MHT CET · Maths · Statistics
The means of 5 observations is 4.4 and variance is 8.24 . If three of the five observations are 1,2 and 6 , then the values of other two observations are
- A 5,7
- B 4,9
- C 3,9
- D 4,8
Answer & Solution
Correct Answer
(B) 4,9
Step-by-step Solution
Detailed explanation
Let the two observation be \(\mathrm{x}\) and \(\mathrm{y}\)
\(
\begin{aligned}
& 1+2+6+x+y=5 \times 4.4=22 \\
& \Rightarrow x+y=13
\end{aligned}
\)
Also \(\frac{1^2+2^2+6^2+x^2+y^2}{5}-(4.4)^2=8.24\)
\(
\begin{aligned}
& \Rightarrow \frac{41+x^2+y^2}{5}=8.24+19.36=27.60 \\
& \Rightarrow x^2+y^2=138-41 \\
& \Rightarrow x^2+(13-x)^2=97 \\
& \Rightarrow x^2-13 x+36=0 \\
& \Rightarrow x=4,9
\end{aligned}
\)
\(\Rightarrow y=9,4\)
\(
\begin{aligned}
& 1+2+6+x+y=5 \times 4.4=22 \\
& \Rightarrow x+y=13
\end{aligned}
\)
Also \(\frac{1^2+2^2+6^2+x^2+y^2}{5}-(4.4)^2=8.24\)
\(
\begin{aligned}
& \Rightarrow \frac{41+x^2+y^2}{5}=8.24+19.36=27.60 \\
& \Rightarrow x^2+y^2=138-41 \\
& \Rightarrow x^2+(13-x)^2=97 \\
& \Rightarrow x^2-13 x+36=0 \\
& \Rightarrow x=4,9
\end{aligned}
\)
\(\Rightarrow y=9,4\)
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