Question

If the standard deviation of the binomial distribution ${\left( {q + p} \right)^{16}}$ is 2, then mean is
A. $2$
B. $4$
C. $8$
D. $7$

Answer 1

A binomial distribution is an experiment with only two possible outcomes and is performed several times. The possible outcomes are independent and the number of trials are definite. The probability of two outcomes are denoted by $p$and $q$, where $p$represents success or pass and $q$represents failure and $n$ denotes the number of trials.
The relation between $p$and $q$ is given by $q = 1 - p$.
The standard deviation of the binomial distribution is $\sqrt {np\left( {1 - p} \right)}$ and the mean is $np$.

Complete step by step solution:
Given, the binomial distribution is ${\left( {q + p} \right)^{16}}$ and the standard deviation is 2.
From the exponent part of the distribution, the number of trials that is $n = 16$.
The standard deviation is 2 which implies $\sqrt {np\left( {1 - p} \right)} = 2$ …(1)
Substitute $n = 16$ in equation (1) and solve .
$\sqrt {16p\left( {1 - p} \right)} = 2$
Squaring both sides of the equation.
${\left( {\sqrt {16p(1 - p)} } \right)^2} = {(2)^2}$
$16p(1 - p) = 4$ …(2)
Dividing both sides of the equation (2) by $4$.
$\dfrac{{16p(1 - p)}}{4} = \dfrac{4}{4}$
$4p(1 - p) = 1$
$4p - 4{p^2} = 1$
$- 4{p^2} + 4p - 1 = 0$ …(3)
Multiplying both sides of equation (3) by $- 1$
$( - 1) \times ( - 4{p^2} + 4p - 1) = ( - 1) \times (0)$
$4{p^2} - 4p + 1 = 0$ …(4)
Equation (4) is a quadratic equation.
Roots of a general quadratic equation $a{x^2} + bx + c = 0$ are given by $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
On comparing equation (4) with general quadratic equation.
$a = 4,b = - 4,c = 1$
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 4 \times 1} }}{{2 \times 4}}$
$= \dfrac{{4 \pm \sqrt {16 - 16} }}{8}$
$= \dfrac{{4 \pm 0}}{8}$
$= \dfrac{4}{8}$
$= \dfrac{{4 \times 1}}{{4 \times 2}}$
$= \dfrac{1}{2}$
The value of $p$obtained is $\dfrac{1}{2}$.
Substitute $\dfrac{1}{2}$for $p$and 16 for $n$ into the formula for mean, that is $np$and solve to obtain the value of mean.
$np = \dfrac{1}{2}\left( {16} \right) = 8$

Therefore, the mean of the binomial distribution is 8. So, Option C is correct.

Note:
In these types of questions, the value of $p$ can also be determined using a hit and trial method where you may take up one value at a time and try to get the required answer.
Here using the hit and trial method, from equation $\sqrt {np\left( {1 - p} \right)} = 2$, $p = \dfrac{1}{2}$ .
Substitute the value$p = \dfrac{1}{2}$ , $n = 16$ in the formula for mean.
Mean $= np = 16 \times \dfrac{1}{2} = 8$
Reject any negative value of $p$as probability can never be negative.
Some other examples of a binomial distribution are tossing of a coin (having two outcome heads and tails), rolling of a dice (having total six outcomes) etc.