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Question: If the mean and standard deviation of a binomial distribution are 12 and 2 respectively, then the va...

If the mean and standard deviation of a binomial distribution are 12 and 2 respectively, then the value of its parameter p is
A. 12\dfrac{1}{2}
B. 13\dfrac{1}{3}
C. 23\dfrac{2}{3}
D. 14\dfrac{1}{4}

Explanation

Solution

Hint: In this question it is given that the mean and standard deviation of a binomial distribution are 12 and 2 respectively, we have to find the value of the parameter p. So to find the solution we need to know that the mean and standard deviation of a binomial distribution is npnp and np(1−p)\sqrt{np\left( 1-p\right) } respectively, where n is the number of trials in a binomial experiment and p is the probability of success on an individual trial. So by using these we will get our required solution.

Complete step-by-step solution:
given that,
Mean(m) = 12 and standard deviation(sd) = 2
Als as we know that for any binomial distribution the mean and standard deviations are,
m = np and sd = np(1−p)\sqrt{np\left( 1-p\right) }
Therefore, we can write,
np = 12………….........(1) and
np(1−p)=2\sqrt{np\left( 1-p\right) } =2
np(1−p)=22np\left( 1-p\right) =2^{2}
np(1−p)=4np\left( 1-p\right) =4………………..(2)
Now putting the value of ‘np’ in equation (2) we get,
np(1−p)=4np\left( 1-p\right) =4
⇒12(1−p)=4\Rightarrow 12\left( 1-p\right) =4
⇒(1−p)=412\Rightarrow \left( 1-p\right) =\dfrac{4}{12} [dividing both side by 12]
⇒(1−p)=13\Rightarrow \left( 1-p\right) =\dfrac{1}{3}
⇒−1+p=−13\Rightarrow -1+p=-\dfrac{1}{3} [multiplying ‘-1’ in the both side of the equation]
⇒p=−13+1\Rightarrow p=-\dfrac{1}{3} +1
⇒p=1−13\Rightarrow p=1-\dfrac{1}{3}
⇒p=3−13\Rightarrow p=\dfrac{3-1}{3}
⇒p=23\Rightarrow p=\dfrac{2}{3}
Therefore the value of the parameter p is 23\dfrac{2}{3}.
Hence the correct option is option C.

Note: While solving this type of question you need to know that in probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes/no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q = 1 − p).
In general, if the random variable X follows the binomial distribution with parameters n∈Nn\in \mathbf{N} and p∈[0,1]p\in \left[ 0,1\right] , we write X∼B(n,p)\mathrm{X} \sim B\left( n,p\right) . The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:
P(X=k)= nCk pkqn−k\mathrm{P} \left( \mathrm{X} =k\right) =\ ^{n} C_{k}\ p^{k}q^{n-k}.