FZahir_Assignment07
Farhana Zahir
3/15/2020
library(tidyverse)## Warning: package 'ggplot2' was built under R version 3.6.3## Warning: package 'dplyr' was built under R version 3.6.3library(ggplot2)Problem 1
Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y.
Solution:
\(k^n\) represents the total number of combinations. This will form the denominator to calculate probability.
\((k-1)^n\) represents all combinations where none of the \(x_i\) is equal to 1.
\[P(x=1)=\frac{k^n-(k-1)^n}{k^n}\] \[P(x=2)=\] If x=3, then there are \(k^n-(k-2)^n-[k^n-(k-1)^n]\) different combinations We can simplify this to \[k^n-(k-2)^n-[k^n-(k-1)^n] =k^n-k^n-(k-2)^n+(k-1)^n=(k-1)^n-(k-2)^n\]
\[P(x=2)=\frac{(k-2+1)^n-(k-2)^n}{k^n}\]
Therefore, we can generalize
\[P(x=y)=\frac{(k-y+1)^n-(k-y)^n}{k^n}\]
Problem 2 part a
Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.)
a)What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)
P(machine will fail) = .10 P(machine will not fail) = .90
\[P(x=n)=p*(1−p)^{n−1}\] Here, the success factor is the machine will fail, so p=0.10, q=0.90 The probabilty that the machine will fail after 8 years means that the machine will not fail for each of the first 8 years. This is caulated as \(q^8\) and is denoted by p below. I calculated the same using the R function pgeom and find out I have to use an n parameter that is less by 1. This is because as per the source code of pgeom, the formula used is \[p(x)=p(1-p)^n\] Therefore, to find the probability that the machine is working by the end of the 8th day, I need to use n parameter of 7.
https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/Geometric
https://rstudio-pubs-static.s3.amazonaws.com/327096_819b7905fc8e4cca83ad7a8bbff24723.html
p<-(0.9^8)
print('the probablity using formula is')## [1] "the probablity using formula is"p## [1] 0.4304672prob<-pgeom(7, 0.1, lower.tail = F)
print('The probablity using pgeom is')## [1] "The probablity using pgeom is"prob## [1] 0.4304672Let us plot the distribution
#still working on 8th day
still_working <-1-pgeom(0:10,.1)
still_working## [1] 0.9000000 0.8100000 0.7290000 0.6561000 0.5904900 0.5314410 0.4782969
## [8] 0.4304672 0.3874205 0.3486784 0.3138106Plotting the Cumulative Density Function
n<-7
p<-0.10
data.frame(x = 0:10,
cdf = pgeom(q = 0:10, prob = p, lower.tail = F)) %>%
mutate(Failures = ifelse(x <= n, n, "other")) %>%
ggplot(aes(x = factor(x), y = cdf, fill = Failures)) +
geom_col() +
geom_text(
aes(label = round(cdf,2), y = cdf + 0.01),
position = position_dodge(0.9),
size = 3,
vjust = 0
) +
labs(title = "Cumulative Probability of X = 8 Failures.",
subtitle = "Geometric Distribution",
x = "Failures prior to first success (x)",
y = "probability") 
Expected value
In probability theory and statistics, the geometric distribution is either of two discrete probability distributions:
The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, … } The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, … }
the following form of the geometric distribution is used for modeling the number of failures until the first success:
\[P(Y=k)=(1-p)^{k}p\] for k = 0, 1, 2, 3, …. https://en.wikipedia.org/wiki/Geometric_distribution
\[E(X)=q/p=\frac{.90}{.10}=9\]
\[StDev(X)=\sqrt\frac{1-p}{p^2}=\sqrt\frac{0.9}{0.1^2}=9.49\]
Problem 2 part b
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Solution:
\[E(x)=\frac{1}{\lambda}\]
\[\sigma=\sqrt\frac{1}{\lambda^2}\]
lambda<-0.10
exp_value<-1/lambda
st_dev<-sqrt(1/lambda^2)
exp_value## [1] 10st_dev## [1] 10Probability that machine will fail after eight years
#no of trials
lambda<-0.10
n<-8
p_exp<-pexp(8, lambda, lower.tail = F)
p_exp## [1] 0.449329#using formula
e<-exp(1)
p1_exp<-e^(-lambda*n)
p1_exp## [1] 0.449329Plotting the distribution
n<-8
lambda<-0.10
data.frame(x = 0:10,
cdf = pexp(q = 0:10, rate = lambda, lower.tail = F)) %>%
mutate(Failures = ifelse(x <= n, n, "other")) %>%
ggplot(aes(x = factor(x), y = cdf, fill = Failures)) +
geom_col() +
geom_text(
aes(label = round(cdf,2), y = cdf + 0.01),
position = position_dodge(0.9),
size = 3,
vjust = 0
) +
labs(title = "Cumulative Probability of X = 8 Failures.",
subtitle = "Exponential Distribution",
x = "Failures prior to first success (x)",
y = "probability") 
Problem 2 part c
What is the probability that the machine will fail after 8 years? Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years).
Solution:
\[E(x)=np\]
\[\sigma=\sqrt(npq)\]
exp_binom<-n*p
stdev_binom<-sqrt(n*p*(1-p))
exp_binom## [1] 0.8stdev_binom## [1] 0.8485281Prob machine fails after 8yrs=prob machine does not fail for the first 8 yrs
p_binom<-pbinom(0, 8, 0.1)
p_binom## [1] 0.4304672#Using formula P(x<=n)=p^0*q^n
p1_binom<-(p^0)*((1-p)^8)
p1_binom## [1] 0.4304672Problem 2 part d
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Solution:
\[E(x)=\lambda, \sigma=\sqrt\lambda\] \[\lambda=\frac{np}{t}=\frac{8*0.1}{1}=0.8\]
lambda<-0.8
exp_pois<-lambda
stdev_pois<-sqrt(lambda)
exp_pois## [1] 0.8stdev_pois## [1] 0.8944272p_pois<-ppois(0,lambda)
p_pois## [1] 0.449329#Checking with the formula
p1_pois<-exp(-0.8)
p1_pois## [1] 0.449329