SOLUTION: A random variable x has the following probability mass function defined in tabular form as shown below x values are -1,1,2 and p(x=x) 2c,3c ,4c Find the value of c and computer p(

Algebra ->  Finance -> SOLUTION: A random variable x has the following probability mass function defined in tabular form as shown below x values are -1,1,2 and p(x=x) 2c,3c ,4c Find the value of c and computer p(      Log On


   

Question 1202695: A random variable x has the following probability mass function defined in tabular form as shown below x values are -1,1,2 and p(x=x) 2c,3c ,4c
Find the value of c and computer p(x=-1),find the standard deviation of x
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

xP(x)
-12c
13c
24c


To have a valid probability distribution, the p(x) values must add to 1.
2c+3c+4c = 1
9c = 1
c = 1/9

So,
2c = 2*(1/9) = 2/9
3c = 3*(1/9) = 3/9 = 1/3
4c = 4*(1/9) = 4/9

We get this updated table.
xP(x)
-12/9
11/3
24/9

We see that P(x=-1) = 2/9
2/9 = 0.222 approximately

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This video goes over an example how to calculate the standard deviation of a probability distribution.
https://www.youtube.com/watch?v=YMbt5rYzp-Q

The professor mentions these two formulas
sigma+=+sqrt%28+sum%28%28x-mu%29%5E2%2AP%28x%29%2C%22%22%2C%22%22%29+%29
and
sigma+=+sqrt%28+%28sum%28x%5E2%2AP%28x%29%2C%22%22%2C%22%22%29%29+-+mu%5E2+%29

where,
mu = mu = greek letter to represent the mean
lowercase sigma = sigma = greek letter for the standard deviation

The fancy looking "E" represents a summation. It's the greek uppercase letter sigma.
Something like sum%28x%5E2%2AP%28x%29%29 tells us to add up all of the x^2*P(x) terms.

Both formulas involve mu, so let's determine that value first.

mu = sum%28x%2AP%28x%29%5E%22%22%29
mu = add up the x*P(x) terms
xP(x)x*P(x)
-12/9-2/9
11/31/3
24/98/9

The sum of the x*P(x) values is:
-2/9 + 1/3 + 8/9
= -2/9 + 3/9 + 8/9
= (-2 + 3 + 8)/9
= 9/9
= 1

Therefore, mu = 1.


Let's say we used the formula
sigma+=+sqrt%28+sum%28%28x-mu%29%5E2%2AP%28x%29%2C%22%22%2C%22%22%29+%29
What we'll need to do is
  1. Subtract mu from each x value to get x-mu.
  2. Square the difference to get (x-mu)^2
  3. Multiply those squares with P(x) to get (x-mu)^2*P(x).
This is what your table should look like
xP(x)x-mu(x-mu)^2(x-mu)^2*P(x)
-12/9-248/9
11/3000
24/9114/9

Add up the values in the (x-mu)^2*P(x) column.
8/9 + 0 + 4/9 = 12/9 = 4/3
That's the variance, so the standard deviation is the square root of that
sqrt%284%2F3%29+=+sqrt%284%29%2Fsqrt%283%29

sqrt%284%2F3%29+=+2%2Fsqrt%283%29

sqrt%284%2F3%29+=+%282%2Asqrt%283%29%29%2F%28sqrt%283%29%2Asqrt%283%29%29

sqrt%284%2F3%29+=+%282%2Asqrt%283%29%29%2F%283%29

sqrt%284%2F3%29+=+%282%2F3%29%2Asqrt%283%29
The standard deviation is exactly (2/3)*sqrt(3)
When using a calculator, (2/3)*sqrt(3) = 1.15470053837926 approximately.

If you wanted to use the formula sigma+=+sqrt%28+%28sum%28x%5E2%2AP%28x%29%2C%22%22%2C%22%22%29%29+-+mu%5E2+%29, then this is what your table could look like
xP(x)x*P(x)x^2x^2*P(x)
-12/9-2/912/9
11/31/311/3
24/98/9416/9

Add up the x^2*P(x) values
sum%28x%5E2%2AP%28x%29%29+=+2%2F9+%2B+1%2F3+%2B+16%2F9

sum%28x%5E2%2AP%28x%29%29+=+2%2F9+%2B+3%2F9+%2B+16%2F9

sum%28x%5E2%2AP%28x%29%29+=+21%2F9

sum%28x%5E2%2AP%28x%29%29+=+7%2F3

Then plug that result into the formula I mentioned earlier
sigma+=+sqrt%28+%28sum%28x%5E2%2AP%28x%29%2C%22%22%2C%22%22%29%29+-+mu%5E2+%29

sigma+=+sqrt%28+7%2F3+-+1%5E2+%29

sigma+=+sqrt%28+7%2F3+-+1+%29

sigma+=+sqrt%28+7%2F3+-+3%2F3+%29

sigma+=+sqrt%28+4%2F3+%29

sigma+=+2%2Fsqrt%283%29

sigma+=+%282%2Asqrt%283%29%29%2F%283%29

sigma+=+%282%2F3%29%2Asqrt%283%29

This calculator can be used to check your work
https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php

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Summary:

c = 1/9
P(x = -1) = 2/9
Standard deviation = (2/3)*sqrt(3) = 1.15470053837926 approximately