What is the largest number of families that can have at least 4 members according to Markov’s Inequality?

I was doing a question on Markov's Inequality where I have to give the answer and the question is stated as follows :

A town of 30 families, the average household size is 2.5. What is the largest number of families that can have at least 4 members according to Markov’s Inequality?

According to this formula P(X >= a) = µ/a

here µ=2.5 and and a=4 thus P(X >= a) = 0.625

that is 30*0.625=18.75 and since 18.75 is not a integer as number of family can't be in fraction therefore answer should be 18 but the question is showing incorrect answer.

I don't understand what am I doing wrong.

• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Aug 4 '18 at 21:35
• – Shaun Aug 4 '18 at 21:36
• @Shaun Thank you I will use this next time. – Mohammad Asif Hashmi Aug 5 '18 at 3:39

You don't need Markov for this. Since the average is $2.5$ the total number of people must be $30\times 2.5 = 75$. If you assume that "family" must mean at least $1$ person then, if there are $N$ members with at least $4$ then, even assuming all the rest have only $1$ we compare $4N+(30-N)$ to $75$ to see that $N≤15$. After all, if $N=16$ we'd have at least $4\times 16+1\times (30-16)>75$ people.
Is $N=15$ possible? Sure. If all the other families have exactly $1$ member and each of those $15$ have exactly $4$ then we'd have $4\times 15+1\times (30-15)=75$ people as desired.
Of course, if you define a family to have at least $2$ people then the answer drops considerably. In the same spirit, your answer of $18$ would be possible if families were allowed to have $0$ members.
Note: if you insist on using Markov, note that you can subtract $1$ from each family in order to accommodate the "at least one person per family" rule. Then we get $\mu=1.5, a=3$ and we conclude that $P(X≥a)≤ \frac {1.5}3=.5$ (Where, now, $X$ denotes the number of people in a family, less one). That leads us to consider $30\times .5=15$ as before.