Question

1) Drawt (b) The normal f. with -50, ơ-10 (d) The expogEntial Ad.f with parameter λ raphs ofthe p.d.f. of the following distr
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Answer #1

2a) We assume that the distribution has rate \lambda=2. Then

E(X)=1/\lambda,~~Var(X)=1/\lambda^2

Thus from CLT we get

μ = E(X) = 1/λ, σ = V Var(X ) = 1/λ

2b) R Program

##Question 2(b)

la=2
rep=10000
w=numeric(rep)
f=function(n)
{
for(i in 1:rep)
{
x=rexp(n,rate=2)
w[i]=sqrt(n)*(mean(x)-(1/2))/(1/2)
}
hist(w,main="Histogram",col="red",xlab="")
text(4,1225,labels=n)
}
par(mfrow=c(2,2))
f(5)
f(10)
f(20)
f(50)

R OUTPUT

Histogram Histogram 10 -20246 -20246 Histogram Histogram 20 50 2 0 2 4 2 02 4

From the histogram (the numerical figure on the graph indicates value of n), we find that as n increases, histogram becomes closer to that of a standard normal distribution.

3. R Program

## Question 3

la=2
rep=10000
w=numeric(rep)
g=function(n)
{
for(i in 1:rep)
{
x=rexp(n,rate=2)
w[i]=sqrt(n)*(mean(x)-(1/2))/(1/2)
}
qqnorm(w,main="Q-Q Plot",col="green",xlab="")
text(-3,3,labels=n)
}
par(mfrow=c(2,2))
g(5)
g(10)
g(20)
g(50)

R OUTPUT

Q-Q Plot Q-Q Plot 寸 10 4 2024 4 2 02 4 Q-Q Plot Q-Q Plot 寸 20 50 4 2 024 4 2 02 4

From the Q-Q plot(the numerical figure on the graph indicates value of n), we find that as n increases, more and more points are concentrating on a straightline. Hence the distribution becomes close to normal with increase in n.

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