B:)) "Determine the mean and standard deviation of the random variable X. Round your results to the nearest penny."
do you know the formula?
Yes it's like... Standard Deviation over the sample...
Unless I'm wrong...
its same what do you get for mean?
\(\mu = mean = \sum p_ix_i \\ \sigma = \sqrt{\dfrac{1}{N}\sum (x_i-\mu )^2}\)
N would be what tho?...
number of samples, here it would be = 2
Ahh ok cuz we had 2 samples or whatever right?
for reference of others: http://assets.openstudy.com/updates/attachments/56fe8aabe4b0261a951e1fdf-yanasidlinskiy-1459522239651-1.png
yes. mean = 35*P(35) + (-1)P(-1) = ... ?
use the values we got from part (a)
Yea, that's what I'm doing rn
Okay, I'm not sure if I'm doing this right but I'm just gonna plug in the numbers, cuz I'm not confident about it..... 35*.0263(35)+(-1).9737(-1)
why 35 and -1 are multiplied twice?
Oh my bad lol
ohh.. P(35) is probability of 35, not probability*35
so, mean is 35*(0.0263) + (-1)(0.9737) = ...
So, 1.9?
i get -0.0526...
What?
Desktop calculator... My laptop lol
now start calculating std deviation
Okay, my sample would be the same, right?
yes same
Same formula as well, right?
Like.... Wait. Actually how would I calculate it.
This: \( \\ \sigma = \sqrt{\dfrac{1}{N}\sum (x_i-\mu )^2}\)
N = 2, x1 = 35 x2 = -1 mu = -0.0526
\(\sqrt {\dfrac{1}{2} [(35-(-0.0526))^2+(-1-(0.0526))^2]}\) just substituted the values
24.8?
you should get 24.797 try it http://www.wolframalpha.com/input/?i=%5Csqrt+%7B%5Cdfrac%7B1%7D%7B2%7D+%5B(35-(-0.0526))%5E2%2B(-1-(0.0526))%5E2%5D%7D oh yes, correct!
Gee, Okay, I may need to remember this ahha
the formulas, yes.
No, I know the formulas just messing with the numbers is really insane.
good luck!
Ughgh, yes thank you
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