Let the mean and the variance of 6 observation a, b, 68, 44, 48, 60$a, b, 68, 44, 48, 60$ be 55$55$ and 194$194$, respectively if a > b$a > b$, then a + 3b$a + 3b$ is

Related Formula

Core Logic

Mean is 55:

Variance is 194:

Expand the squares using a+b=110$a+b=110$: a² + b² - 110(a+b) + 2(3025) = 800$a^2 + b^2 - 110(a+b) + 2(3025) = 800$ a² + b² - 110(110) + 6050 = 800 a² + b² = 6850$a^2 + b^2 - 110(110) + 6050 = 800 \implies a^2 + b^2 = 6850$ Using (a+b)² = 12100 a²+b²+2ab = 12100 2ab = 12100 - 6850 = 5250$(a+b)^2 = 12100 \implies a^2+b^2+2ab = 12100 \implies 2ab = 12100 - 6850 = 5250$. (a-b)² = a²+b² - 2ab = 6850 - 5250 = 1600 a-b = 40$(a-b)^2 = a^2+b^2 - 2ab = 6850 - 5250 = 1600 \implies a-b = 40$ (since a>b$a>b$).

Solving a+b=110$a+b=110$ and a-b=40$a-b=40$:

Finally, evaluate a + 3b$a + 3b$:

Chapter Mix

Class 11 Maths: Statistics