IPMAT Indore 2019 (MCQ) - In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to | PYQs + Solutions

IPMAT Indore 2019

Arithmetic

Ratio, Proportion & Variation

Hard

In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

4.667

4.333

3.667

Correct Option: 2

Let us assume that there are a total of

3m

families. Since the proportion of families with 2, 4 and 6 members is roughly equal, we can say that there are:

m

number of 2-membered families,

\newline

m

number of 4-membered families,

\newline

m

number of 6-membered families.

Therefore, the total number of family members becomes:

\newline

2m

4m

and

6m

respectively for 2, 4 and 6 member families.

Total family members =

2m + 4m + 6m = 12m

Chances of choosing a:

\newline

I. 2-membered family member =

\dfrac{2m}{12m} = \dfrac{1}{6}

\Rightarrow

This means that one-sixth of the people in the village say that they belong to a 2-member family.

II. 4-membered family member =

\dfrac{4m}{12m} = \dfrac{1}{3}

\Rightarrow

This means that one-third of the people in the village say that they belong to a 4-member family.

III. 6-membered family member =

\dfrac{6m}{12m} = \dfrac{1}{2}

\Rightarrow

This means that half of the people in the village say that they belong to a 6-member family.

Clearly, the chances of choosing a 6-member family is higher than the rest of the two cases.

Average family size =

\dfrac{\frac{1}{6} \times 2 + \frac{1}{3} \times 4 + \frac{1}{2} \times 6}{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}}

(here, the fractions are multiplied with their respective family size)

Average family size =

\dfrac{1}{3} + \dfrac{4}{3} + 3 = 4.667

Hence, the answer is

4.667