#### Examples from Activity #10

##### 8) Suppose you find the life expectancy of the batteries are modeled by an exponential distribution with Î» = 0.44. Graph the cdf for X.

Letâ€™s define the cdf and plot it:

```
# We define the function and then set L=0.44
f <- makeFun(1 - exp(-L*x) ~ x, L=0.44)
# Plot the function
plotFun(1 - exp(-L*x) ~ x, L=0.44, x.lim=range(0,11))
```

##### 9) Calculate P(X<3), P(23).

We can do this in a couple ways. Since we already have the CDF defined, we can simply plug in the endpoints:

```
# P(X<3)
f(x=3)-f(x=0)
```

`## [1] 0.7329`

```
# P(2<x<8)
f(x=8)-f(x=2)
```

`## [1] 0.3852`

To get the last one, P(X>3), we can do either of the following:

```
# P(X>3) from 3 to infinity
f(x=Inf)-f(x=3)
```

`## [1] 0.2671`

```
# The complement of P(X<3)
1-(f(x=3)-f(x=0))
```

`## [1] 0.2671`

##### Thereâ€™s a faster way to calculate probabilities from an exponential distribution, though. We can use the built-in exponential distribution function:

pexp(q, rate = 1, lower.tail = TRUE, log.p = FALSE)

Letâ€™s find P(X<3) using *pexp()*:

`pexp(3, rate=0.44, lower.tail=TRUE, log.p=FALSE)`

`## [1] 0.7329`

We can find the other probabilities just as easily:

```
# P(2<x<8)
pexp(8, rate=0.44, lower.tail=TRUE, log.p=FALSE) -
pexp(2, rate=0.44, lower.tail=TRUE, log.p=FALSE)
```

`## [1] 0.3852`

```
# P(X>3) using lower.tail=FALSE
pexp(3, rate=0.44, lower.tail=FALSE, log.p=FALSE)
```

`## [1] 0.2671`

##### 10) On average, a bus arrives at a terminal every 10 minutes. What is the probability that a person must wait for more than 20 minutes for the next bus?

Letâ€™s use the built-in exponential distribution function:

```
# P(x>20) with lambda = 1/10
pexp(20, rate=1/10, lower.tail=FALSE, log.p=FALSE)
```

`## [1] 0.1353`

```
# Plot this distribution and shade-in the probability
x=seq(0,40,length=150)
y=dexp(x,rate=1/10)
plot(x,y,type="l",lwd=2,col="steelblue",ylab="p")
x=seq(20,40,length=150)
y=dexp(x,rate=1/10)
polygon(c(20,x,40),c(0,y,0),col="lightgray")
```

##### 12) According to the National Transportation Safety Board, there were 14 plane crashes (not all fatal) in 2007. Using this number, we can assume a plane crashes, on average, every 26 days. Given we have a plane crash on July 1, whatâ€™s the probability that we will have another crash in July??

```
# P(x<30) with lambda = 1/26
pexp(30, rate=1/26, lower.tail=TRUE, log.p=FALSE)
```

`## [1] 0.6846`

```
# Plot this distribution and shade-in the probability
x=seq(0,60,length=150)
y=dexp(x,rate=1/26)
plot(x,y,type="l",lwd=2,col="steelblue",ylab="p")
x=seq(0,30,length=150)
y=dexp(x,rate=1/26)
polygon(c(0,x,30),c(0,y,0),col="lightgray")
```