Question

Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

 

Answer 1

Number of trials in the binomial distribution = 5
If p is the probability for success, then
np + npq = 4.8  

Or 5p+5p (1-p) = 4.8

\[\Rightarrow 5p + 5p - 5 p^2 = 4 . 8 \]
\[\text{ Or } p^2 - 2p + 0 . 96 = 0\]
\[\text{ By factorising, we get } \]
\[ (p - 0 . 8)(p - 1 . 2) = 0\]
\[\text{ As p cannot exceed 1, }  \]
\[p = 0 . 8 \text{ or} \  \frac{ 4}{5}\]
\[\text{ and } q = 1 - p = \frac{1}{5}\]
\[ \therefore P(X = r) = ^{5}{}{C}_r \left( \frac{4}{5} \right)^r \left( \frac{1}{5} \right)^{5 - r} , r = 0, 1, 2, . . . . 5\]