Question

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

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.)

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

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

µx2.80

σx^21.23

σx1.1100

(i) Calculate the interval [μx ± 2σx]. 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

Solutione:

P(X<3)

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

0.3885753 0.3886

Solutionf:

Rcode is

library(tigerstats)

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

0.2714321 0.2714

Solutiong:

P(X>2)

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

0.6114247 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 P(1<=X<=5)=0.9835

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