The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and 9/4 respectively. Then x⁴ + y² is equal to :

Solution

Maths sir · 5 steps

With the mean 6 known, measure every value as a distance from 6 and the squares stay small.

1. Use the mean. Eight values with mean 6 add up to 48. The six known values add up to 4+5+6+6+7+8 = 36, so x + y = 12.
2. Use the variance. Variance is the average squared deviation from the mean. So the eight squared deviations from 6 must add up to 8 × 9/4 = 18.
3. Known deviations. For 4, 5, 6, 6, 7, 8 the squared deviations are 4, 1, 0, 0, 1, 4, which total 10. That leaves 18 - 10 = 8 for x and y.

   (x-6)² + (y-6)² = 8

4. Find x and y. Since x + y = 12, write x = 6 - d and y = 6 + d with d > 0. Then 2d² = 8 gives d = 2, so x = 4 and y = 8.
5. So. x⁴ + y² = 4⁴ + 8² = 256 + 64 = 320, which is option (b).

The idea to keep. When the mean is given, total squared deviation = n × variance. Subtract what the known values contribute, and the unknowns get a small, easy equation.

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