Let the mean and variance of 8 numbers x, y, 10,12,6,12,4,8 be 9 and 9.25 respectively. If x > y, then 3 x-2 y is equal to _____________.

Solution

Maths sir · 4 steps

The mean gives you x + y and the variance gives you x² + y². Together they fix both numbers.

1. Use the mean. The sum of all 8 numbers is 8 × 9 = 72. The six known numbers add to 10 + 12 + 6 + 12 + 4 + 8 = 52, so x + y = 20.

   σ² = (Σ xᵢ²)/n - x̄² ⇒ 9.25 = (Σ xᵢ²)/8 - 81

2. Use the variance. This gives Σ xᵢ² = 8 × 90.25 = 722. The known squares add to 100 + 144 + 36 + 144 + 16 + 64 = 504, so x² + y² = 218.
3. Find x and y. From (x + y)² = x² + y² + 2xy we get 400 = 218 + 2xy, so xy = 91. Then x and y are the roots of t² - 20t + 91 = 0, which are 13 and 7. Since x > y, x = 13 and y = 7.
4. So. 3x - 2y = 39 - 14 = 25.

The common slip: -5. Ignoring the condition x > y and taking x = 7, y = 13 gives 21 - 26 = -5, not 25.

The idea to keep. Two unknowns in a data set? The mean gives their sum and the variance gives their sum of squares. Then use (x+y)² = x² + y² + 2xy to get the product.

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