Question 1206814
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Answers:
mean = <font color=red>2.3</font>
standard deviation = <font color=red>1.04</font>


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Work Shown


<table border = "1" cellpadding = "5"><tr><td>x</td><td>P(x)</td></tr><tr><td>0</td><td>0.1</td></tr><tr><td>1</td><td>0.15</td></tr><tr><td>2</td><td>0.15</td></tr><tr><td>3</td><td>0.6</td></tr></table>


Multiply the x and P(x) values to form a new column.
Spreadsheet software is recommended.
<table border = "1" cellpadding = "5"><tr><td>x</td><td>P(x)</td><td>x*P(x)</td></tr><tr><td>0</td><td>0.1</td><td>0</td></tr><tr><td>1</td><td>0.15</td><td>0.15</td></tr><tr><td>2</td><td>0.15</td><td>0.3</td></tr><tr><td>3</td><td>0.6</td><td>1.8</td></tr></table>
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 <font color=red>2.3</font> 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.
<table border = "1" cellpadding = "5"><tr><td>x</td><td>P(x)</td><td>x*P(x)</td><td>(x-mu)^2*P(x)</td></tr><tr><td>0</td><td>0.1</td><td>0</td><td>0.50625</td></tr><tr><td>1</td><td>0.15</td><td>0.15</td><td>0.234375</td></tr><tr><td>2</td><td>0.15</td><td>0.3</td><td>0.009375</td></tr><tr><td>3</td><td>0.6</td><td>1.8</td><td>0.3375</td></tr></table>
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 = <font color=red>1.04</font> 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)
<table border = "1" cellpadding = "5"><tr><td>x</td><td>P(x)</td><td>x*P(x)</td><td>x^2*P(x)</td></tr><tr><td>0</td><td>0.1</td><td>0</td><td>0</td></tr><tr><td>1</td><td>0.15</td><td>0.15</td><td>0.15</td></tr><tr><td>2</td><td>0.15</td><td>0.3</td><td>0.6</td></tr><tr><td>3</td><td>0.6</td><td>1.8</td><td>5.4</td></tr></table>
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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