A binomial distribution has mean of $9.6$ and standard deviation of $2.4.$ Find $n$ (no. of trials) and $p$ (probability of success in each trial)

Question

A binomial distribution has a mean of $9.6$ and a standard deviation of $2.4$.

Find $n$, the number of trials and $p$, the probability of success in each trial.

What I have tried so far:

$\text{mean} = 9.6 = np.$

$\text{S.D} = 2.4.$

$\text{S.D} = \sqrt{np(1-p)}.$

$2.4 = \sqrt{np(1-p)}$

(squaring both sides of equation) $5.76 = np(1-p)$

sub in $np=9.6$, $5.76 = 9.6(1-p)$

But I'm not sure how to get the value of $p$.

This is a question from a statistics textbook on Bernoulli & Binomial Distributions.

Any help would be much appreciated!

Answer 1

Looks like you are stuck on algebra: $$ 5.76=9.6(1-p) $$ $$ 1-p=\frac{5.76}{9.6} $$ $$ p = 1-\frac{5.76}{9.6} $$