Probability Question - Random Variable
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Probability Question - Random Variable

[From: ] [author: ] [Date: 11-12-22] [Hit: ]
2)+15.9(0.5)+19.1(0.3)E(x)=2.7+7.......

E(x)=13.5(0.2)+15.9(0.5)+19.1(0.3)
E(x)=2.7+7.95+5.73
E(x)=16.38

So that's the expected value. This means that the average freezer sold has a storage capacity of approximately 16.38 cubic feet, within some interval of confidence.

Next, we have what you refer to as E(X2), which I'm honestly not sure what it is. I'm assuming, though, that it's calculated the same way E(x) was, except you square all of the units for x first. So then...

E(x^2)=13.5^2(0.2)+15.9^2(0.5)+19.1^2(…
E(x^2)=182.25(0.2)+252.81(0.5)+364.81(…
E(x^2)=36.45+126.405+109.443
E(x^2)=272.298

So there you have it. I'm not sure what this tells you, but...here you go.

The next one, I do know how to calculate for sure, as well as what it means. The variance, or Var(x), is equivalent to the sum of the product between the squared deviations of each x-value from the expected value with their respective probabilities. That means it looks like this:

Var(x)=(13.5-E(x))^2(0.2)+(15.9-E(x))^…
Var(x)=(-2.88)^2(0.2)+(-0.48)^2(0.5)+(…
Var(x)=8.2944(0.2)+(0.2304)(0.5)+(7.39…
Var(x)=1.65888+0.1152+2.21952
Var(x)=3.9936

So the variance of this set of data is 3.9936, pretty much close enough to be considered 4. This value is the square of the standard deviation, used to predict the probability of a given consumer's deviation from the norm. I would say this concludes the first part of your question. I'll probably finish the rest tomorrow...sometime.
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