R Programming Task on Distribution

• 11th Feb, 2022
• 17:52 PM

# i)

x <- rep(1, 100000)
for (i in 1:100000) {
  x[i] = (sum(runif(3)) - 3/2)/(sqrt(3/12))
}

qqnorm(x)
mean(x)
sd(x)
# For 95% CI data should lies between

qnorm(0.025)

# and 
qnorm(0.975)

#now from the data we need to see the proportion of observations in this interval

length(x[x < 1> -1.959964])/100000

# So proportion of data between given range is given above

# Similarly for 0.5 and 99.5 percentile is
length(x[x < qnorm> qnorm(0.005)])/100000

# now for (ii)

x1 <- rep(1, 100000)
for (i in 1:100000) {
  x1[i] = qnorm(runif(1))
}

qqnorm(x1)
mean(x1)
sd(x1)
# For 95% CI data should lies between

qnorm(0.025)

# and 
qnorm(0.975)

#now from the data we need to see the proportion of observations in this interval

length(x1[x1 < 1> -1.959964])/100000

# So proportion of data between given range is given above

# Similarly for 0.5 and 99.5 percentile is
length(x1[x1 < qnorm> qnorm(0.005)])/100000

# So as we can see that both the distribution are following normal distribution

# Q 2

x <- rnorm(1000)

y <- x^2

#A)

# So here we will find the probability from sample in such a way that number of observation less than 1.2 divided by total number of observations
length(y[y<=1.2])/1000

#B)

pchisq(1.2, 1)

#C)

# There is a difference between both of them because one of them is determining the probabilty based on the random sample drawn. And another one is determining the exact probability.

#D)
#Estimate of m is given by :

mean(y)

#E)

# As we know that here m = 1 accorording to theoritical distribution
# So if we increase N then value of m will close to 1