- #1

- 24

- 0

a) Z= X^2 + 1

h(y)= f(g^-1(y))

Attempted Solution

X= -1 0 1 2 3 4

Z= 2 1 2 5 10 17

How do i find Pr(Z)?

The book gave an answer of

1/12 Z=1,10

1/4 Z= 5,17

1/3 Z=2

0 elsewhere.

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- Thread starter iHeartof12
- Start date

- #1

- 24

- 0

a) Z= X^2 + 1

h(y)= f(g^-1(y))

Attempted Solution

X= -1 0 1 2 3 4

Z= 2 1 2 5 10 17

How do i find Pr(Z)?

The book gave an answer of

1/12 Z=1,10

1/4 Z= 5,17

1/3 Z=2

0 elsewhere.

- #2

- 768

- 4

When X = 0, Z= 1. So the probability of Z=1 is Pr(x=0)= 1/12

and so on

Notice that when X= 1, Z=2 again. Thus, the probability of Z=2 is Pr(x= -1) plus Pr(x=1) which equals 1/3.

- #3

- 768

- 4

Incidently, are you using Tsokos' new book? The masterpiece of typos?

- #4

- 24

- 0

Let me know if I'm thinking about this correctly.

Z=5 only X=2 can produce this so Pr= 1/4

Z=10 only X=3 can produce this so Pr=1/12

Z= 17 only X=4 can produce this so Pr=1/4.

- #5

- 24

- 0

Lol yes i am.

- #6

- 768

- 4

Let me know if I'm thinking about this correctly.

Z=5 only X=2 can produce this so Pr= 1/4

Z=10 only X=3 can produce this so Pr=1/12

Z= 17 only X=4 can produce this so Pr=1/4.

Yes, this is correct.

Do you have a test Thursday? I'm wondering if you are in my class, since this book is so new. 6:30 Tuesdays and Thursdays?

- #7

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

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a) Z= X^2 + 1

h(y)= f(g^-1(y))

Attempted Solution

X= -1 0 1 2 3 4

Z= 2 1 2 5 10 17

How do i find Pr(Z)?

The book gave an answer of

1/12 Z=1,10

1/4 Z= 5,17

1/3 Z=2

0 elsewhere.

I hope your book and your instructor do not use the terminology "probability density" in this case. Discrete random variables do not have probability densities; they have distribution functions and probability mass functions, but not densities.

RGV

- #8

- 24

- 0

Yes, this is correct.

Do you have a test Thursday? I'm wondering if you are in my class, since this book is so new. 6:30 Tuesdays and Thursdays?

No I have an exam on Friday. I take it MWF 12:55 but it may be for the same professor our exam covers Chpts. 3 and 4!

- #9

- 768

- 4

Yup, must be the same guy. :)

- #10

- 768

- 4

I hope your book and your instructor do not use the terminology "probability density" in this case. Discrete random variables do not have probability densities; they have distribution functions and probability mass functions, but not densities.

RGV

Oh he uses the term "probability density" all right. This book is a joke.

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