# help please: probability and probability density functionsWatch

Thread starter 13 years ago
#1
hi guys

i need a bit of help to do this question:
"A random variable X has the probability density function fx (x)=2x, where 0<x<1, with mean µ=2/3"

"four independent observations are made on this variable. What is the probability that"
(a) at least three of the observations are greater than the mean
(b) all of the observations lie between the mean and the median

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13 years ago
#2
a) the area to the left of x=2/3 is 4/9 and to the right is 5/9

the distribution of observations to the right of the mean is Bin(4,5/9)

so P(at least 3) = P(3) + P(4) =4C3 *(5/9)3 *(4/9)1 + 4C4 *(5/9)4 *(4/9)0

b) the area between the mean and the median is 1/18 in size ( 1/2 - 4/9) so the probability of landing in that region 4 times is

4C4 *(1/18)4*(17/18)0
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