Wk 7 Live Lab, Probability Distribution
1. 1. Binomial Distribution.
R makes the following functions available:
- dbinom(x, size, prob)
- probability of x successes in size trials when the probability of success is
- probability of x successes in size trials when the probability of success is
- pbinom(q, size, prob, lower.tail)
- cumulative probability (lower.tail = TRUE for left tail, lower.tail = FALSE for right tail) of less than or equal to q successes.
- rbinom(n, size, prob) returns n random numbers from the binomial distribution
- A function which randomly generates numbers which follow a binomial distribution with given parameters
rbinom(n = 10, size = 1, prob = 0.5)## [1] 1 0 0 1 1 0 0 1 1 1rbinom(1, 100, 0.3)## [1] 33A<-rbinom(10000,2,0.08)
head(A)## [1] 0 0 0 0 0 0table(A)## A
## 0 1 2
## 8505 1424 71table(A)/10000## A
## 0 1 2
## 0.8505 0.1424 0.0071dbinom(0,2,0.08)#three different lines ## [1] 0.8464dbinom(1,2,0.08)## [1] 0.1472dbinom(2,2,0.08)## [1] 0.0064round(dbinom(0:2,2,0.08),4) #all 3 probabilites in one line## [1] 0.8464 0.1472 0.0064# Axis titles can be added via the xlab and ylab arguments.
barplot(table(A)/10000,main="PDF",xlab="X",
ylab="Probability",col=2)
cumsum(table(A))/10000 ## 0 1 2
## 0.8505 0.9929 1.0000round(pbinom(0:2,2,0.08),4)## [1] 0.8464 0.9936 1.0000plot(0:2,cumsum(table(A))/10000,type='b',
xlab="X",ylab="Probability", main="CDF")
#change in shape when you incease n to 10
barplot(round(dbinom(0:10,10,0.08),4),main="PDF",xlab="X",
names.arg = 0:10, ylab="Probability",col=2,ylim=c(0,1))
1.1.1 Worked example
A fair coin is tossed 60 times, if we treat heads as a success, answer the following:
- Display the probability distribution function on a graph
A<-rbinom(100000,60,0.5)
head(A)## [1] 27 24 30 33 34 26table(A)## A
## 15 16 17 18 19 20 21 22 23 24 25 26
## 5 12 41 72 157 396 693 1251 2118 3210 4547 6110
## 27 28 29 30 31 32 33 34 35 36 37 38
## 7626 9011 9900 10279 9773 8816 7648 6120 4491 3100 2019 1222
## 39 40 41 42 43 44 45
## 715 349 184 84 34 10 7table(A)/100000 #get prob ## A
## 15 16 17 18 19 20 21 22 23
## 0.00005 0.00012 0.00041 0.00072 0.00157 0.00396 0.00693 0.01251 0.02118
## 24 25 26 27 28 29 30 31 32
## 0.03210 0.04547 0.06110 0.07626 0.09011 0.09900 0.10279 0.09773 0.08816
## 33 34 35 36 37 38 39 40 41
## 0.07648 0.06120 0.04491 0.03100 0.02019 0.01222 0.00715 0.00349 0.00184
## 42 43 44 45
## 0.00084 0.00034 0.00010 0.00007barplot(table(A)/100000,main="PDF", xlab="X",
ylab="Probability",col=2)
p<-round(dbinom(0:60,60,0.5),8)
barplot(p,main="PDF",names.arg = 0:60, xlab="X",
ylab="Probability",col=2)
b. What is the probability heads comes up 20 times?
dbinom(20,60,0.5)## [1] 0.003635846- What is the probability heads comes up 20, 25 or 30 times?
# prob 20 or 25 or 30 heads out of 60
dbinom(20,60,0.5)+
dbinom(25,60,0.5)+
dbinom(30,60,0.5)## [1] 0.1512435- What is the probability heads comes up less than 20 times?
pbinom(19,60,0.5) #gives P(X<=19)## [1] 0.003108801- What is the probability heads comes up between 20 and 30 times?
pbinom(30,60,0.5) #gives P(X<=30)## [1] 0.5512891pbinom(19,60,0.5) #gives P(X<=19)## [1] 0.003108801#p(20<=X<=30)
pbinom(30,60,0.5) - pbinom(19,60,0.5)## [1] 0.5481803cum_p<-round(pbinom(0:60,60,0.5),8)
plot(0:60,cum_p,type='l',
xlab="X",ylab="Probability", main="CDF",ylim=c(0,1))
1.1.1 Assessment 1
Q 3a
- A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. Find the probability that the virus enters at least 10 computers.
\(P(x) = \binom{n}{x} p^{x}(1-p)^{n-x}\)
To find the probability of exactly 10 computers being infected
\(P(x) = (0.184756) 0.6^{10}(0.4)^{10} = 0.11714155\)
dbinom(10, size=20, prob=0.4) ## [1] 0.1171416Now find find the probability that the virus enters at least 10 computers are infected:
dbinom(10, size=20, prob=0.4) +
dbinom(11, size=20, prob=0.4) +
dbinom(12, size=20, prob=0.4) +
dbinom(13, size=20, prob=0.4) +
dbinom(14, size=20, prob=0.4) +
dbinom(15, size=20, prob=0.4) +
dbinom(16, size=20, prob=0.4) +
dbinom(17, size=20, prob=0.4) +
dbinom(18, size=20, prob=0.4) +
dbinom(19, size=20, prob=0.4) +
dbinom(20, size=20, prob=0.4) ## [1] 0.2446628Alternatively, we can use the cumulative probability function for binomial distribution pbinom:
1 - pbinom(9, size=20, prob=0.4) ## [1] 0.2446628Q 5a
- Ten percent of computer parts produced by a certain supplier are defective. What is the probability that a sample of 10 parts contains more than 3 defective ones?
$P(x > 3) - 1 - P(X 3) $
1 - pbinom(3, size=10, prob=0.1)## [1] 0.01279521. 2. Geometric Distribution.
1.2.1 Assessment 1
3b
- A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. A computer manager checks the lab computers, one after another, to see if they were infected by the virus. What is the probability that she must test at least 6 computers to find the first infected one?
\(P(x) = p(1-p)^{x-1}\)
The geometric distributon in R is zero based so to get P(6):
dgeom(5, prob=0.4)## [1] 0.031104But it’s “at least” so we need to sum P(6)..P(20), alternatively:
\(P(x \ge 6) = 1 - P(X < 6)\)
1 - (dgeom(0, prob=0.4) +
dgeom(1, prob=0.4) +
dgeom(2, prob=0.4) +
dgeom(3, prob=0.4) +
dgeom(4, prob=0.4))## [1] 0.077761 - pgeom(4,0.4)## [1] 0.07776However, the argument, lower.tail defaults to TRUE, meaning probabilities are P[X <= x], otherwise, P[X > x]. So, we can use lower.tail = FALSE, meaning we compute the probability of an observation greater than 5 (at least 6):
pgeom(4,0.4,lower.tail = FALSE)## [1] 0.077761. 3. Normal Distribution.
1.3.1. Worked example
- A random variable comes from a normal distribution with mean 22 and variance 25.
#var is 25
#parameters describe a normal dist are mean and sd
mu<-22
sigma<-sqrt(25)
sigma## [1] 5- Display the probability density function on a graph.
X<-rnorm(100000,mean=mu,sd=sigma)
hist(X)
plot(density(X),main="Normal PDF",xlab="X")
b. Display the cumulative density function on a graph.
x<-seq(0,44,length=1000) #create a sequence of numbers
x[1:10]## [1] 0.00000000 0.04404404 0.08808809 0.13213213 0.17617618 0.22022022
## [7] 0.26426426 0.30830831 0.35235235 0.39639640x[990:1000]## [1] 43.55956 43.60360 43.64765 43.69169 43.73574 43.77978 43.82382
## [8] 43.86787 43.91191 43.95596 44.00000dnorm(x[500:510],mu,sigma)## [1] 0.07978768 0.07978768 0.07978149 0.07976911 0.07975054 0.07972579
## [7] 0.07969487 0.07965777 0.07961452 0.07956511 0.07950957plot(x,dnorm(x,mu,sigma), type='l',
main="Normal PDF")
plot(x,pnorm(x,mu,sigma),type='l',
main="Normal CDF",xlab="X",
ylab="Cumulative Density")
- What is the probability that the random variable lies between 16.2 and 27.5?
pnorm(27.5,mu,sigma)-pnorm(16.2,mu,sigma)## [1] 0.7413095- What is the probability that the random variable is greater than 29?
pnorm(29,mu,sigma) #p(x<29)## [1] 0.91924331- pnorm(29,mu,sigma)#p(x>29)## [1] 0.08075666- What is the probability that the random variable is less than 17?
pnorm(17,mu,sigma) #P(X<=17)=P(X<17)## [1] 0.1586553- What is the probability that the random variable is less than 15 or greater than 25?
1-pnorm(25,mu,sigma) #P(x>25)## [1] 0.2742531pnorm(15,mu,sigma) #P(x<15)## [1] 0.080756661-pnorm(25,mu,sigma)+pnorm(15,mu,sigma) ## [1] 0.3550098#P(X<15 or X>25) = P(X<15)+P(X>25)
pnorm(15,mu,sigma)+(1-pnorm(25,mu,sigma))## [1] 0.3550098