So my textbook has this in it:
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---------------------------------------------------
freq. | 12 | 28 | 28 | 17 | 7 | 4 | 2 | 2 | 0 |
And its claiming that this is roughly distributed over $B(10, 0.2)$
I want to try and get these values so I calculated sample mean and variance:
- $X$ bar = 2.09
- $S^2$ = 2.41
The mean seems ok since $10 * 0.2 ≈ 2.09$ but a problem happens when I try to estimate the proportion:
- Assume $X$ ~ $B(n, p)$
- $2.09 ≈ np$
- $2.41 ≈ np(1-p)$
- therefore: $np(1-p)/(np) ≈ 2.41/2.09$
- therefore: $1-p ≈ 1.15$
- therefore: $p ≈ -0.15$ ???????
Is it just possible for the estimator to get it completely wrong or have I done something wrong?
The value of $n$ is $n=10$ due to $Bin(n,p)=Bin(10,0.2)$.
$10\cdot p=2.09 \Rightarrow p=0.209 \Rightarrow 1-p=0.791$
The theoretical frequency is
$$f(x)= 100\cdot \binom{10}{x}\cdot 0.209^x\cdot 0.791^{10-x}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x&0&1&2&3&4&5&6&7&8&9&10 \\ \hline f(x)&9.59&25.33&30.12&&&&&0.1&0.01 \\ \hline \end{array}$$
I have left some blanks for you.
Finally you have to round the numbers to the nearest integer I would say. But as you can read in the comments BruceET has a different opinion.