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.)

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

(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 
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
mean2sd=2.82*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
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