Compare binomial distribution with the sample

Question

You are given the sample.

\begin{array}{|c|c|c|c|c|c|c|} \hline x_i & 12 & 14 & 16 & 18 & 20& 22 \\ \hline f_i & 5 & 15 & 50 & 16 & 10 & 4 \\ \hline \end{array}

Compare relative frequency distribution to binomial distirubtion.

Answer 1

You have observations $x = (12,14,16,18,20,22)$ with six respective frequencies $f = (5,15,50, 16, 10, 4),$ which total $\sum_i f_i = 100,$ so that relative frequencies are $r = f/100 = (.05, .15, .50, .16, .10, .04).$

It is unclear what binomial distribution you should use for the 'comparison.' Ordinarily, one expects a binomial random variable to take consecutive integer values $0$ through some number of trials $n$. For example, the random variable $Y \sim \mathsf{Binom}(n = 5, p=0.5)$ has probabilities $p = (0.03125, 0.15625, 0.31250, 0.31250, 0.15625, 0.03125)$ corresponding to respective values $y = (0, 1, 2, 3, 4, 5).$

However, your random variable shows only even values from 12 through 22. So the underlying distribution cannot be binomial in any direct sense. Perhaps the problem has in mind something like $X = 2W + 12 = 2(W + 6),$ where $W$ has a binomial distribution based on 5 trials.

It might be possible to find a binomial random variable $W \sim \mathsf{Binom}(5, p),$ for an appropriate 'success probability' $p$ that comes close to matching the relative frequencies $r$ in my first paragraph. But that seems rather far from your original question. Perhaps you can find such a binomial random variable (by matching means and standard deviations).

In any case, your question needs some clarification before I am willing to spend more time on it. (While I have been typing this, votes to close your question have been accumulating, so others may be as puzzled as I am with the current formulation.)