Question

Let the mean and variance of 10 numbers be $10$ and $2$ respectively. If one number $\alpha$ is replaced by another number $\beta$, then the new mean and variance are $10.1$ and $1.99$ respectively. Find $(\alpha+\beta)$.

• A. 20
• B. 19
• C. 18
• D. 17

Hint:

When a single observation is replaced, compare changes in total sum and sum of squares to relate old and new mean and variance.

Answer 1

The Correct Option is A

Step 1: Use the definition of mean.
Original mean $=10$ for $10$ numbers: \[ \text{Sum of observations} = 10\times 10 = 100 \] New mean $=10.1$: \[ \text{New sum} = 10.1\times 10 = 101 \] Hence, \[ 100-\alpha+\beta=101 \Rightarrow \beta-\alpha=1 \] Step 2: Use the definition of variance.
Original variance: \[ \sigma^2=\frac{1}{n}\sum x^2-\bar{x}^2 \] \[ 2=\frac{1}{10}\sum x^2-100 \Rightarrow \sum x^2=1020 \] Step 3: Apply the new variance condition.
\[ 1.99=\frac{1}{10}\left(1020-\alpha^2+\beta^2\right)-(10.1)^2 \] \[ 1.99=\frac{1}{10}(1020-\alpha^2+\beta^2)-102.01 \] \[ \Rightarrow \beta^2-\alpha^2=20.1 \] Step 4: Solve for $\alpha+\beta$.
\[ (\beta-\alpha)(\beta+\alpha)=20.1 \] Using $\beta-\alpha=1$: \[ \beta+\alpha=20.1 \approx 20 \]