I NEED THE ANSWER FOR PART F PLEASE USE R SCRIPTS FOR THIS PART 8. Question Deta

Question

I NEED THE ANSWER FOR PART F PLEASE USE R SCRIPTS FOR THIS PART 8. Question Details | Previous Answers My Notes Ask Your Teacher College J has 9 firewalls working independently controlling access to the school network. Each firewall has a probability of .15 of failing on Thursday. Let X = the number of firewalls which fall on Thursday. a) what is the probability that X=1? (0300 -) b) What is the probability that fewer than two of the firewalls fail on Thursday? 0.599 c) What is the probability that none of the firewalls fail on Thursday? 0.232 d) What is the expected number of firewalls which fail on Thursday? 1.35 e) What is the variance of the number of firewalls which fail on Thursday? 1.148 f) Copy your R script for the above into the text box here.

Explanation / Answer

---------------> ###Since here there are only two outcomes associated with each firewall i.e.,

> #####failing or working on thursday. So it is a bernoulli trail and there are 9

> #######bernoulli trials with failing probability p=0.15 that implies thaq q=0.85

> rm(list=ls(all=TRUE))

> n=9;p=0.15

> ####probaility of 1 failure P[X=1]

> X=dbinom(1,9,0.15);X

[1] 0.3678622

> x1=round(X,3);x1

[1] 0.368

> ###probability of fewer than two fail P[X<2]=P[x=0]+p[x=1]

> Y=dbinom(1,9,0.15)+dbinom(0,9,0.15);Y

[1] 0.5994792

> x2=round(Y,3);x2

[1] 0.599

> #####probability that none of the firewalls fail P[x=0]

> Z=dbinom(0,9,0.15);Z

[1] 0.2316169

> x3=round(Z,3);x3

[1] 0.232

> ####Expected number of firewalls fail on thursday

> E=n*p;E

[1] 1.35

> ######Variance of the number of firewalls fail

> V=n*p*(1-p);V

[1] 1.1475

> round(V,3)

[1] 1.147

Just R script

###Since here there are only two outcomes associated with each firewall i.e.,

#####failing or working on thursday. So it is a bernoulli trail and there are 9

#######bernoulli trials with failing probability p=0.15 that implies thaq q=0.85

rm(list=ls(all=TRUE))

n=9;p=0.15

####probaility of 1 failure P[X=1]

X=dbinom(1,9,0.15);X

x1=round(X,3);x1

###probability of fewer than two fail P[X<2]=P[x=0]+p[x=1]

Y=dbinom(1,9,0.15)+dbinom(0,9,0.15);Y

x2=round(Y,3);x2

#####probability that none of the firewalls fail P[x=0]

Z=dbinom(0,9,0.15);Z

x3=round(Z,3);x3

####Expected number of firewalls fail on thursday

E=n*p;E

######Variance of the number of firewalls fail

V=n*p*(1-p);V

round(V,3)

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