Explanation
From the question, the mean has already being given as 11.4 and n as 5. Other necessary value have also being given hence we can then find the sample variance and deviation below.
[tex]\begin{gathered} sample\text{ variance}=\frac{\sum_{n\mathop{=}1}^5(\left(Value-Mean\right)^2}{n-1} \\ sample\text{ standard deviation =}\sqrt{sample\text{ }variance}\frac{}{} \end{gathered}[/tex]
Therefore, we will have the sample variance and and sample standard deviation as
[tex]\begin{gathered} sample\text{ }variance=\frac{91.2}{5-1}=\frac{91.2}{4}=22.8 \\ sample\text{ }standard\text{ }deviation=\sqrt{22.8}=4.7749 \end{gathered}[/tex]
Answer: sample variance = 22.8 and sample standard deviation is 4.7749
