The Mean and Standard Deviation for random variable Z is 19 and 4.74.
Explanation:
The Renowned Soccer player scores 30% goals of 50 attempts. Its random variable is X.
P(X) = 0.3.
n(X) = 50.
The Renowned gambler wins 25% and its random variable is Y.
P(Y) = 0.25 or
.
A random variable is Z is the total of random variables X and Y.
Thus the mean and Standard deviation of random variable Z,
mean (Z) = mean (X) + mean (Y).
SD (Z) = SD(X) + SD(Y).
Let us find the Mean and SD of X and Y.
For random variable X,
mean (X) = np.
=50(0.30).
mean (X) = 15.
SD (X) =
.
=
.
=
.
SD (X) =
.
For random variable Y,
mean (Y) =
. (probability at first time)
=
.
mean (Y) = 4.
SD (Y) =
.
=
.
=
.
=
.
SD (Y) =
.
mean (Z) = 15+4.
mean (Z)= 19.
SD (Z) =

=
.
SD (Z) =4.74.