Question

A couple plans to have children until they get a​ girl, but they agree they will not have more than three​ children, even if all are boys. Assume that the probability of having a girl is 0.52. Let X be a random variable indicating how many children the couple will have. Find the standard deviation of the random variable X. x 1 2 3 ​P(Xequals​x) 0.520 0.250 0.230

Answer 1

Answer:

$E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71$

Now we can calculate the second moment with the following formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 1^2*0.52+ 2^2*0.25 +3^2*0.23= 3.59$

And the variance is given by:

$Var(X) = E(X^2) -E(X)$

And replacing we got:

$Var(X) = 3.59 -[1.71]^2 = 0.6659$

And the standard deviation is just the square root of the variance:

$Sd(X) = \sqrt{0.6659}= 0.816$

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:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71$

Now we can calculate the second moment with the following formula:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 1^2*0.52+ 2^2*0.25 +3^2*0.23= 3.59$

And the variance is given by:

$Var(X) = E(X^2) -E(X)$

And replacing we got:

$Var(X) = 3.59 -[1.71]^2 = 0.6659$

And the standard deviation is just the square root of the variance:

$Sd(X) = \sqrt{0.6659}= 0.816$