Answers
Answer
we have the following data
| x | P(x) |
| 8 | 0.1 |
| 90 | 0.2 |
| 125 | 0.5 |
| 180 | 0.1 |
| 350 | 0.1 |
First, we need to find the mean
we know mean = 
setting the values, we get
Mean= 
Now, the formula for standard deviation is given as
![sigma= sqrt{[sum (x^2*P(x))] - mu^2}](http://img.homeworklib.com/questions/dfff5820-71c9-11ea-abe6-4be145ee76e3.png?x-oss-process=image/resize,w_560)
setting the given values, we get
![νΊδ2 * 0.1) + (902 * 0.2) + (125? * 0.5) + (1802 * 0.1) (3502 * 0.1)]-(134.3)2 σ](http://img.homeworklib.com/questions/e0602ff0-71c9-11ea-a74e-dbf6c57f6eb6.png?x-oss-process=image/resize,w_560)
on solving, we get
So, the required standard deviation for this salary distribution is approximately is 83.02 (rounded to 2 decimals)
r P(z)
μ (8 * 0.1) + (90 * 0.2) + (125 * 0.5) + (180 * 0.1) + (350 * 0.1) = 134.3
We were unable to transcribe this image
νΊδ2 * 0.1) + (902 * 0.2) + (125? * 0.5) + (1802 * 0.1) (3502 * 0.1)]-(134.3)2 σ
σ = V/24928,9| _ 18036.49 = V689241-83.02
.
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