Probability and statistics expected value
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > Probability and statistics expected value

Probability and statistics expected value

[From: ] [author: ] [Date: 11-06-21] [Hit: ]
16/25 outer. (by area)So:1/2 * 1/25 = 1/50 of the time he scores 101/2 * 8/25 = 8/50 of the time he scores 51/2 * 16/25 = 16/50 of the time he scores 31/2 = 25/50 of the time he scores 0Then E(x) = sum x p(x)= 10 * 1/50 + 5 * 8/50+ 3 * 16/50 + 0 * 25/50EDIT: Removed a stray * 5 from the last line. Total of 98/50 = 1.96 agrees with Ian.......

It's actually quite simple, really. All you do is multiply each value by it's respective probability of outcome and add them all up. So it looks like this:
E=10(1/6) + 5(1/6) + 3(1/6) + 0(1/2)

So now you just do the math:
E=10(1/6) + 5(1/6) + 3(1/6) + 0(1/2)
E=10/6 + 5/6 + 3/6 + 0/6
E=18/6
E=3

So the expected number of points the archer is expected to make on any given shot is 3 points.





It would appear as though I have made a mistake. You said "...equally likely to hit one point on the target as any other...". This means that the areas must be taken into account for the probability. I could do this, but I don't care enough to, so just check the other answers. It appears as though they've got it right.

-
If the the archer has a probability of 1/2 of hitting the target at all, then half the time the archer will receive 0 points and the other half of the time they will receive a positive amount of points. Since it is mentioned that one point value on the target is equally likely to another point value, then we can ignore the area of the each point value and deviate the remaining 1/2 probability equally between the point values. So, you should have determined that:

Points---------------Probability
0---------------------------1/2
3---------------------------1/6
5---------------------------1/6
10-------------------------1/6

Therefore, the expected value of points on a single throw would be

3(1/6)+5(1/6)+10(1/6)
= 3 points <--------------------Answer

-
Notice that the target is 1/25 bullseye, 8/25 centre, 16/25 outer. (by area)

So:
1/2 * 1/25 = 1/50 of the time he scores 10
1/2 * 8/25 = 8/50 of the time he scores 5
1/2 * 16/25 = 16/50 of the time he scores 3
1/2 = 25/50 of the time he scores 0

Then E(x) = sum x p(x)
= 10 * 1/50 + 5 * 8/50 + 3 * 16/50 + 0 * 25/50

EDIT: Removed a stray "* 5" from the last line. Total of 98/50 = 1.96 agrees with Ian.

-
What Class is this for
12
keywords: expected,value,and,statistics,Probability,Probability and statistics expected value
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .