Question

Find the variance and the standard deviation of a random variable with the
following distribution:

-3 1 2 3
.5 .2 .1 .2

Answer 1

variance relies on the mean; so to determine the mean we have to add together all the products of the variable and it probability.

-1.5 + .2 + .2 + .6 = -.5

the SD = sqrt(sum[(x-mean)^2 * P(x)]

the variance is the square of the SD which is then sum[(x-mean)^2*P(x)]

Answer 2

(-3+.5)^2 * .5: 3.125
+ (1-.5)^2*.2 : .05
+ (2-.5)^2*.1 : .225
+ (3-.5)^2*.2 : 1.25
------------------------
sum = 3.525 = variance

sqrt(3.525) = SD

if I recall it correctly

Answer 3

for some reason my calculator gives me a different answer for the variance and sd; but we agree on the mean ....

Answer 4

3.125 shoulda been 6.125 ....

Answer 5

if I could learn to add, this would turn out better....

Answer 6

when you subtract a negative, its like adding a positive...

(-3+.5)^2 * .5: 3.125
+ (1+.5)^2*.2 : .45
+ (2+.5)^2*.1 : .625
+ (3+.5)^2*.2 : 2.45
------------------------
sum = 6.65 = variance
sqrt(6.65) = sd = 2.57875....

now it agrees with the calculator :)

Answer 7

so 6.65 is the variance and 2.5787 is the sd?

Answer 8

yes, after much deliberation and retraining in the art of addition; I came to the same conclusion as my TI-83.