Answer:
![E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%201%2A0.52%2B%202%2A0.25%20%2B3%2A0.23%3D%201.71)
Now we can calculate the second moment with the following formula:
And replacing we got:
![E(X^2) = 1^2*0.52+ 2^2*0.25 +3^2*0.23= 3.59](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%201%5E2%2A0.52%2B%202%5E2%2A0.25%20%2B3%5E2%2A0.23%3D%203.59)
And the variance is given by:
![Var(X) = E(X^2) -E(X)](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-E%28X%29)
And replacing we got:
![Var(X) = 3.59 -[1.71]^2 = 0.6659](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%203.59%20-%5B1.71%5D%5E2%20%3D%200.6659)
And the standard deviation is just the square root of the variance:
![Sd(X) = \sqrt{0.6659}= 0.816](https://tex.z-dn.net/?f=Sd%28X%29%20%3D%20%5Csqrt%7B0.6659%7D%3D%200.816)
Step-by-step explanation:
Previous concepts
For this case we define the random variable X =" how many children the couple will have" and we know the following distribution:
X 1 2 3
P(X) 0.52 0.250 0.230
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we can find the expected value with the following formula:
![E(X) = \sum_{i=1}^n X_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20X_i%20P%28X_i%29)
And replacing we got:
![E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%201%2A0.52%2B%202%2A0.25%20%2B3%2A0.23%3D%201.71)
Now we can calculate the second moment with the following formula:
And replacing we got:
![E(X^2) = 1^2*0.52+ 2^2*0.25 +3^2*0.23= 3.59](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%201%5E2%2A0.52%2B%202%5E2%2A0.25%20%2B3%5E2%2A0.23%3D%203.59)
And the variance is given by:
![Var(X) = E(X^2) -E(X)](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-E%28X%29)
And replacing we got:
![Var(X) = 3.59 -[1.71]^2 = 0.6659](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%203.59%20-%5B1.71%5D%5E2%20%3D%200.6659)
And the standard deviation is just the square root of the variance:
![Sd(X) = \sqrt{0.6659}= 0.816](https://tex.z-dn.net/?f=Sd%28X%29%20%3D%20%5Csqrt%7B0.6659%7D%3D%200.816)