#### Examples from Activity #9

##### 8. Given f(x)=6x(1-x), find P(x≤.25)

We first must define the function:

```
# We define the function and then set a=6
f <- makeFun(a*x - a*x^2 ~ x, a=6)
# Plot the function
plotFun(a*x - a*x^2 ~ x, a=6, x.lim=range(0,1))
```

We can then integrate this function to find F(0.25)-F(0)

```
F = antiD(a*x - a*x^2 ~ x, a=6)
F(x=0.25)-F(x=0)
```

`## [1] 0.1562`

##### 11. Find P(0.25 ≤ x ≤ 0.75)

```
F = antiD(a*x - a*x^2 ~ x, a=6)
F(x=0.75)-F(x=0.25)
```

`## [1] 0.6875`

##### 11. Find the median

```
F = antiD(a*x - a*x^2 ~ x, a=6)
F(x=0.75)-F(x=0.25)
```

`## [1] 0.6875`

F = antiD(a*x - a*x^2 ~ x, a=6) F(x=0.5)-F(x=0)

We need to find b, such that F(b)-F(0) = 0.50. We already know:

`F(x) = a * 1/2 * x^2 - a * 1/3 * x^3 + C`

`F(0) = 0`

So we’re, in essence, trying to solve:

`0.50 = a * 1/2 * x^2 - a * 1/3 * x^3`

`where a = 6`

We can subtract 0.50 from both sides to get:

`a * 1/2 * x^2 - a * 1/3 * x^3 - 0.50 = 0`

`where a = 6`

We now need to find the zero(s) of this function. The *Mosaic Package* has a function to find zeros, but we need to tell the computer where to look:

```
# We define the function and then set a=6
g <- makeFun(a*1/2 * x^2 - a * 1/3 * x^3 - 0.50 ~ x, a=6)
# Plot the function to estimate the zero visually
plotFun(a*1/2 * x^2 - a * 1/3 * x^3 - 0.50 ~ x, a=6, x.lim=range(0,1))
plotFun(0 ~ x, add=TRUE, col="black")
```

It looks like the zero is between 0.4 and 0.6 (ok, we all know it’s 0.5… play along anyway).

To find the zero, we use:

```
# We tell the computer to look somewhere between 0 and 1
findZeros(a*1/2 * x^2 - a * 1/3 * x^3 - 0.50 ~ x, a=6, x.lim = range(0, 1))
```

```
## x
## 1 0.5
```

##### 11. Find the 25th percentile

```
# We tell the computer to look somewhere between 0 and 1
findZeros(a*1/2 * x^2 - a * 1/3 * x^3 - 0.25 ~ x, a=6, x.lim = range(0, 1))
```

```
## x
## 1 0.3263
```