1. Discrete Distribution

1.(a) 

1 - (.7 + .2 + .05 + .03 + .01)
## [1] 0.01

f(5)=.01

1.(b) 

x <- c(0,1,2,3,4,5)
Fx <- c(.7,.9,.95,.98,.99,1)
CDF <- data.frame(x,Fx)
CDF

1.(c)

Using the cdf, it appears as if it would be .98 probability that at most 3 grafts fail.

1.(d)

At least two grafts fail:
1-F(2)
1-.95
p=.05

1.(e)

X=3
F(3)-F(2)
.98-.95
Probability of exactly 3 failures = 0.03

  1. Expectation and Variance

2.(a)

E|X|=1(.25)+2(.25)+3(.25)+4(.25)+5(.25)

(1 * .25) + (2 * .25) + (3 * .25) + (4 * .25) + (5 *.25)
## [1] 3.75

E|X|=3.75

2.(b)

First thing I tried resulted in a negative number for variance.

((1 ^ 2) * .25) + ((2 ^ 2) * .25) + ((3 ^ 2)* .25) + ((4 ^ 2)* .25) + ((5 ^ 2)*.25)
## [1] 13.75
13.75 - (3.75 ^ 2)
## [1] -0.3125



That didn’t seem right, and I don’t think a negative variance is feasible. However, the next thing that I tried, the variance seems large given the 1 minute window (additionally, there can’t be a negative time that it takes to process, so the variance still seems off to me). 

((1 ^ 2) * .25) + ((3 ^ 2) * .25) + ((6 ^ 2)* .25) + ((10 ^ 2)* .25) + ((15 ^ 2)*.25)
## [1] 92.75
92.75 - (3.75 ^ 2)
## [1] 78.6875
sqrt(78.6875)
## [1] 8.870597

2.(c) 

.75 ^ 5
## [1] 0.2373047

The probability that none of these programs are accepted within a minute is .24

  1. Binomial Distribution

3.(a)

E|X| = 0 x (1-p) + 1 x (p)

E|X| = 0 + 1p

E|X| = p

3.(b)

E|X^2| = 0 ^ 2 x (1-p) + 1 ^ 2 x (p)

E|X^2| = 0 + 1p

E|X^2| = p

3.(c)

VarX = (0 ^ 2 x (1-p) + 1 ^ 2 x (p)) - (0 x (1-p) + 1 x (p) ^ 2)

VarX = (0 + 1p) - (0 + 1p)^2

VarX = p

  1. Normal Distribution

Note: normal dist, mean is 82, std is 10

4.(a)

27% of students scored 88% or higher on the exam.

4.(b) 

1 - pnorm(88, mean = 82, sd = 10)
## [1] 0.2742531

  1. Normal Distribution

5.(a) 

1 - pnorm(8, mean = 6.8, sd = .6)
## [1] 0.02275013

The probability that a randomly selected person sleeps more than 8 hours is 0.02.

5.(b) 

pnorm(6, mean = 6.8, sd = .6)
## [1] 0.09121122

The probability that a randomly selected person sleeps less than 6 hours is 0.09.

5.(c)

{r}pnorm(9, mean = 6.8, sd = .6) - pnorm(7, mean = 6.8, sd = .6)

37% of the population gets between 7 and 9 hours of sleep