Question

I have a statistics class of 20 students, of whom 12 are female. I randomly select three students to make a group presentation to the class. What is the probability of the following occurring?

a. All of the students in the group are male.

b. One of the students in the group is male.

c. Two of the students in the group are male.

d. All three of the students in the group are female.

e. Calculate the mean and standard deviation for the distribution which defines selecting a male as a success.

Answer 1

The sample size, n =3

Success, or selecting males, p = 8/20= 0.4

Failure, or selecting females, q = 12/20= 0.6

This is a binomial distribution with p = 0.4 and n = 3.

Binomial distribution formula: ⁿC_r * p^r * q^n-r

a.

P(X=3)

We must find the probability that all 3 students in the group are male.

r= 3; n=3; p=0.4; q=0.6

Substituting in the equation, we get

P(X=3) = 0.064 or 6.4%.

b.

P(X=1)

We must find the probability that any 1 student in the group is male.

r= 1; n=3; p=0.4; q=0.6

Substituting in the equation, we get

P(X=1) =0.432 or 43.2%.

c.

P(X=2)

We must find the probability that any 2 students in the group are male.

r= 2; n=3; p=0.4; q=0.6

Substituting in the equation, we get

P(X=2) =0.288 or 28.8%.

d.

P(X=0)

We must find the probability that 0 students in the group are male.

r= 0; n=3; p=0.4; q=0.6

Substituting in the equation, we get

P(X=0) =0.216 or 21.6%.

e.

mean of binomial distribution= n*p = 3*0.4= 1.2

variance of binomial distribution = n*p*q = 3*0.4*0.6= 0.72

standard deviation= sqrt(variance) = sqrt(0.72) = 0.8485281