1. **Problem Statement:** We are given a probability distribution for the number of defective bulbs $X$ produced in a day: $$ \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & 0.10 & 0.20 & 0.30 & 0.25 & 0.15 \end{array} $$ We need to find: - (a) The expected value (mean) and variance of $X$. - (b) The expected number of defective bulbs if 1,000 bulbs are produced. 2. **Formulas and Important Rules:** - The expected value (mean) of a discrete random variable $X$ is: $$E(X) = \sum x_i P(x_i)$$ - The variance of $X$ is: $$Var(X) = E(X^2) - [E(X)]^2$$ where $$E(X^2) = \sum x_i^2 P(x_i)$$ 3. **Calculate the Expected Value $E(X)$:** $$E(X) = 0 \times 0.10 + 1 \times 0.20 + 2 \times 0.30 + 3 \times 0.25 + 4 \times 0.15$$ $$= 0 + 0.20 + 0.60 + 0.75 + 0.60 = 2.15$$ 4. **Calculate $E(X^2)$:** $$E(X^2) = 0^2 \times 0.10 + 1^2 \times 0.20 + 2^2 \times 0.30 + 3^2 \times 0.25 + 4^2 \times 0.15$$ $$= 0 + 0.20 + 4 \times 0.30 + 9 \times 0.25 + 16 \times 0.15$$ $$= 0 + 0.20 + 1.20 + 2.25 + 2.40 = 6.05$$ 5. **Calculate the Variance $Var(X)$:** $$Var(X) = E(X^2) - [E(X)]^2 = 6.05 - (2.15)^2 = 6.05 - 4.6225 = 1.4275$$ 6. **Interpretation:** - The expected number of defective bulbs per day is $2.15$. - The variance, which measures the spread of defective bulbs around the mean, is approximately $1.43$. 7. **Part (b) - Expected Defective Bulbs in 1,000 bulbs:** Since the expected defective bulbs per bulb is $2.15$ defective bulbs per day (assuming the distribution is per day for the total production), if the factory produces 1,000 bulbs, the expected number of defective bulbs is: $$1000 \times \frac{2.15}{\text{total bulbs produced per day}}$$ However, the problem states the distribution is for the number of defective bulbs produced in a day, so the expected defective bulbs in a day is already $2.15$ regardless of total bulbs produced. If the question means the factory produces 1,000 bulbs and the distribution is per bulb, then: $$\text{Expected defective bulbs} = 1000 \times 2.15 = 2150$$ But this is unrealistic since $X$ is number of defective bulbs per day, not per bulb. Assuming the distribution is for the whole day's production, the expected defective bulbs in 1,000 bulbs is $2.15$. **Final answers:** - Expected value (mean) $E(X) = 2.15$ - Variance $Var(X) = 1.4275$ - Expected defective bulbs in 1,000 bulbs produced in a day = $2.15$