A table gives the probability of 0, 1, 2, 3, 4 and 5 daily absences as 0.20, 0.18, 0.10, 0.25, 0.25 and 0.02. Find the mean and variance of the number of daily absences, and the probability that more than two people are absent on a given day.

Step-by-step solution

1. 1

   Validate the table first. A probability distribution must have probabilities summing to $1$:

   $0.20 + 0.18 + 0.10 + 0.25 + 0.25 + 0.02 = 1.00 \;\checkmark$

   If this had failed, every later number would be meaningless, so it is worth the ten seconds.

2. 2

   Compute the mean as $\sum x P(x)$.

   $\mu = 0(0.20) + 1(0.18) + 2(0.10) + 3(0.25) + 4(0.25) + 5(0.02)$

   $= 0 + 0.18 + 0.20 + 0.75 + 1.00 + 0.10 = 2.23$

   So on an average day $2.23$ employees are absent.

3. 3

   Compute $E(X^2)$, then use the shortcut for the variance. Squaring each $x$ before weighting:

   $E(X^2) = 0(0.20) + 1(0.18) + 4(0.10) + 9(0.25) + 16(0.25) + 25(0.02)$

   $= 0 + 0.18 + 0.40 + 2.25 + 4.00 + 0.50 = 7.33$

4. 4

   Subtract the square of the mean. The computational formula $\sigma^2 = E(X^2) - \mu^2$ avoids six separate squared deviations:

   $\sigma^2 = 7.33 - (2.23)^2 = 7.33 - 4.9729 = 2.3571$

   $\sigma = \sqrt{2.3571} = 1.5353$

   Note the order: square the mean, do not take the mean of the squares as the variance.

5. 5

   (b) Add the probabilities above 2. "More than two" means $x = 3, 4, 5$ — it excludes $x = 2$:

   $P(X > 2) = 0.25 + 0.25 + 0.02 = 0.52$

6. 6

   Cross-check the variance the long way. Summing $(x-\mu)^2 P(x)$: $(0-2.23)^2(0.20) = 0.99458$, $(1-2.23)^2(0.18) = 0.27222$, $(2-2.23)^2(0.10) = 0.00529$, $(3-2.23)^2(0.25) = 0.148225$, $(4-2.23)^2(0.25) = 0.783225$, $(5-2.23)^2(0.02) = 0.153458$. Total $= 2.3571$ ✓, matching the shortcut exactly. Also $P(X \le 2) = 0.20+0.18+0.10 = 0.48$ and $0.48 + 0.52 = 1$ ✓.