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

Solution

Maths sir · 4 steps

Write the variance as a function of k, then turn variance < 10 into a bound on (k − 1).

1. Two sums. The data are nine 1s and one k. So Σ xᵢ = 9 + k and Σ xᵢ² = 9 + k², with n = 10.

   σ² = (9 + k²)/10 - ((9 + k)/10)² = (9k² - 18k + 9)/100 = 9(k-1)²/100

2. Apply the condition. 9(k-1)²/100 < 10 gives (k-1)² < 1000/9 ≈ 111.1. Since 10² = 100 < 111.1 < 121 = 11², the largest whole number k - 1 can be is 10.
3. Check the edge. For k = 11, σ² = (9 × 100)/100 = 9 < 10, so it works. For k = 12, σ² = (9 × 121)/100 = 10.89, which is too big.
4. So. The maximum possible value of k is 11.

The common slip: 10. Dividing by n - 1 = 9 instead of n = 10 gives (k-1)² < 100 and a maximum of 10, not 11. In JEE, variance means dividing by n.

The idea to keep. When all values but one are equal, the variance depends only on how far the odd value sits from the rest. Here it is 9(k-1)²/100. Get that formula first, then bound it.

Answer from the NTA final answer key. Solution by PrepWiser.