MHT CET · Maths · Statistics
A student scores the following marks in five tests : \(54,45,41,43,57\). His score is not known for the sixth test. If the mean score is 48 in six tests, then the standard deviation of marks in six tests is
- A \(\frac{100}{\sqrt{3}}\)
- B \(\frac{10}{\sqrt{3}}\)
- C \(\frac{100}{3}\)
- D \(\frac{10}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{10}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Let students' sixth test score be \(x\),
\(\begin{aligned}
\therefore \quad & \text { Mean }=48 \\
& \Rightarrow \frac{54+45+41+43+57+x}{6}=48 \\
& \Rightarrow x=48
\end{aligned}\)
\(\begin{aligned} \therefore \quad & \text { Standard deviation } \\ & =\sqrt{\frac{1}{\mathrm{n}} \sum\left(x_{\mathrm{i}}-\bar{x}\right)^2} \\ = & \sqrt{\frac{1}{6} \times\left[\left(6^2+(-3)^2+(-7)^2+(-5)^2+(9)^2\right)\right]} \\ = & \sqrt{\frac{200}{6}} \\ = & \frac{10}{\sqrt{3}}\end{aligned}\)
\(\begin{aligned}
\therefore \quad & \text { Mean }=48 \\
& \Rightarrow \frac{54+45+41+43+57+x}{6}=48 \\
& \Rightarrow x=48
\end{aligned}\)
\(\begin{aligned} \therefore \quad & \text { Standard deviation } \\ & =\sqrt{\frac{1}{\mathrm{n}} \sum\left(x_{\mathrm{i}}-\bar{x}\right)^2} \\ = & \sqrt{\frac{1}{6} \times\left[\left(6^2+(-3)^2+(-7)^2+(-5)^2+(9)^2\right)\right]} \\ = & \sqrt{\frac{200}{6}} \\ = & \frac{10}{\sqrt{3}}\end{aligned}\)
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