Harold Nelson

October 13, 2015

```
v = MyNormProb(lb=100,ub=110,mean=100,sd=10,
MyLabel = "Non-Standard Normal RV")
```

`v`

`## [1] 0.3413447`

```
par(mfrow=c(3,1))
MyNormProb(ub=1)
```

`## [1] 0.8413131`

`MyNormProb(ub = -1)`

`## [1] 0.1586236`

`MyNormProb(lb=-1,ub=1)`

`## [1] 0.6826895`

`pnorm(1) - pnorm(-1)`

`## [1] 0.6826895`

`par(mfrow=c(1,1))`

Find the probability that a single random number drawn from a set of numbers with a standard normal distribution is < .7. Do this three ways.

- Use pnorm()
- Use MyNormProb() to get a picture.
- Create a set of 1,000,000 numbers with this distribution and find the fraction that meet the requirement.

```
# Method 1.
pnorm(.7)
```

`## [1] 0.7580363`

```
# Method 2
MyNormProb(ub=.7)
```

`## [1] 0.7580047`

```
# Method 3
x = rnorm(1000000)
mean(x<.7)
```

`## [1] 0.758655`

Find the probability that a single random number drawn from a set of numbers with a standard normal distribution is > .7. Do this three ways.

- Use pnorm()
- Use MyNormProb() to get a picture.
- Create a set of 1,000,000 numbers with this distribution and find the fraction that meet the requirement.

```
# Method 1
1 - pnorm(.7)
```

`## [1] 0.2419637`

```
# Method 2
MyNormProb(lb=.7)
```

`## [1] 0.241932`

```
# Method 3
x = rnorm(1000000)
mean(x>.7)
```

`## [1] 0.241482`

Find the probability that a single random number drawn from a set of numbers with a standard normal distribution is > .3 and < .9. Do this three ways.

- Use pnorm()
- Use MyNormProb() to get a picture.
- Create a set of 1,000,000 numbers with this distribution and find the fraction that meet the requirement.

Exercise 3 Answers

`pnorm(.9) - pnorm(.3)`

`## [1] 0.1980285`

```
# Method 2
MyNormProb(lb=.3,ub=.9)
```

`## [1] 0.1980285`

```
# Method 3
x = rnorm(1000000)
mean(x>.3 & x < .9)
```

`## [1] 0.198896`

Find the probability that a single random number drawn from a set of numbers with a normal distribution having a mean of 100 and a standard deviation of 10 is > 90 and < 115. Do this three ways.

- Use pnorm()
- Use MyNormProb() to get a picture.

`pnorm(115,mean=100,sd=10) - pnorm(90,mean=100,sd=10)`

`## [1] 0.7745375`

```
# Method 2
MyNormProb(lb=90,ub=115,mean=100,sd=10,
MyLabel = "Non-Standard Normal RV (100,10)")
```

`## [1] 0.7745375`

```
# Method 3
x = rnorm(1000000,mean=100,sd=10)
mean(x> 90 & x < 115)
```

`## [1] 0.774659`

90% of a standard normal distribution lies to the left of x. What is the value of x?

`qnorm(.9)`

`## [1] 1.281552`

10% of a standard normal distribution lies to the right of x. What is the value of x?

`qnorm(1 - .1)`

`## [1] 1.281552`

A bank’s loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find P60, the score which separates the lower 60% from the top 40%.

`qnorm(.6,mean=200,sd=50)`

`## [1] 212.6674`

The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected.

```
Top = qnorm(1-.03,mean = 8.5, sd = .09)
Top
```

`## [1] 8.669271`

```
Bottom = qnorm(.03,mean = 8.5, sd = .09)
Bottom
```

`## [1] 8.330729`

The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. What is the probability that a randomly selected component weighs more than 8.6 g.

`1 - pnorm(8.6,mean=8.5,sd=.09)`

`## [1] 0.1332603`