SOLUTION: x P(x) 0 0.1 1 0.15 2 0.15 3 0.6 Find the mean of this probability distribution. Round your answer to one decimal place. Find the standard deviation of this probabil

Question 1206814: x P(x)
0 0.1
1 0.15
2 0.15
3 0.6
Find the mean of this probability distribution. Round your answer to one decimal place.

Find the standard deviation of this probability distribution. Give your answer to 2 decimal places.

Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

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Find the mean of this probability distribution. Round your answer to one decimal place
to find the mean (expectation) of the given distribution, we will use the following formula:

Find the standard deviation of this probability distribution. Give your answer to decimal places.

since , and

we have

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Answers:
mean = 2.3
standard deviation = 1.04

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Work Shown

x P(x)
0 0.1
1 0.15
2 0.15
3 0.6

Multiply the x and P(x) values to form a new column.
Spreadsheet software is recommended.

x P(x) x*P(x)
0 0.1 0
1 0.15 0.15
2 0.15 0.3
3 0.6 1.8

Add up the values in that new column to get the mean aka expected value.
The spreadsheet command called "SUM" is useful for adding up numbers quickly.
mu = 0 + 0.15 + 0.3 + 1.8 = 2.25
This rounds to 2.3 when rounding to one decimal place.

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mu = 2.25 = mean

Introduce a new column called (x-mu)^2*P(x)
The naming of this column should be self-explanatory. If not then please let me know.

x P(x) x*P(x) (x-mu)^2*P(x)
0 0.1 0 0.50625
1 0.15 0.15 0.234375
2 0.15 0.3 0.009375
3 0.6 1.8 0.3375

variance = sum of the (X-mu)^2*P(X) values
variance = 0.50625+0.234375+0.009375+0.3375
variance = 1.0875

Then,
SD = standard deviation
SD = sqrt( variance )
SD = sqrt( 1.0875 )
SD = 1.0428327 approximately
SD = 1.04 when rounding to 2 decimal places.

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Another way to find the variance is to introduce a column labeled x^2*P(x)

x P(x) x*P(x) x^2*P(x)
0 0.1 0 0
1 0.15 0.15 0.15
2 0.15 0.3 0.6
3 0.6 1.8 5.4

Adding up all of the x^2*P(x) values will get us
0 + 0.15 + 0.6 + 5.4 = 6.15
Subtract off mu^2 = (2.25)^2 = 5.0625 and you'll get:
6.15 - 5.0625 = 1.0875 which is the variance we calculated in the previous section above.

So there are two versions we could use
variance = Sum of (x-mu)^2*P(x) values
or
variance = [Sum of x^2*P(x) values] - mu^2

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x	P(x)	x*P(x)	(x-mu)^2*P(x)
0	0.1	0	0.50625
1	0.15	0.15	0.234375
2	0.15	0.3	0.009375
3	0.6	1.8	0.3375