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Numerade Educator



Problem 15 Medium Difficulty

Refer to Exercise 3. Calculate and interpret the standard deviation of the random variable X. Show your work.




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Video Transcript

So in this problem, we have a probability distribution. Our random variable X and we have the corresponding probabilities of 012 and three also four. And this is 0.1 point 2.3 point 3.1. And just to make sure 6789 10 Yes, the probability all add up to one, just double checking that. So we know we have to find the standard deviation. But first of all, we have to find the mean of that random variable. So we're gonna have to take zero times 00.1 plus one times point to plus two times 20.3 plus three times 30.3 and four times 40.1. So I should be able to do most of that in my head. Um, but we'll we'll accumulated on the calculators that zero plus 00.2 plus that will give me 0.6. That will give me plus 0.9. And that will give me plus 0.4. And those add up to 2.1. So we know the mean is 2.1, and now we want to find the standard deviation. So the standard deviation we know of that random variable is we're going to have to take the difference between each of our elements up here. So we're going to have zero minus 2.1 quantity squared to find that deviation. And then we waited by its corresponding probability. Uh, next one is one minus 2.1. Again, quantity squared, and we wait. At times, it's probability next one to minus 2.1 quantity squared and wait. At times it's probability, and we're getting there. We have three minus 2.1 quantity squared times. It's probability, and then the last one and I should be able to squeeze it in there and we'll go for minus 2.1 quantity squared times its corresponding probability, which is 0.1. So now I'm ready to put this into my calculator, and I know that difference is 2.1. Absolute value is 2.1 squared times 0.1 plus, and I know that difference is negative 1.1, but I'll put it in my calculator is 1.1 squared times that 0.2 plus. That difference is absolute. Value is 0.1 squared times 0.3. That difference is 0.9 point nine squared times 0.3 plus last one, I believe. And that difference is 1.9 squared times. That's corresponding probability times 0.1. And so I get square root of and underneath the radical we have that. And so now we can take the square root and find that that answer is 1.1358 Yeah, now. So that means that on the average, if they think this is a spell check or finding some airs on the average, If you did this over and over and over again, you would have a mean number of lots and lots of trials of obviously all of them will be. The trials will all be integers from 0 to 4. However, the mean of those values would be 2.1 if you do it for a very, very long period of time and the standard deviation of those values would come out to be this theoretically, so they're all going to be about 68% of them or so will be within one standard deviation. And so on, depending again, on how many trials you do