1. Interpret the given sums

The question clearly intends $\sum_{i=1}^{10}(x_i-5)=10$ and $\sum_{i=1}^{10}(x_i-5)^2=40.$

We need the mean and variance of the observations $x_1-3,\ x_2-3,\ \dots,\ x_{10}-3.$

2. Find the mean of the original observations $x_i$

From $\sum_{i=1}^{10}(x_i-5)=10,$ we get $\sum_{i=1}^{10}x_i-10\cdot 5=10$ $\sum_{i=1}^{10}x_i-50=10$ $\sum_{i=1}^{10}x_i=60.$

Hence the mean of $x_1,x_2,\dots,x_{10}$ is $\bar x=\frac{60}{10}=6.$

Now the new observations are $x_i-3$, so their mean is $\mu=\bar x-3=6-3=3.$

3. Find the variance

Let $y_i=x_i-3.$ Then variance of $y_i$ is same as variance of $x_i$, because subtracting a constant does not change variance.

So we first find variance of the original $x_i$.

Given $\sum_{i=1}^{10}(x_i-5)^2=40.$ But note that $x_i-5=(x_i-6)+1.$ So, $\sum (x_i-5)^2=\sum \big((x_i-6)+1\big)^2.$ Expanding, $\sum (x_i-5)^2=\sum (x_i-6)^2+2\sum (x_i-6)+\sum 1.$

Since mean is $6$, $\sum_{i=1}^{10}(x_i-6)=0,$ and $\sum 1=10.$ Therefore, $40=\sum_{i=1}^{10}(x_i-6)^2+0+10.$ So, $\sum_{i=1}^{10}(x_i-6)^2=30.$

Hence variance of $x_i$ is $\sigma^2=\frac{1}{10}\sum_{i=1}^{10}(x_i-6)^2=\frac{30}{10}=3.$

Therefore variance of $y_i=x_i-3$ is also $\lambda=3.$

4. Final ordered pair

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

So the correct option is B.