Let sets A and B have 5 elements each.

Question

Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 fr...

Accepted Answer

To solve this problem, let's break it down step by step :

Determine the mean of set : 1 : Determine the mean of set C

The mean of set A = $5$

The mean of set B = $8$

After subtracting 3 from each element in set A, the new mean becomes $5-3=2$.
After adding 2 to each element in set B, the new mean becomes $8+2=10$.

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:

Mean of C = $\frac{5	imes 2+5	imes 10}{10}=\frac{10+50}{10}=6$

Determine the variance of set : 2 : Determine the variance of set C

Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one :

Variance of C

= $\frac{{n}_{1}	imes +{n}_{1}	imes (-{)}^{2}+{n}_{2}	imes +{n}_{2}	imes (-{)}^{2}}{{n}_{1}+{n}_{2}}$

Given :

${n}_{1}={n}_{2}=5$

Variance of A = $12$, Variance of B = $20$

Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$

Plugging in the values, we get :

Variance of C = $\frac{5	imes 12+5	imes (2-6{)}^{2}+5	imes 20+5	imes (10-6{)}^{2}}{10}$

Variance of C = $\frac{60+80+100+80}{10}=\frac{320}{10}=32$

Sum of the mean and variance of set : 3 : Sum of the mean and variance of set C

Sum = Mean of C + Variance of C = $6+32=38$

So, the correct answer is :

Option C : 38.

Solutions · 1