Question

Find the probability of 3 success for the binomial experiment with 7 trial and the success probability of 0.3. Then find the mean and standard deviation. Write the formula substitute

the values.

Answer 1

`P(x=3)=0.2269`

Mean=2.1

Standard deviation=1.21

Explanation:

We are given that

n=7

Probability of success, p=0.3

q=1-p=1-0.3=0.7

We have to find the probability of 3 success for the binomial experiment and find the mean and standard deviation.

Binomial distribution formula

`P(X=x)=nC_xp^(x)q^(n-x)`

Using the formula

`P(x=3)=7C_3(0.3)^3(0.7)^(7-3)`

`P(x=3)=7C_3(0.3)^3(0.7)^(4)`

`P(x=3)=(7!)/(3!4!)(0.3)^3(0.7)^(4)`

`P(x=3)=(7* 6* 5* 4!)/(3* 2* 1* 4!)(0.3)^(3)(0.7)^(4)`

Using the formula

`nC_r=(n!)/(r!(n-r)!)`

`P(x=3)=0.2269`

Now,

Mean,
`\mu=np=7* 0.3=2.1`

Standard deviation,
`\sigma=√(np(1-p))`

Standard deviation,
`\sigma=√(7* 0.3* 0.7)`

Standard deviation,
`\sigma=1.21`