p=80% (chances of winning a game)
n=5 (number of games)
expected value of x=4
How do I find the variance and standard deviation of x?
And also, how do I find the probability that my opponent will win at least one game
Thanks!
p=80% (chances of winning a game)
n=5 (number of games)
expected value of x=4
How do I find the variance and standard deviation of x?
And also, how do I find the probability that my opponent will win at least one game
Thanks!
well, variance is found by using V = n*p*(1-p)p=80% (chances of winning a game)
n=5 (number of games)
expected value of x=4
How do I find the variance and standard deviation of x?
V = 5*.80*.20 = 0.80
The standard deviation is found by taking the square root of the variance.
sigma = sqrt(V) = sqrt(0.80) = .8944
I'm gonna try to figure out the second part of the question now.
p=80% (chances of winning a game)
n=5 (number of games)
expected value of x=4
How do I find the variance and standard deviation of x?
And also, how do I find the probability that my opponent will win at least one game
I think you would model your opponent winning at least one game as the probability of you winning at most 4 games:
P total = P(win 0) + P(win 1) + P(win 2) + P(win 3) + P(win4)
$\displaystyle
P(x\!=\!0) \;=\;{5\choose0}(0.8)^0(0.2)^5 \;=\;0.0003
$
$\displaystyle
P(x\!=\!1) \;=\;{5\choose1}(0.8)^1(0.2)^4 \;=\;0.0064
$
$\displaystyle
P(x\!=\!2) \;=\;{5\choose2}(0.8)^2(0.2)^3 \;=\;0.0512
$
$\displaystyle
P(x\!=\!3) \;=\;{5\choose3}(0.8)^3(0.2)^2 \;=\;0.2048
$
$\displaystyle
P(x\!=\!4) \;=\;{5\choose4}(0.8)^4(0.2)^1 \;=\;0.4096
$
Add all of these values to get: P total = 0.6723
Please quote what you are responding to, so we do not have the present situation where we do not know what the original question was because the OP has deleted it.
Sorry I see that you have quoted it (if you use the quote button rather than the reply it will put the quoted text in a nice box).
Thanks
CB