Question

For this week's homework, we will revisit some old homework problems. But this time we'll use R to solve them (you must use R to do problems 1 - 5). We'll also collect a little data in class. 1) Find the following probabilities: a) Pr{Z < 4.3} b) Pr{Z > -1.69} c) Pr{-1.12 < Z < 1.12} d) Pr{-1.3 < Z < 0.98} e) Pr{Z < -1.12 or Z > 1.12} (you want the probability outside the range -2.67 to 2.67)

Answer 1

`P(Z < 4.3) = 0.9999`

`P(Z> -1.69) = 0.9545`

`P(-1.12 < Z < 1.12) = 0.7373`

`P(-1.3 < Z < 0.98) = 0.7397`

`P(Z < -1.12\ or\ Z > 1.12) = 0.2628`

Explanation:

To answer these questions, we will make use of a z score table

Solving (a):
`P(Z < 4.3)`

`P(Z < 4.3) = 0.9999`

Solving (b):
`P(Z> -1.69)`

First, we rewrite this using complement rule

`P(Z> -1.69) = 1 - P(Z < -1.69)`

From the z score table:

`P(Z< -1.69) = 0.045514`

So, we have:

`P(Z> -1.69) = 1 - 0.045514`

`P(Z> -1.69) = 0.9545`

Solving (c)
`P(-1.12 < Z < 1.12)`

This is equivalent to:

`P(-1.12 < Z < 1.12) = P(Z<1.12) - P(Z<-1.12)`

Use the z score table

`P(-1.12 < Z < 1.12) = 0.86864 - 0.13136`

`P(-1.12 < Z < 1.12) = 0.7373`

Solving (d)
`P(-1.3 < Z < 0.98)`

This is equivalent to

`P(-1.3 < Z < 0.98) = P(Z<0.98)- P(Z<-1.3)`

`P(-1.3 < Z < 0.98) = 0.83646- 0.0968`

`P(-1.3 < Z < 0.98) = 0.7397`

Solving (e)
`P(Z < -1.12 or Z > 1.12)`

This is interpreted as:

`P(Z < -1.12\ or\ Z > 1.12) = P(Z < -1.12) + P(Z > 1.12)`

Apply complement rule

`P(Z < -1.12\ or\ Z > 1.12) = P(Z < -1.12) + 1 - P(Z < 1.12)`

`P(Z < -1.12\ or\ Z > 1.12) = 0.1314 + 1 - 0.8686`

`P(Z < -1.12\ or\ Z > 1.12) = 0.2628`