Question

Find n and p for a binomial distribution if mean = 6 and standard deviation = 2. Also find probability at $\left( x=1 \right)$?

Answer 1

Hint: We know that from the binomial distribution that “p” is the probability of success and “q” is the probability of failure (which is 1 – p) and “n” is the number of trials in a binomial experiment. We have given mean as 6 and from the binomial distribution the mean is equal to the multiplication of n and p so substitute these values and mark this as eq. (1) then we have given standard deviation as 2 and variance is the square of standard deviation and from the binomial distribution, variance is equal to the multiplication of n, p and q. Mark this as equation (2). Solve the eq. (1) and eq. (2) and find the value of n and p. Then we have given $\left( x=1 \right)$ which is the number of success in the binomial experiment and the probability of it is equal to ${}^{n}{{C}_{1}}{{\left( p \right)}^{1}}{{\left( q \right)}^{n-1}}$. Now, substitute the values of n, p and q and get the value of probability.

Complete step by step answer:
We have given a binomial distribution with mean as 6 and standard deviation as 2. We are asked to find the value of n and p.
In binomial distribution, n are the number of trials, “p” is the probability of success and “q” is the probability of failure.
We know that total probability is equal to 1 so adding p and q then equating to 1 we get,
$\begin{align}
  & p+q=1 \\
 & \Rightarrow q=1-p \\
\end{align}$
From the binomial distribution,
$Mean=np$
Substituting mean as 6 in the above equation we get,
$6=np$………… Eq. (1)
In the above problem, standard deviation is given as 2 so variance is the square of standard deviation.
$Variance=4$
From the binomial distribution, variance is equal to multiplication of n, p and q.
$Variance=npq$
Substituting variance as 4 and q as 1 – p we get,
$4=np\left( 1-p \right)$……… Eq. (2)
Dividing eq. (2) by eq. (1) we get,
$\dfrac{4}{6}=\dfrac{np\left( 1-p \right)}{np}$
In the above equation, “np” will be cancelled out and we get,
$\begin{align}
  & \dfrac{4}{6}=1-p \\
 & \Rightarrow \dfrac{2}{3}=1-p \\
\end{align}$
Rearranging the above equation we get,
$\begin{align}
  & p=1-\dfrac{2}{3} \\
 & \Rightarrow p=\dfrac{3-2}{3}=\dfrac{1}{3} \\
\end{align}$
Substituting p as $\dfrac{1}{3}$ in eq. (1) we get,
$\begin{align}
  & 6=n\left( \dfrac{1}{3} \right) \\
 & \Rightarrow n=18 \\
\end{align}$
Hence, we have got the value of n as 18 and p as $\dfrac{1}{3}$.
Now, substituting p as $\dfrac{1}{3}$ in $q=1-p$ we get,
$\begin{align}
  & q=1-\left( \dfrac{1}{3} \right) \\
 & \Rightarrow q=\dfrac{3-1}{3}=\dfrac{2}{3} \\
\end{align}$
We are also asked to find the probability when the number of successes is 1$\left( x=1 \right)$. We know that the probability when number of success is 1 is as follows:
${}^{n}{{C}_{1}}{{\left( p \right)}^{1}}{{\left( q \right)}^{n-1}}$
Substituting the values of n, p and q that we have calculated above in the above equation we get,
$\begin{align}
  & {}^{18}{{C}_{1}}{{\left( \dfrac{1}{3} \right)}^{1}}{{\left( \dfrac{2}{3} \right)}^{18-1}} \\
 & ={}^{18}{{C}_{1}}{{\left( \dfrac{1}{3} \right)}^{1}}{{\left( \dfrac{2}{3} \right)}^{17}} \\
\end{align}$
We know that ${}^{n}{{C}_{1}}=n$ so using this relation in the above equation we get,
\[\begin{align}
  & 18{{\left( \dfrac{1}{3} \right)}^{1}}{{\left( \dfrac{2}{3} \right)}^{17}} \\
 & =18\left( \dfrac{1}{3} \right)\left( \dfrac{{{2}^{17}}}{{{3}^{17}}} \right) \\
 & =\dfrac{{{2}^{17}}\left( 6 \right)}{{{\left( 3 \right)}^{17}}}=\dfrac{{{2}^{17}}\left( 6 \right)}{3.{{\left( 3 \right)}^{16}}} \\
 & =\dfrac{{{2}^{18}}}{{{3}^{16}}} \\
\end{align}\]

Hence, the probability when $x=1$ is $\dfrac{{{2}^{18}}}{{{3}^{16}}}$.

Note: Usually, in the hurry of solving the problem in exams, students tend to forget to convert standard deviation into the variance by not squaring the standard deviation. So, make sure you will first square the standard deviation which is variance and then equate variance to npq.
This question demands that you should know the formulas for mean, variance in binomial distribution and how to calculate probability of some number of successes in binomial distribution. Failing to know these formulas will cost you marks in examination.