The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :

Solution

Maths sir · 4 steps

Turn the mean into a + b and the variance into a² + b², and then the cube identity finishes the job.

1. Use the mean. Call the missing observations a and b. The mean is 5, so all five add to 5 × 5 = 25. That gives a + b = 25 - (1 + 3 + 5) = 16.

   σ² = (Σ xᵢ²)/n - x̄² ⇒ 8 = (Σ xᵢ²)/5 - 25

2. Use the variance. This gives Σ xᵢ² = 5 × 33 = 165. Remove the known squares 1 + 9 + 25 = 35, which leaves a² + b² = 130.
3. Find ab. 2ab = (a+b)² - (a² + b²) = 256 - 130 = 126, so ab = 63. As a check, a and b are 7 and 9.
4. So. a³ + b³ = (a+b)³ - 3ab(a+b) = 4096 - 3 × 63 × 16 = 4096 - 3024 = 1072. You get the same total from 7³ + 9³ = 343 + 729. The answer is (d).

The idea to keep. Mean gives you the sum, and variance gives you the sum of squares. With a+b and a²+b² known, you can get ab, and then any symmetric expression, like a³+b³.

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