For the discrete random variable X, the probability distribution is given by
P(X=x)= kx x=1,2,3,4,5
=k(10-x) x=6,7,8,9
Find the value of the constant k
E(X)
I am lost , it is the bonus question in my homework on random variables so it must be hard.
Many thanks for any help.
Alicia
Thanks a lot. I understood. What is E (X) ?
graph p
symmetric about x=5 where it is 5k
suspect zero at x = 0 and x = 10 :)
k + 2k + 3k + 4k + 5k + k(10-6)+ k(10-7) + k(10-8) + k(10-9) = 1 if that is the whole distribution
2k + 4k +6k + 8k + 5k = 1
k = 1/25
multiply each x value by its probability (kx)
E(x) is the sum of them all -- the expected value of x.
Well Alicia, let's tackle this bonus question together. Don't worry, with a little humor added in, it won't seem as hard as you think!
To find the value of the constant k, we can use the fact that the sum of all probabilities should add up to 1. So, let's sum up all the probabilities.
For x = 1, 2, 3, 4, 5:
P(X = x) = kx
For x = 6, 7, 8, 9:
P(X = x) = k(10 - x)
Now, let's calculate the sum:
P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
= k(1) + k(2) + k(3) + k(4) + k(5) + k(10 - 6) + k(10 - 7) + k(10 - 8) + k(10 - 9)
= k(1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1)
= k(25)
Since this sum should be equal to 1, we have:
k(25) = 1
Now, we can solve for k:
k = 1/25
So, the value of the constant k is 1/25.
Now, let's find E(X), which represents the expected value. To do this, we need to multiply each value of X by its corresponding probability and then sum them up.
E(X) = (1)(P(X = 1)) + (2)(P(X = 2)) + (3)(P(X = 3)) + (4)(P(X = 4)) + (5)(P(X = 5)) + (6)(P(X = 6)) + (7)(P(X = 7)) + (8)(P(X = 8)) + (9)(P(X = 9))
E(X) = (1)(1/25) + (2)(2/25) + (3)(3/25) + (4)(4/25) + (5)(5/25) + (6)(4/25) + (7)(3/25) + (8)(2/25) + (9)(1/25)
E(X) = (1 + 4 + 9 + 16 + 25 + 24 + 21 + 16 + 9)/25
E(X) = 125/25
E(X) = 5
So, the expected value E(X) is 5.
I hope that helps, Alicia! If you have any more questions, feel free to ask. Just remember, math can be funny too!
To find the value of the constant k in the given probability distribution, we need to use the property that the sum of the probabilities for all possible values of X must equal 1.
First, let's calculate the probabilities for the given values of X:
P(X=1) = k(1) = k
P(X=2) = k(2) = 2k
P(X=3) = k(3) = 3k
P(X=4) = k(4) = 4k
P(X=5) = k(5) = 5k
P(X=6) = k(10-6) = 4k
P(X=7) = k(10-7) = 3k
P(X=8) = k(10-8) = 2k
P(X=9) = k(10-9) = k
To find the value of k, we can sum up all these probabilities:
P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9)
= k + 2k + 3k + 4k + 5k + 4k + 3k + 2k + k
= 25k
Since the sum of all probabilities must equal 1, we have:
25k = 1
To find the value of k, we divide both sides of the equation by 25:
k = 1/25
So, the value of the constant k is 1/25.
Now, let's calculate the expected value, E(X), which represents the average value of the random variable X.
The expected value, E(X), is calculated by summing up the product of each possible value of X and its corresponding probability, and in this case, we have values of X ranging from 1 to 9.
E(X) = (1 * P(X=1)) + (2 * P(X=2)) + (3 * P(X=3)) + (4 * P(X=4)) + (5 * P(X=5)) + (6 * P(X=6)) + (7 * P(X=7)) + (8 * P(X=8)) + (9 * P(X=9))
Substituting the values of the probabilities we calculated earlier:
E(X) = (1 * k) + (2 * 2k) + (3 * 3k) + (4 * 4k) + (5 * 5k) + (6 * 4k) + (7 * 3k) + (8 * 2k) + (9 * k)
Substituting the value of k we found earlier:
E(X) = (1 * 1/25) + (2 * 2/25) + (3 * 3/25) + (4 * 4/25) + (5 * 5/25) + (6 * 4/25) + (7 * 3/25) + (8 * 2/25) + (9 * 1/25)
Now, calculate each term and multiply:
E(X) = 1/25 + 4/25 + 9/25 + 16/25 + 25/25 + 24/25 + 21/25 + 16/25 + 9/25
E(X) = 125/25
E(X) = 5
Therefore, the value of the constant k is 1/25 and the expected value E(X) is 5.