1. **State the problem:** We have data representing the number of cars in households of classmates. We need to construct the frequency and probability distribution for the variable $x$ (number of cars), and then find the mean, variance, and standard deviation. 2. **Organize the data:** The data points are: 1, 5, 5, 1, 5, 6, 1, 3, 0, 6, 0, 6, 0, 3, 6, 0, 5, 6, 0, 0, 4, 0, 1, 5, 6, 5, 0, 6, 1, 0. 3. **Frequency distribution:** Count how many times each value of $x$ appears. - $x=0$: 8 times - $x=1$: 4 times - $x=3$: 2 times - $x=4$: 1 time - $x=5$: 5 times - $x=6$: 8 times 4. **Probability distribution:** Probability $P(x) = \frac{\text{frequency of } x}{\text{total data points}}$. Total data points $n=30$. - $P(0) = \frac{8}{30} = 0.267$ - $P(1) = \frac{4}{30} = 0.133$ - $P(3) = \frac{2}{30} = 0.067$ - $P(4) = \frac{1}{30} = 0.033$ - $P(5) = \frac{5}{30} = 0.167$ - $P(6) = \frac{8}{30} = 0.267$ 5. **Mean calculation:** Use formula $$\mu = \sum x_i P(x_i)$$ $$\mu = 0\times0.267 + 1\times0.133 + 3\times0.067 + 4\times0.033 + 5\times0.167 + 6\times0.267$$ $$= 0 + 0.133 + 0.201 + 0.133 + 0.833 + 1.602 = 2.902$$ 6. **Variance calculation:** Use formula $$\sigma^2 = \sum (x_i - \mu)^2 P(x_i)$$ Calculate each term: - $(0 - 2.902)^2 \times 0.267 = 7.02 \times 0.267 = 1.874$ - $(1 - 2.902)^2 \times 0.133 = 3.62 \times 0.133 = 0.481$ - $(3 - 2.902)^2 \times 0.067 = 0.01 \times 0.067 = 0.001$ - $(4 - 2.902)^2 \times 0.033 = 1.21 \times 0.033 = 0.040$ - $(5 - 2.902)^2 \times 0.167 = 4.41 \times 0.167 = 0.736$ - $(6 - 2.902)^2 \times 0.267 = 9.59 \times 0.267 = 2.561$ Sum these: $$\sigma^2 = 1.874 + 0.481 + 0.001 + 0.040 + 0.736 + 2.561 = 5.693$$ 7. **Standard deviation:** $$\sigma = \sqrt{\sigma^2} = \sqrt{5.693} = 2.386$$ **Final answers:** - Mean $\mu = 2.902$ - Variance $\sigma^2 = 5.693$ - Standard deviation $\sigma = 2.386$