Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, where x>y, be 8 and 16 respectively. Two numbers are chosen from $\{1, 2, 3, x-4, y, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:

Question

If the mean and the variance of the data

are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is

Answer 1

To solve this problem, we need to follow these steps:

Find the values of $ x $ and $ y $ using the given mean and variance of the observations.
Determine the probability that the smaller number among the chosen two numbers is less than 4.

Step 1: Calculate values of $ x $ and $ y $

The mean of 7 observations is given as 8.

$$\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$$

Solving this equation:

$$42 + x + y = 56$$

This simplifies to:

$$x + y = 14 \quad \text{(Equation 1)}$$

The variance is given as 16. The formula for variance is:

$$\frac{(2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2)}{7} - 8^2 = 16$$

Expanding and simplifying:

$$\frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7} - 64 = 16$$

Simplifying further:

$$\frac{460 + x^2 + y^2}{7} - 64 = 16$$

$$460 + x^2 + y^2 = 560$$

Thus,

$$x^2 + y^2 = 100 \quad \text{(Equation 2)}$$

Using equations 1 & 2, we solve for $ x $ and $ y $ using:

$$(x + y)^2 = x^2 + y^2 + 2xy $$

$$14^2 = 100 + 2xy$$

$$196 = 100 + 2xy$$

$$2xy = 96 \Rightarrow xy = 48$$

Simplifying $ x $ and $ y $ with Quadratic:

$$t^2 - 14t + 48 = 0$$

Finding roots:

$$t = \frac{14 \pm \sqrt{196 - 192}}{2} = \frac{14 \pm 2}{2}$$

This results in $ t = 8 $ or $ t = 6 $. Since $ x > y $, $ x = 8 $ and $ y = 6 $.

Step 2: Calculate the probability

The set of numbers is $\{1, 2, 3, 4, 6, 5\}$. We are choosing two numbers without replacement. The total number of ways to choose 2 numbers is:

$$\binom{6}{2} = 15$$

Favorable outcome: The smaller number among the two is less than 4. Consider the favorable pairs:

(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
(3, 4)
(3, 5)
(3, 6)

Total favorable outcomes = 12.

Therefore, the probability is:

$$\frac{12}{15} = \frac{4}{5}$$

Thus, the correct answer is $ \frac{4}{5} $.

Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, where x>y, be 8 and 16 respectively. Two numbers are chosen from \(\{1, 2, 3, x-4, y, 5\}\) one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is: