A scratch-card costs 5¤ and has two possible prizes. There is a 10% chance you get 5¤ back
and
a 1% chance you get 100¤. In all other cases, you get nothing. Let X denote your net winnings,
i.e, (X = prize - cost). What are the expectation and standard deviation of X?
To calculate the expectation and standard deviation of X, we will use the concept of expected value.
First, let's calculate the expected value of X, denoted as E[X]. The expected value can be found by multiplying each possible outcome by its probability and summing them up.
Let's break down the possibilities:
- There is a 10% chance of winning 5¤, which means the net winnings would be 5¤ - 5¤ (cost) = 0¤.
- There is a 1% chance of winning 100¤, which means the net winnings would be 100¤ - 5¤ (cost) = 95¤.
- In all other cases, the net winnings would be 0 - 5¤ = -5¤.
Now we can calculate the expected value:
E[X] = (0¤ * 0.10) + (95¤ * 0.01) + (-5¤ * 0.89)
E[X] = 0¤ + 0.95¤ - 4.45¤
E[X] = -3.50¤
So, the expected value of X is -3.50¤.
Next, let's calculate the standard deviation of X, denoted as σ[X]. The standard deviation measures the dispersion of the possible outcomes around the expected value.
To do this, we need to calculate the variance first. The variance is the average squared difference between each possible outcome and the expected value. Then, we take the square root of the variance to find the standard deviation.
The variance can be calculated using the formula:
Var(X) = Σ [(X - E(X))^2 * P(X)]
Using the same possibilities as before:
Var(X) = [(0¤ - (-3.50¤))^2 * 0.10] + [(95¤ - (-3.50¤))^2 * 0.01] + [(-5¤ - (-3.50¤))^2 * 0.89]
Var(X) = (3.50¤^2 * 0.10) + (98.50¤^2 * 0.01) + (1.50¤^2 * 0.89)
Var(X) = 12.25¤ * 0.10 + 9702.25¤ * 0.01 + 2.25¤ * 0.89
Var(X) = 1.225¤ + 97.0225¤ + 2.0025¤
Var(X) = 100.25¤
Now, we can find the standard deviation by taking the square root of the variance:
σ[X] = √(Var(X))
σ[X] = √(100.25¤)
σ[X] = 10.02¤
So, the standard deviation of X is 10.02¤.
In summary:
- The expectation of X (E[X]) is -3.50¤.
- The standard deviation of X (σ[X]) is 10.02¤.