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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)

Respuesta :

MrRoyal

Answer:

[tex]P(Z < 4.3) = 0.9999[/tex]

[tex]P(Z> -1.69) = 0.9545[/tex]

[tex]P(-1.12 < Z < 1.12) = 0.7373[/tex]

[tex]P(-1.3 < Z < 0.98) = 0.7397[/tex]

[tex]P(Z < -1.12\ or\ Z > 1.12) = 0.2628[/tex]

Step-by-step explanation:

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

Solving (a): [tex]P(Z < 4.3)[/tex]

[tex]P(Z < 4.3) = 0.9999[/tex]

Solving (b): [tex]P(Z> -1.69)[/tex]

First, we rewrite this using complement rule

[tex]P(Z> -1.69) = 1 - P(Z < -1.69)[/tex]

From the z score table:

[tex]P(Z< -1.69) = 0.045514[/tex]

So, we have:

[tex]P(Z> -1.69) = 1 - 0.045514[/tex]

[tex]P(Z> -1.69) = 0.9545[/tex]

Solving (c) [tex]P(-1.12 < Z < 1.12)[/tex]

This is equivalent to:

[tex]P(-1.12 < Z < 1.12) = P(Z<1.12) - P(Z<-1.12)[/tex]

Use the z score table

[tex]P(-1.12 < Z < 1.12) = 0.86864 - 0.13136[/tex]

[tex]P(-1.12 < Z < 1.12) = 0.7373[/tex]

Solving (d) [tex]P(-1.3 < Z < 0.98)[/tex]

This is equivalent to

[tex]P(-1.3 < Z < 0.98) = P(Z<0.98)- P(Z<-1.3)[/tex]

[tex]P(-1.3 < Z < 0.98) = 0.83646- 0.0968[/tex]

[tex]P(-1.3 < Z < 0.98) = 0.7397[/tex]

Solving (e) [tex]P(Z < -1.12 or Z > 1.12)[/tex]

This is interpreted as:

[tex]P(Z < -1.12\ or\ Z > 1.12) = P(Z < -1.12) + P(Z > 1.12)[/tex]

Apply complement rule

[tex]P(Z < -1.12\ or\ Z > 1.12) = P(Z < -1.12) + 1 - P(Z < 1.12)[/tex]

[tex]P(Z < -1.12\ or\ Z > 1.12) = 0.1314 + 1 - 0.8686[/tex]

[tex]P(Z < -1.12\ or\ Z > 1.12) = 0.2628[/tex]

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