^2 - 5\bar{x}^2,$ we get $\sum_{i=1}^5 x_i^2 - 5\left(\frac{24}{5}\right)^2 = \frac{194}{5}.$

Now, $5\left(\frac{24}{5}\right)^2 = 5\cdot \frac{576}{25} = \frac{576}{5}.$ So, $\sum_{i=1}^5 x_i^2 = \frac{194}{5} + \frac{576}{5} = \frac{770}{5} = 154.$

5. Relate this to the first 4 observations

Since $x_5=10$, $\sum_{i=1}^4 x_i^2 = 154 - 10^2 = 154-100 = 54.$

But from step 3, $\sum_{i=1}^4 x_i^2 = 4a+49.$ So, $4a+49 = 54 \implies 4a=5.$

Therefore, $4a+x_5 = 5+10 = 15.$

6. Check options

• A: $13$ ❌
• B: $15$ ✅
• C: $17$ ❌
• D: $18$ ❌

Therefore, the correct answer is B.