1. Given data for 5 observations

Let the five observations be $x_1,x_2,x_3,x_4,x_5$.

Their mean is $\bar x=\frac{24}{5}$ and variance is $\sigma^2=\frac{194}{25}.$

Using $\sigma^2=\frac{\sum x_i^2}{n}-\bar x^2,$ for $n=5$, $\frac{194}{25}=\frac{\sum x_i^2}{5}-\left(\frac{24}{5}\right)^2.$

Now, $\left(\frac{24}{5}\right)^2=\frac{576}{25}.$ So, $\frac{194}{25}=\frac{\sum x_i^2}{5}-\frac{576}{25}$ $\frac{\sum x_i^2}{5}=\frac{194+576}{25}=\frac{770}{25}=\frac{154}{5}.$ Hence, $\sum x_i^2=154.$

Also, sum of all 5 observations 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 four observations is $\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}.$ Thus, $x_1+x_2+x_3+x_4=4\cdot \frac{7}{2}=14.$

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

3. Find sum of squares of the first 4 observations

Since $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=154,$ and $x_5=10$, we get $x_1^2+x_2^2+x_3^2+x_4^2=154-10^2=154-100=54.$

4. Compute variance of the first 4 observations

Mean of first four observations is $\bar y=\frac{7}{2}.$

If their variance is $v$, then $v=\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\left(\frac{7}{2}\right)^2.$

Substitute values: $v=\frac{54}{4}-\frac{49}{4}$ $v=\frac{5}{4}.$

5. Check options

The required variance is $\boxed{\frac{5}{4}}.$ So the correct option is A.