A university exercise Statistics Learning and data analysis.
This is the problem:
Given that $ X=x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9, x_{10} = \left(2,3,7,8,2,4,7,5,5,7\right) $
- What is the mean $ {\overline{x}} $ ?
Answer: ${\overline{x}}=\frac{\left(2+3+7+8+2+4+7+5+5+7\right)}{10}=5 $
- What is the probability of $ P\left(X\le\overline{x}\right) $? with help from $ x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9, x_{10}$ from a $ NegBin\left(2,\frac{1}{3}\right) $ ?
My answer (That is wrong would be): $$ P\left(X\le\overline{x}\right)=\sum_{i=0}^5\left(\frac{1}{3}\right)^{^2}\left(\frac{2}{3}\right)^i\binom{i+2-1}{i}=\sum_{i=0}^5\left(\frac{1}{3}\right)^{^2}\left(\frac{2}{3}\right)^i\binom{i+1}{i}=\sum_{i=0}^5\left(\frac{1}{3}\right)^{^2}\left(\frac{2}{3}\right)^{^i}\left(i+1\right)=0.73663 $$
In the answer they stated that they drew $ x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9, x_{10}$ 100 times and got the answer $ P\left(X\le5\right) = 0.19$
My questions are:
- How do you come to the same conclusion as the "right answer"?
- Is there a different way of solving this problem?
- How would you answer the problem (2)?
Thanks!
Update:
The data X are the number of times something is observed in a day. Meaning there are 10 days and therefore 10 observed data in X.
NegBin(r,p) = $$ NegBin(r,p): fx(x) = (p)^{r}(p-1)^i\binom{i+r-1}{i}, E[X] = r(1 − p)/p $$
Best Answer
To answer the question as the "real answer" you will have to: