You must be logged in to reply to this topic.
- E2P 1
Viewing 8 posts - 1 through 8 (of 8 total)
You must be logged in to reply to this topic.
You must be logged in to reply to this topic.
A random variable has the following probability distribution:
X -2 -1 1 2
P(X) 0.3 0.2 0.3 ?
Find and interpret a) probability when X=2 b) mean c) standard deviation.
a) P[X=2] = 1 – (0.3+0.2+0.3) = 0.2
The probability that outcome would be equal 2 is 20 %
here is Melecia Lee’s answer to b)
Let E[X] represent the Mean.
E[X] = -2(0.3) + (-1)(0.2) + 1(0.3) + 2(0.2)
= -0.6 -0.2 +0.3+0.4
= -0.1
This means that the weighted average of the probabily distribution is -0.1
What you say is fine. I would say “The average for this random variable is -0.1.”
Jonathan Ciabotaru
My resolution for exercise 1
1)
-2 = 0.3 -1 = 0.2 1 = 0.3 2 = x
a) 0.2
b)E[X]= -0.6 – 0.2 + 0.3 + 0.4 = -0.1
E[X^2]= 1.2 + 0.2 + 0.3 + 0.8 = 2.5
V[X] = E[X^2] – E[X]^2 = 2.5 – 0.01 = 2.49
c)SD[X] = Square root of V[X] = (2.49)^0.5
SD[X] = 1.57797
when it asks for interpret the standard deviation and mean, we are supposed to write something in words like this “the average outcome is -0.1 and on average, an outcome will be about 1.58 of a unit from the mean”, right?
Jonathan’s resolution is correct and yes Ying, how you wrote your interpretation is correct
V[x]= sum [ ( x-u)^2 – P(x)
V[x}= [-2(-0.1)]^2 . 0.3 + [-1(-0.1)]^2 . 0.2 + [1(-0.1)]^2 . 0.3 + [2(-0.1)]^2 . 0.2
= 0.025
SD[x}= 0.158
#1 fully done, all steps shown.
You must be logged in to reply to this topic.
Ursula C. Schwerin Library
New York City College of Technology, C.U.N.Y
300 Jay Street, Library Building - 4th Floor
Our goal is to make the OpenLab accessible for all users.