Question

Determine the binomial distribution whose mean is 9 and standard deviation is 3/2

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The binomial distribution whose mean is 9 and standard deviation is 3/2 is $P(X=x)=^nC_x(\dfrac{3}{4})^x(\dfrac{1}{4})^{n-x}$ .

Explanation:

Formula for binomial distribution :

$P(X=x)=^nC_xp^xq^{n-x}$

, where x is a binomial random variable .

P(X=x) is the probability of getting x success in n trials .

p= probability of getting each success.

q= probability of getting each failure.

Mean and standard deviation for binomial is :

$Mean = np$

$Standrad \ deviation =\sqrt{npq}$

AS per given , $np =9$         (1)

and $\sqrt{npq}=\dfrac{3}{2}$                     (2)

Put value of np from (1) in (2) , we get

$\sqrt{(9)q}=\dfrac{3}{2}\\\\ 3\sqrt{q}=\dfrac{3}{2}\\\\ \sqrt{q}=\dfrac{1}{2}\\\\ q=(\dfrac{1}{2})^2=\dfrac{1}{4}$

$\Rightarrow\ p=1-q=\dfrac{3}{4}$

Now , the formula binomial distribution becomes : $P(X=x)=^nC_x(\dfrac{3}{4})^x(\dfrac{1}{4})^{n-x}$

, where x is a binomial random variable .

# Learn more :

In a binomial distribution the probability of getting a success is 1/4 and standard deviation is 3 , then its mean is

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