A random variable has aprobability function p(x)=kx; x=1,2,3,4,5.find the value of k, p(x>=3),compute mean and variance of x.
k(1+2+3+4+5) = 1
k = 1/15
p(3) = 3/15
p(4) = 4/15
p(5) = 5/15
so p (x>/=3) = 12/15 = 4/5
etc
To find the value of k, we can use the fact that the sum of the probabilities for all possible values of x must be equal to 1. In other words, the sum of p(x) for all x must be equal to 1.
Therefore, we can calculate k as follows:
p(1) + p(2) + p(3) + p(4) + p(5) = 1
k(1) + k(2) + k(3) + k(4) + k(5) = 1
k(1 + 2 + 3 + 4 + 5) = 1
k(15) = 1
k = 1/15
Now let's compute p(x >= 3):
p(x >= 3) means the probability of getting a value of x that is greater than or equal to 3. In this case, the possible values of x that satisfy this condition are 3, 4, and 5.
p(x >= 3) = p(x = 3) + p(x = 4) + p(x = 5)
= (1/15)(3) + (1/15)(4) + (1/15)(5)
= 3/15 + 4/15 + 5/15
= 12/15
= 4/5
Next, let's compute the mean of x:
The mean of a random variable is the weighted average of all possible values of x, with the probabilities serving as weights.
Mean (µ) = Σ (x * p(x)) for all possible values of x
= (1/15)(1) + (2/15)(2) + (3/15)(3) + (4/15)(4) + (5/15)(5)
= 1/15 + 4/15 + 9/15 + 16/15 + 25/15
= 55/15
= 11/3
Finally, let's compute the variance of x:
Variance is a measure of how spread out the values of a random variable are.
Variance (σ²) = Σ((x - µ)² * p(x)) for all possible values of x
= (1 - 11/3)²(1/15) + (2 - 11/3)²(2/15) + (3 - 11/3)²(3/15) + (4 - 11/3)²(4/15) + (5 - 11/3)²(5/15)
= (2/3)²(1/15) + (-1/3)²(2/15) + (4/3)²(3/15) + (7/3)²(4/15) + (10/3)²(5/15)
= 4/9 * 1/15 + 1/9 * 2/15 + 16/9 * 3/15 + 49/9 * 4/15 + 100/9 * 5/15
= 4/135 + 2/135 + 48/135 + 196/135 + 500/135
= 250/135
= 50/27
So, the value of k is 1/15, p(x >= 3) is 4/5, the mean of x is 11/3, and the variance of x is 50/27.