Q.52 - A First Course in Probability

Question

A population is made up of $r$ disjoint subgroups. Let $p_i$ denote the proportion of the population that is in subgroup $i,i=1,\dots ,r$. If the average weight of the members of subgroup $i$ is $w_i,i=1,\dots ,r$, what is the average weight of the members of the population?

Step-by-Step Solution

Answer

The average weight of the members of the population is $E\left(W\right)=\sum W_i{p}_i$

Step 1: Given information

Given in the question that, A population is made up of $r$ disjoint subgroups. Let $p_i$ denote the proportion of the population that is in subgroup $i,i=1,\dots ,r$. If the average weight of the members of subgroup $i$s $wi, i = 1, ... ,r,$

Step 2: Explanation

Allow $W$ to mean the pay of an arbitrary part. We are given that

$E\left(W\mid A_i\right)=w_i$

where $A_i$ is the occasion that an irregular part comes from subgroup $i$. Utilizing the law of the total expectation, we have that

$E\left(W\right)=\sum _{i}E\left(W\mid A_i\right)P\left(A_i\right)=\sum _i{w}_i{p}_i$

Step 3: Final answer

The average weight of the members of the population is$E\left(W\right)=\sum w_i{p}_i$