Question

Question 4 options: Find the mean and standard deviation for a binomial distribution with 680 trials and a probability of success of 0.55 and enter the answers below. Round the mean to the nearest integer and the standard deviation to two decimal places.

Answer 1

Mean for a binomial distribution = 374

Standard deviation for a binomial distribution = 12.97

Explanation:

We are given a binomial distribution with 680 trials and a probability of success of 0.55.

The above situation can be represented through Binomial distribution;

`P(X=r) = \binom{n}{r}p^(r) (1-p)^(n-r) ; x = 0,1,2,3,.....`

where, n = number of trials (samples) taken = 680 trials

r = number of success

p = probability of success which in our question is 0.55

So, it means X ~
`Binom(n=680, p=0.55)`

Now, we have to find the mean and standard deviation of the given binomial distribution.

• Mean of Binomial Distribution is given by;

E(X) = n
`*` p

So, E(X) = 680
`*` 0.55 = 374

• Standard deviation of Binomial Distribution is given by;

S.D.(X) =
`√(n * p * (1-p))`

=
`√(680 * 0.55 * (1-0.55))`

=
`√(680 * 0.55 * 0.45)` = 12.97

Therefore, Mean and standard deviation for binomial distribution is 374 and 12.97 respectively.