Question

Find the mean, μ, and standard deviation, σ, for a binomial random variable X. (Round all answers for σ to three decimal places.)

(a) n = 5, p = .50.

μ =

σ =

(b) n = 1, p = 0.25.

μ =

σ =

(c) n = 100, p = 0.95.

μ =

σ =

(d) n = 20, p = .01.

μ =

σ =

Answer 1

Solution :

Given that,

p = 0.50

q = 1 - p =1-0.50=0.50

n = 5

Using binomial distribution,

Mean = = n * p = 5*0.50=2.5

Standard deviation = = n * p * q =  5*0.50*0.50=1.118

(B)

p = 0.25

q = 1 - p =1-0.25=0.75

n = 1

Using binomial distribution,

Mean = = n * p = 1*0.25=0.25

Standard deviation = = n * p * q =  1*0.25*0.75=0.433

(C)

p = 0.95

q = 1 - p =1-0.95=0.05

n = 100

Using binomial distribution,

Mean = = n * p = 100*0.95=95

Standard deviation = = n * p * q =  100*0.95*0.05=2.179

(D)

p = 0.01

q = 1 - p =1-0.01=0.99

n =20

Using binomial distribution,

Mean = = n * p =20*0.01=02

Standard deviation = = n * p * q =  20*0.01*0.99=0.445