Calculate the mean, variance, and standard deviation of the following discrete probability distribution.(Negative values should be indicated by a minus sign. Round your intermediate calculations to 4 decimal places and final answers to 2 decimal places.)

x -33 -25 -14 -4
P(X = x) 0.5 0.24 0.18 0.08

A.) Mean???? for the question above and B below this is a seperate question!!*****

B.) Assume that X is a binomial random variable with n = 20 and p = 0.87. Calculate the following probabilities.(Round your intermediate calculations and final answers to 4 decimal places.)

Probability
a. P(X = 19)
b. P(X = 18)
c. P(X > 18)

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).

We can calculate the mean with the following formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And if we replace we got:

[tex] E(X) = -33*0.5 - 25*0.24 -14*018 -4*0.08= -25.34[/tex]

Now we can calculate the second moment given by:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we have:

[tex] E(X^2) = (-33)^2*0.5 + (-25)^2*0.24 +(-14)^2*018+ (-4)^2*0.08= 731.06[/tex]

And the variance is given by:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 731.06 -(-25.34)^2 = 88.944[/tex]

And the deviation would be:

[tex] sd(X) = \sqrt{88.944}= 9.431[/tex]

Part B

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

For this case we know that [tex] X \sim Bin (n = 20, p =0.87)[/tex]

a) [tex]P(X=19)=(20C19)(0.87)^{19} (1-0.87)^{20-19}=0.184[/tex]

b) [tex]P(X=18)=(20C18)(0.87)^{18} (1-0.87)^{20-18}=0.262[/tex]

c) [tex] P(X>18) = P(X=19) +P(X=20)[/tex]

[tex]P(X=20)=(20C20)(0.87)^{20} (1-0.87)^{20-20}=0.0617[/tex]

And replacing we have:

[tex] P(X>18) = P(X=19) +P(X=20)= 0.184+0.0617= 0.246[/tex]