Solving for x and y using Mean and Variance
We are given 7 observations: 2, 4, 10, x, 12, 14, y, with mean $\bar{x} = 8$ and variance $\sigma^2 = 16$. We are also given the condition $x > y$.
- Calculate the sum of observations using the mean:
The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$.
Substituting the given values: $8 = \frac{2+4+10+x+12+14+y}{7}$
Simplify the sum: $8 = \frac{42+x+y}{7}$
Multiply by 7: $56 = 42+x+y$
Isolate $x+y$: $x+y = 56 - 42 = 14$. Let this be (Equation 1).
- Calculate the sum of squares of deviations using the variance:
The formula for variance is $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.
Substituting the values: $16 = \frac{(2-8)^2 + (4-8)^2 + (10-8)^2 + (x-8)^2 + (12-8)^2 + (14-8)^2 + (y-8)^2}{7}$
Multiply by 7: $112 = (-6)^2 + (-4)^2 + (2)^2 + (x-8)^2 + (4)^2 + (6)^2 + (y-8)^2$
Calculate squares: $112 = 36 + 16 + 4 + (x-8)^2 + 16 + 36 + (y-8)^2$
Sum the known squares: $112 = 108 + (x-8)^2 + (y-8)^2$
Isolate the terms with x and y: $(x-8)^2 + (y-8)^2 = 112 - 108 = 4$.
- Solve for x and y:
Expand the equation from step 2: $(x^2-16x+64) + (y^2-16y+64) = 4$
Group terms: $x^2+y^2 - 16(x+y) + 128 = 4$.
Substitute $x+y = 14$ (from Equation 1): $x^2+y^2 - 16(14) + 128 = 4$
$x^2+y^2 - 224 + 128 = 4$
$x^2+y^2 - 96 = 4$
$x^2+y^2 = 100$. Let this be (Equation 2).
Use the identity $(x+y)^2 = x^2+y^2+2xy$.
Substitute known values: $14^2 = 100 + 2xy$
$196 = 100 + 2xy \implies 2xy = 96 \implies xy = 48$.
We need two numbers ($x$ and $y$) whose sum is 14 and product is 48. These numbers are the roots of the quadratic equation $t^2 - (\text{sum})t + (\text{product}) = 0$.
$t^2 - 14t + 48 = 0$
Factor the quadratic: $(t-6)(t-8) = 0$. The roots are $t=6$ and $t=8$.
Given the condition $x > y$, we assign $x=8$ and $y=6$.
Calculating the Probability
The set from which two numbers are chosen is $\{1, 2, 3, x-4, y, 5\}$.
Substitute the values of $x$ and $y$: $\{1, 2, 3, 8-4, 6, 5\} = \{1, 2, 3, 4, 5, 6\}$.
The total number of elements in the set is $N=6$.
Two numbers are chosen one after another without replacement. The total number of possible ordered outcomes (sample space size) is $P(6, 2) = 6 \times 5 = 30$.
We need the probability that the smaller of the two chosen numbers is less than 4.
Let A be the event: $\min(\text{chosen pair}) < 4$.
Consider the complement event, A': $\min(\text{chosen pair}) \ge 4$. This means both chosen numbers must be 4 or greater.
The numbers in the set that are greater than or equal to 4 are $\{4, 5, 6\}$. There are 3 such numbers.
The number of ways to choose 2 numbers from $\{4, 5, 6\}$ without replacement, in order, is $P(3, 2) = 3 \times 2 = 6$.
The probability of the complement event A' is:
$P(A') = \frac{\text{Number of outcomes in A'}}{\text{Total number of outcomes}} = \frac{6}{30} = \frac{1}{5}$
The required probability $P(A)$ is $1 - P(A')$:
$P(A) = 1 - \frac{1}{5} = \frac{4}{5}$
Similar Questions
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Important Questions from Arithmetic
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