1. **Stating the problem:** We have a probability distribution for the number of smartphones $Y$ owned by households in a certain income bracket. We need to find probabilities for various events and also calculate the mean and variance of $Y$. 2. **Given data:** $$ \begin{array}{c|cccccc} Y & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(Y) & 0.009 & 0.376 & 0.371 & 0.167 & 0.061 & 0.016 \end{array} $$ 3. **Formulas:** - Probability of event $A$ is sum of probabilities of outcomes in $A$. - Mean (expected value) $E(Y) = \sum y_i P(Y=y_i)$. - Variance $Var(Y) = E(Y^2) - [E(Y)]^2$ where $E(Y^2) = \sum y_i^2 P(Y=y_i)$. 4. **Calculate requested probabilities:** (a) $P(Y \geq 1) = 1 - P(Y=0) = 1 - 0.009 = 0.991$ (b) $P(Y=2) = 0.371$ (c) $P(1 \leq Y \leq 3) = P(Y=1) + P(Y=2) + P(Y=3) = 0.376 + 0.371 + 0.167 = 0.914$ (d) $P(Y=1 \text{ or } 3 \text{ or } 5) = P(Y=1) + P(Y=3) + P(Y=5) = 0.376 + 0.167 + 0.016 = 0.559$ 5. **Calculate mean $E(Y)$:** $$ E(Y) = 0 \times 0.009 + 1 \times 0.376 + 2 \times 0.371 + 3 \times 0.167 + 4 \times 0.061 + 5 \times 0.016 = 0 + 0.376 + 0.742 + 0.501 + 0.244 + 0.08 = 1.943 $$ 6. **Calculate $E(Y^2)$:** $$ E(Y^2) = 0^2 \times 0.009 + 1^2 \times 0.376 + 2^2 \times 0.371 + 3^2 \times 0.167 + 4^2 \times 0.061 + 5^2 \times 0.016 = 0 + 0.376 + 4 \times 0.371 + 9 \times 0.167 + 16 \times 0.061 + 25 \times 0.016 = 0.376 + 1.484 + 1.503 + 0.976 + 0.4 = 4.739 $$ 7. **Calculate variance:** $$ Var(Y) = E(Y^2) - [E(Y)]^2 = 4.739 - (1.943)^2 = 4.739 - 3.775 = 0.964 $$ **Final answers:** - $P(Y \geq 1) = 0.991$ - $P(Y=2) = 0.371$ - $P(1 \leq Y \leq 3) = 0.914$ - $P(Y=1 \text{ or } 3 \text{ or } 5) = 0.559$ - Mean $= 1.943$ - Variance $= 0.964$