Question

The mean of 5 observations is \(4.4\) and their variance is \(8.24\). If three of the observations are \(1, 2\) and \(6\), then the other two observations are

• A. \(4, 9\)
• B. \(5, 8\)
• C. \(3, 10\)
• D. \(4, 10\)

Hint:

For MCQs involving mean and variance: \[ \sum x_i=n\bar{x} \] and \[ \sum x_i^2=n(\sigma^2+\bar{x}^2) \] These formulas quickly reduce the problem to solving simple equations.

Answer 1

The Correct Option is A

Concept: For \(n\) observations, \[ \text{Mean}=\frac{\sum x_i}{n} \] and \[ \sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^2 \] where \(\sigma^2\) is the variance.

Step 1: Find the sum of all observations. Given, \[ \bar{x}=4.4,\qquad n=5 \] Hence, \[ \sum x_i=5\times4.4=22 \] Let the remaining observations be \(a\) and \(b\). Then, \[ 1+2+6+a+b=22 \] \[ a+b=13 \]

Step 2: Use the variance formula. Given, \[ \sigma^2=8.24 \] \[ 8.24=\frac{\sum x_i^2}{5}-(4.4)^2 \] \[ 8.24=\frac{\sum x_i^2}{5}-19.36 \] \[ \frac{\sum x_i^2}{5}=27.60 \] \[ \sum x_i^2=138 \]

Step 3: Find \(a^2+b^2\). \[ 1^2+2^2+6^2+a^2+b^2=138 \] \[ 1+4+36+a^2+b^2=138 \] \[ a^2+b^2=97 \]

Step 4: Check the options. \[ \begin{aligned} (4,9) &:\quad a+b=13,\qquad a^2+b^2=16+81=97 \\ (5,8) &:\quad a+b=13,\qquad a^2+b^2=25+64=89 \\ (3,10) &:\quad a+b=13,\qquad a^2+b^2=9+100=109 \\ (4,10) &:\quad a+b=14 \end{aligned} \] Only option (A) satisfies both conditions. \[ \boxed{4,\;9} \] Hence, option \(\mathbf{(A)}\) is correct.