Let the observation $$x_i(1 \le i \le 10)$$ satisfy the equations $$\sum_{i=1}^{10}(x_i - 5) = 10$$, $$\sum_{i=1}^{10}(x_i - 5)^2 = 40$$. If $$\mu$$ and $$\lambda$$ are the mean and the variance of the observations, $$x_1 - 3, x_2 - 3, \ldots, x_{10} - 3$$, then the ordered pair $$(\mu, \lambda)$$ is equal to:

We are told that the ten numbers $$x_1,x_2,\ldots ,x_{10}$$ satisfy two conditions:

1. $$\displaystyle \sum_{i=1}^{10}(x_i-5)=10$$

2. $$\displaystyle \sum_{i=1}^{10}(x_i-5)^2=40$$

First we convert the given information into the usual sums of the $$x_i$$.

From the first condition we have

$$$ \sum_{i=1}^{10}(x_i-5)=\sum_{i=1}^{10}x_i-\sum_{i=1}^{10}5 =\sum_{i=1}^{10}x_i-50=10. $$$

So

$$$ \sum_{i=1}^{10}x_i = 50+10 = 60. $$$

The mean of the original observations is therefore

$$$ \overline{x}= \frac{\sum_{i=1}^{10}x_i}{10}=\frac{60}{10}=6. $$$

Now we form the shifted observations

$$$ y_i = x_i-3\qquad (i=1,2,\ldots,10). $$$

The mean $$\mu$$ of the $$y_i$$ is just the old mean minus $$3$$ because subtracting a constant from every observation moves the mean by the same constant:

$$$ \mu = \overline{x}-3 = 6-3 = 3. $$$

Next we determine the variance $$\lambda$$ of the $$y_i$$. A key fact is that adding or subtracting a constant does not change variance. Nevertheless, we shall verify this directly from the data.

By definition, for our ten observations the variance is

$$$ \lambda=\frac1{10}\sum_{i=1}^{10}(y_i-\mu)^2. $$$

Because $$y_i=x_i-3$$ and $$\mu=3$$, we have $$y_i-\mu = (x_i-3)-3 = x_i-6$$. Hence

$$$ \lambda=\frac1{10}\sum_{i=1}^{10}(x_i-6)^2. $$$

We already know $$\displaystyle\sum (x_i-5)^2$$, so we convert $$(x_i-6)^2$$ in terms of $$(x_i-5)$$. Observe that

$$$ x_i-6=(x_i-5)-1. $$$

Squaring and summing gives

$$$ \sum_{i=1}^{10}(x_i-6)^2 =\sum_{i=1}^{10}\left[(x_i-5)-1\right]^2 =\sum_{i=1}^{10}\Big[(x_i-5)^2-2(x_i-5)+1\Big]. $$$

We split this sum term by term:

$$$ \sum_{i=1}^{10}(x_i-6)^2 =\underbrace{\sum_{i=1}^{10}(x_i-5)^2}_{=40} -2\underbrace{\sum_{i=1}^{10}(x_i-5)}_{=10} +\underbrace{\sum_{i=1}^{10}1}_{=10}. $$$

Substituting the known values, we obtain

$$$ \sum_{i=1}^{10}(x_i-6)^2 = 40 -2(10) + 10 = 40 - 20 + 10 = 30. $$$

Therefore the variance is

$$$ \lambda = \frac1{10}\times 30 = 3. $$$

We have found

$$$ (\mu,\lambda) = (3,3). $$$

Hence, the correct answer is Option A.