Answer:
Since, the uniformly distributed integers range from 1 to k that means it is definitely greater than 1. And sum of any variables regardless of distribution would be a normal density function.
So, the distribution of Y will be a right skewed normal density funtion which starts from 0.
According to the question, P(X = x) = (1 - p)^(x-1) * p = (9/10)^7 * 1/10
# we can use dgeom for this
y <- dgeom(8, 0.1)
y
## [1] 0.04304672
p <- 1/10
Expected_value <- 1/p
Expected_value
## [1] 10
st_dev = sqrt((1 - (1/10))/((1/10)^2))
st_dev
## [1] 9.486833
The probability that the machine will fail after 8 years is 9.48
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential. Answer: Here, According to the question, Rate = 1/10
Failure after 8 years: 1 - P(X <= 8)
y <- 1 - pexp(8,0.1)
y
## [1] 0.449329
p <- 1/10
Expected_value <- 1/p
Expected_value
## [1] 10
std_dev <- 1/((0.1)^2)
std_dev
## [1] 100
The probability that the machine will fail after 8 years is 100.
y = dbinom(0, size=8, prob=0.1)
y
## [1] 0.4304672
Expected_value <- 8 * 0.1
Expected_value
## [1] 0.8
std_dev = sqrt((8 * 0.1) * (1 - 0.1))
std_dev
## [1] 0.8485281
The probability that the machine will fail after 8 years is 0.84.
y <- ppois(0,0.8)
y
## [1] 0.449329
# expected value = variance
Expected_value <- sqrt(0.8)
Expected_value
## [1] 0.8944272
The probability that the machine will fail after 8 years is 0.89