Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1, respectively, that 0, 1, 2, or 3 power failures will strike a certain subdivision in any given year. Find the mean and variance of the random variable X representing the number of power failures striking this subdivision.

mean = 1 power failure

variance = 1 (power failure) ²

Step-by-step explanation:

Since the mean is computed as

mean = E (x) = ∑ x * p (x) for all x

then for the random variable x=power failures, we have

mean = ∑ x * p (x) = 0 * 0.4 + 1 * 0.3 + 2*0.2 + 3 * 0.1 = 1 power failure

since the variance can be calculated through

variance = ∑[x-E (x) ]² * p (x) for all x

but easily in this way

variance = E (x²) - [E (x) ]², then

E (x²) = ∑ x² * p (x) = 0² * 0.4 + 1² * 0.3 + 2²*0.2 + 3² * 0.1 = 2 power failure²

then

variance = 2 power failure² - (1 power failure) ² = 1 power failure²

therefore

mean = 1 power failure

variance = 1 power failure²