Question 1206570: Use the probability distribution for the random variable x to answer the question.
x 0 1 2 3 4 5
p(x)
0.25 0.05 0.15 0.2 0.05 0.3
Find 𝜇,
𝜎2,
and 𝜎. (Round your standard deviation to two decimal places.)
𝜇 =
𝜎2 =
𝜎 =
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
𝜇 = mu
𝜎 = sigma
Given table
| x | 0 | 1 | 2 | 3 | 4 | 5 | | p(x) | 0.25 | 0.05 | 0.15 | 0.2 | 0.05 | 0.3 |
Introduce a new row called x*p(x), where you multiply the paired x and p(x) values.
Spreadsheet software is recommended.
| x | 0 | 1 | 2 | 3 | 4 | 5 | | p(x) | 0.25 | 0.05 | 0.15 | 0.2 | 0.05 | 0.3 | | x*p(x) | 0 | 0.05 | 0.3 | 0.6 | 0.2 | 1.5 |
Add up the x*p(x) values to get the expected value.
mu = mean = expected value = E[X]
mu = sum of the x*p(x) values
mu = 0+0.05+0.3+0.6+0.2+1.5
mu = 2.65
We'll use that value of mu to determine the variance, and by extension, the standard deviation as well.
Introduce a new row called (x-mu)^2*p(x)
The naming should be self-explanatory. If not then please let me know.
Example calculation: if x = 0, then (x-mu)^2*p(x) = (0-2.65)^2*0.25 = 1.755625
| x | 0 | 1 | 2 | 3 | 4 | 5 | | p(x) | 0.25 | 0.05 | 0.15 | 0.2 | 0.05 | 0.3 | | x*p(x) | 0 | 0.05 | 0.3 | 0.6 | 0.2 | 1.5 | | (x-mu)^2*p(x) | 1.755625 | 0.136125 | 0.063375 | 0.0245 | 0.091125 | 1.65675 |
sigma^2 = variance
sigma^2 = sum of the (x-mu)^2*p(x) values
sigma^2 = 1.755625+0.136125+0.063375+0.0245+0.091125+1.65675
sigma^2 = 3.7275
Side note: another way to find the variance is to compute E[X^2] - (E[X])^2 aka E[X^2] - mu^2
I'll leave this as an exercise to the reader.
Then,
sigma = standard deviation
sigma = sqrt( variance )
sigma = sqrt( 3.7275 )
sigma = 1.9306735 approximately
When rounding to 2 decimal places we get 1.93
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Answers:
mu = 2.65
sigma^2 = 3.7275
sigma = 1.93
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