1. **State the problem:** We have a probability distribution for the number of customers $X$ who order takeout in a day with probabilities involving $k$. We need to find $k$, then compute the mean and variance of $X$, and interpret the results. 2. **Given table:** $$\begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \ \text{or more} \\ \hline P(X=x) & 0.15 & 3k & 0.25 & 2k & 0.35 + 5k \end{array}$$ 3. **Find $k$:** Since probabilities sum to 1, $$0.15 + 3k + 0.25 + 2k + (0.35 + 5k) = 1$$ Simplify: $$0.15 + 0.25 + 0.35 + 3k + 2k + 5k = 1$$ $$0.75 + 10k = 1$$ Subtract 0.75: $$10k = 1 - 0.75 = 0.25$$ Divide both sides by 10: $$k = \frac{0.25}{10} = 0.025$$ 4. **Compute mean (expected value) $E(X)$:** Formula: $$E(X) = \sum x P(X=x)$$ Calculate each term: $$0 \times 0.15 = 0$$ $$1 \times 3k = 3 \times 0.025 = 0.075$$ $$2 \times 0.25 = 0.5$$ $$3 \times 2k = 3 \times 2 \times 0.025 = 0.15$$ $$4 \times (0.35 + 5k) = 4 \times (0.35 + 5 \times 0.025) = 4 \times (0.35 + 0.125) = 4 \times 0.475 = 1.9$$ Sum: $$E(X) = 0 + 0.075 + 0.5 + 0.15 + 1.9 = 2.625$$ 5. **Compute variance $Var(X)$:** Formula: $$Var(X) = E(X^2) - [E(X)]^2$$ Calculate $E(X^2)$: $$0^2 \times 0.15 = 0$$ $$1^2 \times 3k = 3 \times 0.025 = 0.075$$ $$2^2 \times 0.25 = 4 \times 0.25 = 1$$ $$3^2 \times 2k = 9 \times 2 \times 0.025 = 0.45$$ $$4^2 \times (0.35 + 5k) = 16 \times 0.475 = 7.6$$ Sum: $$E(X^2) = 0 + 0.075 + 1 + 0.45 + 7.6 = 9.125$$ Calculate variance: $$Var(X) = 9.125 - (2.625)^2 = 9.125 - 6.890625 = 2.234375$$ 6. **Interpretation:** - The mean $2.625$ means on average about 2 to 3 customers order takeout daily. - The variance $2.234$ shows the spread of the number of customers around the mean; a moderate variability in daily takeout orders.