Random variable(s) & Probability Distribution(s)

Assignment # 5

1) Variance and Standard Deviation of a Discrete Random Variable

We were given two probability distribution tables and asked to find the variance and standard distribution of each.

Table #1

X	p(x)
0	0.50
1	0.20
2	0.15
3	0.10

Table #2

x	P(x)
1	0.10
3	0.20
5	0.60
4	0.20

  R Studio Coding 

Table #1 is represented by x and Table #2 is represented by y.

x <- c(0.5, 0.2, 0.15, 010)

y <- c(0.10, 0.2, 0.6, 0.2)

Then simply find the variance and standard deviation of each vector by using the var and sd

functions.

varianceX <- var(x)

standardX <- sd(x)

varianceY <- var(y)

standardY <- sd(y)

You can print the results:

varianceX

varianceY

standardX

standardY

R reports that the variance of Table #1 is 23.62729 and Table #2 is 23.62729. The standard deviation of Table #1 is 4.860791 while Table #2 is 0.2217356.

The R Markdown of this script:

x  <- c(0.5, 0.2, 0.15, 010) y <- c(0.10, 0.2, 0.6, 0.2)

varianceX <- var(x) standardX <- sd(x)

varianceY <- var(y) standardY <- sd(y)

varianceX

[1]  23.62729 varianceY

[1]  0.04916667

standardX

[1]  4.860791 standardY

[1]  0.2217356

2) Binomial Distribution:

To find the Binomial Distribution of P(0) in R, we use the dbinom function.

dbinom(x = 0, size = 4, prob = 0.2)

R computes the value to be 0.4096.

3) Poisson Distribution:

The first technique using rpois function. By using the rpois function, we can generate a vector of 20 random numbers.

#  rpois(n, lambda) randomNums <- rpois(20, 4)

Plug the vector into our var function and we can find the variance of all 20 randomly generated numbers.

variance <- var(randomNums)

##Then tell R to show us the results using randomNums

##Here is the compiled R Markdown of this script:

randomNums <- rpois(20, 4)

variance <- var(randomNums)

randomNums

[1]52244384165314274425

variance

[1] 3.536842*

##Poisson random number

ranx <- rpois(20, lambda = 4)

ranx

##expected value (sample mean)

mean(ranx)

##variance

var(ranx)

>  ##Poisson random number

>  ranx <- rpois(20, lambda = 4)

>  ranx

[1]  38812556469343524445

>  mean(ranx)

[1]  4.55

> 

> ##variance > var(ranx)

[1]  4.365789

1. Does the sample proportion p have approximately a normal distribution? Explain.

base  of the Standard Dev sample  . sqrt(.95(1-0.95))/100=0.021 we can say that  we 95 percent sure that p  does have an normal distribution since its fall between the interval.