# Statistics with Probability

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• #13264

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.

#16799

a) P[X=2] = 1 – (0.3+0.2+0.3) = 0.2
The probability that outcome would be equal 2 is 20 %

#16808

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.”

#16809

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

#16830

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?

#16835

Jonathan’s resolution is correct and yes Ying, how you wrote your interpretation is correct

#16889

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

#16890

#1 fully done, all steps shown.

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