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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.

Respuesta :

The Mean and Standard Deviation for random variable Z is 19 and 4.74.

Step-by-step 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 [tex]\frac{1}{4}[/tex].

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) = [tex]\sqrt{np(1-p)}[/tex].

=[tex]\sqrt{(50)(0.3)(1-0.3)}[/tex].

=[tex]\sqrt{(50)(0.3)(0.7)}[/tex].

SD (X) = [tex]\sqrt{10.5}[/tex].

For random variable Y,

mean (Y) = [tex]\frac{1}{p}[/tex].          (probability at first time)

=[tex]\frac{1}{\frac{1}{4} }[/tex].

mean (Y) = 4.

SD (Y) = [tex]\sqrt{\frac{(1-p)}{p^2} }[/tex].

=[tex]\sqrt{\frac{(1-\frac{1}{4} )}{(\frac{1}{4}) ^2} }[/tex].

= [tex]\sqrt{\frac{\frac{3}{4} }{\frac{1}{16} } }[/tex].

=[tex]\sqrt{(\frac{3}{4} )(\frac{16}{1} )}[/tex].

SD (Y) =[tex]\sqrt{12}[/tex].

mean (Z) = 15+4.

mean (Z)= 19.

SD (Z) = [tex]\sqrt{12} +\sqrt{10.5}[/tex]

=[tex]\sqrt{22.5}[/tex].

SD (Z) =4.74.

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