Abstract,
painted by Jackson Pollock
A
Random Variable:
9.
The missing probability value (under the number 4) in the random
variable table above is: 0.15
Steps
to find answer:
1
 (0.10+0.15+0.25+0.10+0.10+0.15) = 0.15
10.
The sum of the probabilites in the second row of any random variable
table like the one above should equal (round to 3 decimal places): 1
Steps
to find answer:
Based
on theory that for any Random Variables, sum of probabilities should
be 1 (one)
11.
Read section 4.4.1 in the book (Yakir, 2011). Do the numbers in
the table above (for the random variable) represent a data sample
(Yes/No)? No
Steps
to find answer:
It
is because the Table above represents the sample space.
Painted
by Jackson Pollock
12.
In the random variable table shown above, the value in the second row
represents the cumulative probability of the corresponding values in
the first row (True/False) : FALSE
Steps
to find answer:
It
is because the row represents the individual probability for each of
the sample space. the probability value for the last entry is 1.
13.
The probability that a randomly selected value from this random value
will be less than or equal to 3 is : 0.6
Steps
to find answer: P(X<=3) = 0.10+0.15+0.25+0.10 = 0.60
14.
What is the probability that a randomly selected value from the
random variable would be exactly 1.5?
Answer:
ZERO
15.
Review section 4.4 in the book (Yakir, 2011), especially pages
57—58. The expectation of the random variable is: 2.95
Steps
to find answer:
E(X)
= 0*0.10 + 1*0.15 + 2*0.25 + 3*0.10 + 4*0.15 + 5*0.10 + 6*0.15 = 2.95
Painted
by Jackson Pollock
16.
To find the expectation of a random variable by using a relative
frequency table, you can add the values in the first row of the table
and divide by the number of columns in the table (True/False): FALSE
Steps
to find answer: We use formula to find the expectation of a random
variable as of to answer question 15
17.
Study Yakir (2011) pp. 5759 and solved problems 4.1.64.1.8.
The (population) standard deviation of the random variable above is
(round to 3 decimal places): 1.909
(hint,
you can not put values from the table into the sd() function because
the sd() function does not adjust for the probabilities).
Steps
to find answer:
So,
Standard deviation = (E(X^2)(E(X))^2)^0.5 = (12.352.952)^0.5 =
1.909
18.
If you have already calculated the standard deviation of a data
sample, what is the next thing to do to find the variance:
Answer:
Variance
is equal to square of standard deviance
E(X^2)
= 0*0.10 + 1*0.15 + 4*0.25 + 9*0.10 + 16*0.15 + 25*0.10 + 36*0.15 =
12.35
_{}^{}
#
Continue to part 3
I don't understand the math one bit. But I like the Pollock paintings!
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

A Random Variable:
9. The missing probability value (under the number 4) in the random variable table above is: 0.15
Steps to find answer:
1  (0.10+0.15+0.25+0.10+0.10+0.15) = 0.15
Ummm, very interesting. This looks like GREEK to me. Thanks to you I now know what GREEK looks like. Please LAUGH...that's just me goofing off being silly!
ReplyDeleteThanks for sharing this informative post. it's very thought provoking.
Thank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

10. The sum of the probabilites in the second row of any random variable table like the one above should equal (round to 3 decimal places): 1
Steps to find answer:
Based on theory that for any Random Variables, sum of probabilities should be 1 (one)
I think it would be better if you provide the data in a table then put the result below the data.
ReplyDeleteno data, just answer the question one by one
DeleteThank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

11. Read section 4.4.1 in the book (Yakir, 2011). Do the numbers in the table above (for the random variable) represent a data sample (Yes/No)? No
Steps to find answer:
It is because the Table above represents the sample space.
i like this :D
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

12. In the random variable table shown above, the value in the second row represents the cumulative probability of the corresponding values in the first row (True/False) : FALSE
Steps to find answer:
It is because the row represents the individual probability for each of the sample space. the probability value for the last entry is 1.
Interesting post :) have a lovely day :)
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

13. The probability that a randomly selected value from this random value will be less than or equal to 3 is : 0.6
Steps to find answer: P(X<=3) = 0.10+0.15+0.25+0.10 = 0.60
Love the paintings, Jackson Pollock is a brave man.
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

14. What is the probability that a randomly selected value from the random variable would be exactly 1.5?
Answer: ZERO
Interesting post!
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

15. Review section 4.4 in the book (Yakir, 2011), especially pages 57—58. The expectation of the random variable is: 2.95
Steps to find answer:
E(X) = 0*0.10 + 1*0.15 + 2*0.25 + 3*0.10 + 4*0.15 + 5*0.10 + 6*0.15 = 2.95
very clever post! Thanks for sharing!
ReplyDeleteDon't Call Me Fashion Blogger
Facebook
Bloglovin'
Thank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

Have a nice weekend
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

16. To find the expectation of a random variable by using a relative frequency table, you can add the values in the first row of the table and divide by the number of columns in the table (True/False): FALSE
Steps to find answer: We use formula to find the expectation of a random variable as of to answer question 15
Thanks for sharing!
ReplyDeleteBest regards
Thank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

17. Study Yakir (2011) pp. 5759 and solved problems 4.1.64.1.8. The (population) standard deviation of the random variable above is (round to 3 decimal places): 1.909
(hint, you can not put values from the table into the sd() function because the sd() function does not adjust for the probabilities).
Steps to find answer:
E(X^2) = 0*0.10 + 1*0.15 + 4*0.25 + 9*0.10 + 16*0.15 + 25*0.10 + 36*0.15 = 12.35
So, Standard deviation = (E(X^2)(E(X))^2)^0.5 = (12.352.952)^0.5 = 1.909
Well, I am lost in math but I enjoyed so much Pollock's paintings.
ReplyDeleteThey are brilliant!
Wish you a beautiful warm weekend!
I see...yes, Pollock is one of brilliant painters
DeleteThank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

18. If you have already calculated the standard deviation of a data sample, what is the next thing to do to find the variance:
Answer:
Variance is equal to square of standard deviance
E(X^2) = 0*0.10 + 1*0.15 + 4*0.25 + 9*0.10 + 16*0.15 + 25*0.10 + 36*0.15 = 12.35

You've tangled my head ... interesting post.
ReplyDeleteNice weekend.
Thank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

A Random Variable:
9. The missing probability value (under the number 4) in the random variable table above is: 0.15
Steps to find answer:
1  (0.10+0.15+0.25+0.10+0.10+0.15) = 0.15
nice abstracts
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

10. The sum of the probabilites in the second row of any random variable table like the one above should equal (round to 3 decimal places): 1
Steps to find answer:
Based on theory that for any Random Variables, sum of probabilities should be 1 (one)
I know those paintings, so pretty in not obvious sense☺
ReplyDeleteThank you to visit:
DeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

11. Read section 4.4.1 in the book (Yakir, 2011). Do the numbers in the table above (for the random variable) represent a data sample (Yes/No)? No
Steps to find answer:
It is because the Table above represents the sample space.
Have I mentioned I suck at math?
ReplyDeleteThank you to visit:
ReplyDeleteMATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 2

12. In the random variable table shown above, the value in the second row represents the cumulative probability of the corresponding values in the first row (True/False) : FALSE
Steps to find answer:
It is because the row represents the individual probability for each of the sample space. the probability value for the last entry is 1.