Can a binomial random variable X have a mean =7 and the variance =11. Why or why not?

Question

Suhadolahbb · Accepted Answer

Step 1
The probability distribution function of binomial random variable x is given by

P (x) = \frac{n!}{x! (n - x)!} p^{x} {(1 - p)}^{n - x}

, where n is the total number of trials, p is the probability of success in each trial and x is the number of successes.
The mean of binomial distribution is given by

μ = n p

and variance of the distribution is given by

V a r (x) = n p (1 - p)

. The probability of success can have any values between 0 and 1.
Disclaimer: Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify question number or post only that question.
Step 2
The mean of the distribution is

μ = 7

and variance

V a r (x) = 11

. Substitute these value in

μ = n p and V a r (x) = n p (1 - p)

.

n p = 7

n p (1 - p) = 11

Substitute the value

n p = 7

in the expression of variance

n p (1 - p) = 11

.

7 (1 - p) = 11

1 - p = \frac{11}{7}

p = 1 - \frac{11}{7}

= - \frac{4}{7}

The value of probability cannot be a negative value thus it is not possible that X has mean 7 and variance 11. Also for a binomial random variable, the variance is always less than the mean value and in this case variance is more than the mean. Thus X cannot be a binomial random variable.

Virginia Palmer · Accepted Answer

Given information:  μ=7  σ2=11  Step 2   The formula for mean and variance for the binomial distribution are:  μ=n⋅p  σ2=n⋅p⋅(1−p)  σ2=μ(1−p)  From the above formula it is clear that the value of variance should be less than or equal to mean because the value of (1−p) is always less or equal to 1.   A binomial random variable X cannot have a mean =7 and variance =11.

Can a binomial random variable X have a mean =7

Answered question

Answer & Explanation

New Questions in High school probability