Question

The difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 11. Find the distribution.

Answer 1

Hint:
Here, we will use the formula of mean and variance of a Binomial Distribution to find the equation with the given conditions. Then we will solve the equations to find the probability of success, probability of failure and number of trials. We will substitute the obtained values in the formula of the binomial distribution to find the distribution.

Formula Used:
We will use the following formula:
1) The binomial distribution is given by the formula \[{\left( {q + p} \right)^n}\]
2) Mean of a Binomial Distribution \[ = np\].
3) Variance of a Binomial Distribution \[ = npq\], where \[p\] is the probability of success, \[q\] is the probability of failure and \[n\] is the number of trials.
4) The difference of the square of two numbers is given by \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\].

Complete Step by step Solution:
We know that Mean of a Binomial Distribution \[ = np\]
Variance of a Binomial Distribution \[ = npq\]
We are given that the difference between mean and variance of a Binomial distribution is 1.
\[np - npq = 1\]
By taking the common factors, we get
\[ \Rightarrow np\left( {1 - q} \right) = 1\] ……………………………………………………………………………………….\[\left( 1 \right)\]
We are given that the difference of squares between mean and variance of a Binomial distribution is 11. Therefore,
\[ \Rightarrow {\left( {np} \right)^2} - {\left( {npq} \right)^2} = 11\]
Applying the exponent on the terms, we get
\[ \Rightarrow \left( {{n^2}{p^2}} \right) - \left( {{n^2}{p^2}{q^2}} \right) = 11\]
By taking the common factors, we get
\[ \Rightarrow {n^2}{p^2}\left( {1 - {q^2}} \right) = 11\] ……………………………………………………………………………………\[\left( 2 \right)\]
Squaring equation\[\left( 1 \right)\], we get
\[ \Rightarrow {\left( {np\left( {1 - q} \right)} \right)^2} = {1^2}\]
\[ \Rightarrow {n^2}{p^2}{\left( {1 - q} \right)^2} = 1\] ………………………………………………………………………………….......\[\left( 3 \right)\]
Now, dividing equation \[\left( 2 \right)\] by equation \[\left( 3 \right)\], we get
\[\dfrac{{{n^2}{p^2}\left( {1 - {q^2}} \right)}}{{{n^2}{p^2}{{\left( {1 - q} \right)}^2}}} = \dfrac{{11}}{1}\]
\[ \Rightarrow \dfrac{{\left( {1 - {q^2}} \right)}}{{{{\left( {1 - q} \right)}^2}}} = \dfrac{{11}}{1}\]
By using the algebraic identity \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], we get
\[ \Rightarrow \dfrac{{\left( {1 + q} \right)\left( {1 - q} \right)}}{{\left( {1 - q} \right)\left( {1 - q} \right)}} = \dfrac{{11}}{1}\]
\[ \Rightarrow \dfrac{{\left( {1 + q} \right)}}{{\left( {1 - q} \right)}} = 11\]
By cross multiplying, we get
\[ \Rightarrow 1 + q = 11\left( {1 - q} \right)\]
By simplifying and rewriting the equation, we get
\[ \Rightarrow 1 + q = 11 - 11q\]
\[ \Rightarrow 11q + q = 11 - 1\]
Adding the like terms, we get
\[ \Rightarrow 12q = 10\]
Dividing both side by 12, we get
\[ \Rightarrow q = \dfrac{{10}}{{12}}\]
\[ \Rightarrow q = \dfrac{5}{6}\]
We know that \[p + q = 1\] , so we have \[p = 1 - q\].
\[p = 1 - \dfrac{5}{6}\]
By taking LCM , we get
\[ \Rightarrow p = 1 \times \dfrac{6}{6} - \dfrac{5}{6}\]
\[ \Rightarrow p = \dfrac{{6 - 5}}{6}\]
Subtracting the terms in the numerator, we get
\[ \Rightarrow p = \dfrac{1}{6}\]
Substituting the values of \[p\] and \[q\] in equation \[\left( 1 \right)\], we get \[np\left( {1 - q} \right) = 1\]
\[ \Rightarrow n\left( {\dfrac{1}{6}} \right)\left( {1 - \dfrac{5}{6}} \right) = 1\]
By taking LCM, we get
\[ \Rightarrow n\left( {\dfrac{1}{6}} \right)\left( {1 \times \dfrac{6}{6} - \dfrac{5}{6}} \right) = 1\]
\[ \Rightarrow n\left( {\dfrac{1}{6}} \right)\left( {\dfrac{{6 - 5}}{6}} \right) = 1\]
Subtracting the terms in the numerator, we get
\[ \Rightarrow n\left( {\dfrac{1}{6}} \right)\left( {\dfrac{1}{6}} \right) = 1\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{n}{{36}} = 1\]
On cross multiplication, we get
\[ \Rightarrow n = 36\]
The binomial distribution is given by the formula \[{\left( {q + p} \right)^n}\]
By substituting the values of \[n,p,q\], we get
Binomial Distribution \[ = {\left( {\dfrac{5}{6} + \dfrac{1}{6}} \right)^{36}}\]

Therefore, the Binomial Distribution is \[{\left( {\dfrac{5}{6} + \dfrac{1}{6}} \right)^{36}}\].

Note:
Binomial Distribution is a distribution of Bernoulli’s experiment to find the number of successes from the number of experiments. The given distribution is a binomial distribution only if the number of observations is fixed. Each observation obtained is independent and the observations would be the one of the two outcomes which can either be the success or failure.