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

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)