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Answer:
Expected gain $25
Standard deviation of the gain $134.63
Step-by-step explanation:
Let C, T the events
C = You have to cancel the flight
T = You can take the flight
P(C) = 30% = 0.3 with a “gain” of -$150
P(T) = 70% = 07 with a gain of $100
Expected gain E
E = -150P(C) + 100P(T) = -150*0.3 + 100*0.7 = -45 + 70 =$25
Standard deviation of your gain
[tex]\large\bf \sqrt{\frac{(-150-E)^2+(100-E)^2}{2}}=\sqrt{\frac{(-150-25)^2+(100-25)^2}{2}}=\\\\=\sqrt{18125}=$134.63[/tex]
Using the discrete distribution, it is found that:
1. The expected gain is of 25 dollars.
2. The standard deviation is of 114.56 dollars.
What are the mean and the standard deviation of a discrete distribution?
- The mean is given by the sum of each outcome multiplied by it's probability.
- The standard deviation is given by square root of the sum of the difference squared of each outcome and the mean, multiplied by it's respective probability.
In this problem, considering the scenarios, the distribution is:
P(X = -150) = 0.3
P(X = 100) = 0.7
Hence:
[tex]E(X) = 0.3(-150) + 0.7(100) = 25[/tex]
[tex]\sqrt{V(X)} = \sqrt{0.3(-150-25)^2 + 0.7(100-25)^2} = 114.56[/tex]
1. The expected gain is of 25 dollars.
2. The standard deviation is of 114.56 dollars.
More can be learned about the mean and the standard deviation of a discrete distribution at https://brainly.com/question/24855677