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

Answer:

Mean for a binomial distribution = 374

Standard deviation for a binomial distribution = 12.97

Step-by-step 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 $\times$ p

       So, E(X) = 680 $\times$ 0.55 = 374

• Standard deviation of Binomial Distribution is given by;

                   S.D.(X) = $\sqrt{n \times p \times (1-p)}$

                               = $\sqrt{680 \times 0.55 \times (1-0.55)}$

                               = $\sqrt{680 \times 0.55 \times 0.45}$ = 12.97

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