Question

If the variance of 10 natural numbers 1, 1, 1, ......., 1, k is less than 10, then the maximum possible value of k is ______

Hint:

For a population with two distinct values $a$ (frequency $n_1$) and $b$ (frequency $n_2$), the variance is $\frac{n_1 n_2 (a-b)^2}{(n_1+n_2)^2}$. Here: $\frac{9 \times 1 \times (k-1)^2}{10^2}$.

Answer 1

Correct Answer: 11

Step 1: Numbers: nine $1$s and one $k$. Total $n=10$.
Step 2: Mean $\bar{x} = \frac{9(1) + k}{10} = \frac{9+k}{10}$.
Step 3: Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{9(1^2) + k^2}{10} - \left(\frac{9+k}{10}\right)^2$.
Step 4: $\sigma^2 = \frac{10(9+k^2) - (81+k^2+18k)}{100} = \frac{9k^2 - 18k + 9}{100} = \frac{9(k-1)^2}{100}$.
Step 5: Given $\sigma^2<10 \Rightarrow \frac{9(k-1)^2}{100}<10 \Rightarrow (k-1)^2<\frac{1000}{9} \approx 111.11$.
Step 6: $k-1<\sqrt{111.11} \approx 10.54 \Rightarrow k<11.54$. The maximum natural number $k$ is $11$.