Question

The renowned soccer player, Levi Gupta scores a goal on 30% of his attempts. The random variable X is defined as the number of goals scored on 50 attempts. The renowned gambler, Mohammed Smith, wins at Blackjack 25% of the time. The random variable Y is defined as the number of games needed to win his first game. Define the random variable Z as the total number of soccer goals scored and blackjack games played. Determine the mean and standard deviation of the random variable Z.

Answer 1

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
`(1)/(4)`.

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) =
`√(np(1-p))`.

=
`√((50)(0.3)(1-0.3))`.

=
`√((50)(0.3)(0.7))`.

SD (X) =
`√(10.5)`.

For random variable Y,

mean (Y) =
`(1)/(p)`. (probability at first time)

=
`(1)/((1)/(4) )`.

mean (Y) = 4.

SD (Y) =
`\sqrt{((1-p))/(p^2) }`.

=
`\sqrt{((1-(1)/(4) ))/(((1)/(4)) ^2) }`.

=
`\sqrt{((3)/(4) )/((1)/(16) ) }`.

=
`\sqrt{((3)/(4) )((16)/(1) )}`.

SD (Y) =
`√(12)`.

mean (Z) = 15+4.

mean (Z)= 19.

SD (Z) =
`√(12) +√(10.5)`

=
`√(22.5)`.

SD (Z) =4.74.