Question

Suppose that x is a binomial random variable with n = 5, p = .56, and q = .44.

(b) For each value of x, calculate p(x). (Round final answers to 4 decimal places.)

p(0) =0.0164

p(1) =0.1049

p(2) =0.2671

p(3) =0.3399

p(4) =0.2163

p(5) =0.0550

(c) Find P(x = 3). (Round final answer to 4 decimal places.)

P(x = 3)

0.3399selected answer correct

(d) Find P(x ≤ 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(x ≤ 3)

0.7285selected answer correct

(e) Find P(x < 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(x < 3) = P(x ≤ 2)

0.7119selected answer incorrect

(f) Find P(x ≥ 4). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(x ≥ 4)0.2714selected answer correct

(g) Find P(x > 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(x > 2)0.8787selected answer incorrect

(h) Use the probabilities you computed in part b to calculate the mean μ_x, the variance, σ2xσx2 , and the standard deviation, σ_x, of this binomial distribution. Show that the formulas for μ_x , σ2xσx2 , and σ_x given in this section give the same results. (Do not round intermediate calculations. Round final answers to µ_x in to 2 decimal places, σ ²_x and σ_x in to 4 decimal places.)

µx2.80

σx^21.23

σx1.1100

(i) Calculate the interval [μ_x ± 2σ_x]. Use the probabilities of part b to find the probability that x will be in this interval. Hint: When calculating probability, round up the lower interval to next whole number and round down the upper interval to previous whole number. (Round your answers to 4 decimal places. A negative sign should be used instead of parentheses.)

The interval is [			not attempted	,	not attempted	].
P(	not attempted		≤ x ≤		not attempted	) =not attempted	not attempted

Answer 1

Solutione:

P(X<3)

pbinomGC(bound=2,region="below", size=5,prob=0.56,graph=TRUE)

0.3885753

ANSWER:

0.3886

Solutionf:

Rcode is

library(tigerstats)

pbinomGC(bound=3,region="above", size=5,prob=0.56,graph=TRUE)

0.2714321

ANSWER:

0.2714

Solutiong:

P(X>2)

pbinomGC(bound=2,region="above", size=5,prob=0.56,graph=TRUE)

0.6114247

ANSWER:

0.6114

Solutionh:

mean=n=5*0.56= 2.8

standard deviation=sqrt(npq)=sqrt(5*0.56*0.44)= 1.11

variance=npq=(5*0.56*0.44)=1.232

mean-2sd=2.8-2*1.11=0.58=1

mean+2sd=2.8+2*1.11= 5.02=5

P(1<X<5) to be found

Rcode:

pbinomGC(bound=c(1,5),region="between", size=5,prob=0.56,graph=TRUE)

=0.9835084

ANSWER:

P(1<=X<=5)=0.9835