The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350.

Question

The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350. Find the number of trials.

Answer: 96

Accepted Answer

$np+npq=82.5$, $np\cdot npq=1350$. These are roots of $x^2-82.5x+1350=0\Rightarrow x=60$ or $22.5$. So mean $np=60$, variance $npq=22.5\Rightarrow q=\dfrac{22.5}{60}=\dfrac{3}{8},\ p=\dfrac58$.

$$n = \frac{60}{5/8}=96$$

Solutions · 1