question_answer

Find the binomial distribution for which the mean is 4 and variance 3.

Answer:

Given, mean = 4, variance = 3 We know that, Mean \[=np\] and variance \[=npq\] \[p+q=1\] and                   \[p=1-q\]                          ?(i) According to the question, Mean\[=np\] \[4=np\]                                      ?(ii)             and                   variance\[=npq\]             \[3=npq\] \[3=4q\][from Eq. (ii)] \[q=\frac{3}{4}\] On putting the value of q in Eq. (i), we get \[p=1-\frac{3}{4}=\frac{1}{4}\]             Now, on putting the value of p in Eq. (ii), we get                                     \[n\times \frac{1}{4}=4\Rightarrow n=16\]             Now, binomial distribution             \[={}^{n}{{C}_{r}}\cdot {{p}^{r}}{{q}^{n-r}}={}^{16}{{C}_{r}}{{\left( \frac{1}{4} \right)}^{r}}{{\left( \frac{3}{4} \right)}^{16-r}}\]