Respuesta :
Answer:
a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.
b) 0.05 = 5% probability that he has four goals and two assists in one game
Step-by-step explanation:
In hockey, a point is counted for each goal or assist of the player.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval. The standard deviation is the square root of the mean.
(a) Find the mean and standard deviation for the total revenue he earns per game.
60% of six are goals, which means that 60% of the time he earned 3K.
40% of six are goals, which means that 40% of the time he earned 1K.
The mean is:
[tex]\mu = 6*0.6*3 + 6*0.4*1 = 13.2[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63[/tex]
The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.
(b) What is the probability that he has four goals and two assists in one game
Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.
Probability of four goals:
60% of 6 are goals, which means that:
[tex]\mu = 6*0.6 = 3.6[/tex]
The probability of scoring four goals is:
[tex]P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122[/tex]
Probability of two assists:
40% of 2 are assists, which means that:
[tex]\mu = 6*0.4 = 2.4[/tex]
The probability of getting two assists is:
[tex]P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127[/tex]
Probability of four goals and two assists:
[tex]P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05[/tex]
0.05 = 5% probability that he has four goals and two assists in one game
