Back
to top
S3 Probability Fair Die
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
This discussion is closed.
How many times must a fair dice be rolled in order for there to be a less than 1% chance that the mean of all scores differs from 3.5 by more than 0.1.
could anyone please help me with this problem.
Thank you.
could anyone please help me with this problem.
Thank you.
0
Report
#3
(Original post by jpepsred)
Bump
Bump
What do you know about the distribution of the sample mean if the die is rolled n times say?
0
Report
#4
Mean(X) is given by the distribution (3.5, 2.92/n), where 2.92 is the variance of X. I get that much.
I end up with p(z < 0.0585sqrt(n)) > 0.995 which, by referring to the tables, implies that 0.0585sqrt(n) >2.6.
The solutionbank, however, gives 0.0585sqrt(n) > 2.5758
2.5758 isn't a value of z which is in the tables, and I can't understand where it comes from.
It makes a huge difference: book gives n => 1936, I get n=> 1976.
I wonder if op appreciates finally getting an answer a decade after their a levels.
I end up with p(z < 0.0585sqrt(n)) > 0.995 which, by referring to the tables, implies that 0.0585sqrt(n) >2.6.
The solutionbank, however, gives 0.0585sqrt(n) > 2.5758
2.5758 isn't a value of z which is in the tables, and I can't understand where it comes from.
It makes a huge difference: book gives n => 1936, I get n=> 1976.
I wonder if op appreciates finally getting an answer a decade after their a levels.
0
Report
#5
(Original post by jpepsred)
Bump
Bump
(Original post by ghostwalker)
9 year old thread - must be a record bump.
9 year old thread - must be a record bump.
0
Report
#6
(Original post by jpepsred)
...
...
0
Report
#8
(Original post by jpepsred)
THANK YOU. I haven't used that since s1. Completely forgot it existed!
THANK YOU. I haven't used that since s1. Completely forgot it existed!
0
X
Page 1 of 1
Go to first unread
Skip to page:
new posts
