Given: 

$\overline{x}=\frac{24}{5}$ and variance is $\frac{194}{25}$

$\Rightarrow \sum _{i=1}^5{x}_i=\frac{24}{5}\times 5$

$\Rightarrow \sum _{i=1}^5{x}_i=24$

Also, variance

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

$\Rightarrow \sum _{i=1}^5{\left(x_i\right)}^2=154$

Now, $5^{\mathrm{th}}$ observation $=24-\text{sum of first} 4 \text{terms}$

$=24-\frac{7}{2}\times 4=10$

So, new variance $=\frac{\sum _{i=1}^4{\left(x_i\right)}^2}{4}-{\left(\frac{7}{2}\right)}^2$

$=\frac{154-100}{4}-\frac{49}{4}$

$=\frac{5}{4}$