Mathematics 22 Online
OpenStudy (anonymous):

Find the variance and the standard deviation of a random variable with the following distribution: -3 1 2 3 .5 .2 .1 .2

OpenStudy (amistre64):

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)]

OpenStudy (amistre64):

(-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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

3.125 shoulda been 6.125 ....

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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 :)

OpenStudy (anonymous):

so 6.65 is the variance and 2.5787 is the sd?

OpenStudy (amistre64):

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

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