Let the observations `x_i(1 le I le 10)` satisfy the equations, `Sigma_(i=1)^(10)(x_i-5)=10` and `Sigma_(i=1)^(10) (x_i-5)^2=40`. If `mu` and `lamda` are the mean and the variance of the observations, `x_1-3, x_2-3, …, -3, ` then the ordered pair `(mu, lamda)` is equal to :

Options

A. `(3,3)`
B. `(6, 3)`
C. `(6,6)`
D. `(3, 6)`

Answer

To solve the problem, we need to find the mean (μ) and variance (λ) of the observations \( x_1 - 3, x_2 - 3, \ldots, x_{10} - 3 \) given the conditions on the observations \( x_i \).

### Step 1: Find \( \Sigma_{i=1}^{10} x_i \)

We are given the equation:
\[ \Sigma_{i=1}^{10} (x_i - 5) = 10 \]
This can be rewritten as:
\[ \Sigma_{i=1}^{10} x_i - 50 = 10 \]
So,
\[ \Sigma_{i=1}^{10} x_i = 10 + 50 = 60 \]

**Hint:** Remember to isolate \( \Sigma_{i=1}^{10} x_i \) by moving constants to the other side of the equation.

### Step 2: Find \( \Sigma_{i=1}^{10} x_i^2 \)

We are also given:
\[ \Sigma_{i=1}^{10} (x_i - 5)^2 = 40 \]
Expanding this, we have:
\[ \Sigma_{i=1}^{10} (x_i^2 - 10x_i + 25) = 40 \]
This simplifies to:
\[ \Sigma_{i=1}^{10} x_i^2 - 10\Sigma_{i=1}^{10} x_i + 250 = 40 \]
Substituting \( \Sigma_{i=1}^{10} x_i = 60 \):
\[ \Sigma_{i=1}^{10} x_i^2 - 10(60) + 250 = 40 \]
This gives:
\[ \Sigma_{i=1}^{10} x_i^2 - 600 + 250 = 40 \]
Thus,
\[ \Sigma_{i=1}^{10} x_i^2 - 350 = 40 \]
So,
\[ \Sigma_{i=1}^{10} x_i^2 = 390 \]

**Hint:** When expanding the square, remember to distribute each term correctly and combine like terms.

### Step 3: Calculate the Mean \( \mu \)

The mean of the observations \( x_i - 3 \) is given by:
\[ \mu = \frac{\Sigma_{i=1}^{10} (x_i - 3)}{10} = \frac{\Sigma_{i=1}^{10} x_i - 30}{10} \]
Substituting \( \Sigma_{i=1}^{10} x_i = 60 \):
\[ \mu = \frac{60 - 30}{10} = \frac{30}{10} = 3 \]

**Hint:** When calculating the mean, remember to adjust for the constant you are subtracting from each observation.

### Step 4: Calculate the Variance \( \lambda \)

The variance is calculated as:
\[ \lambda = \frac{\Sigma_{i=1}^{10} (x_i - 3)^2}{10} = \frac{\Sigma_{i=1}^{10} (x_i^2 - 6x_i + 9)}{10} \]
This expands to:
\[ \lambda = \frac{\Sigma_{i=1}^{10} x_i^2 - 6\Sigma_{i=1}^{10} x_i + 90}{10} \]
Substituting the known values:
\[ \lambda = \frac{390 - 6(60) + 90}{10} \]
Calculating this gives:
\[ \lambda = \frac{390 - 360 + 90}{10} = \frac{120}{10} = 12 \]

**Hint:** When calculating variance, ensure to include the constant adjustments and simplify step by step.

### Final Result

Thus, the ordered pair \( (μ, λ) \) is:
\[ (3, 12) \]