1. **State the problem:** We have a probability distribution with values of $x$ and their probabilities $P(X)$. We need to complete the table, find the mean ($\mu$), variance ($\sigma^2$), and standard deviation ($\sigma$), then interpret the results. 2. **Recall formulas:** - Mean (expected value): $$\mu = \sum x_i P(x_i)$$ - Variance: $$\sigma^2 = \sum (x_i - \mu)^2 P(x_i)$$ - Standard deviation: $$\sigma = \sqrt{\sigma^2}$$ 3. **Given data:** | $x$ | $P(X)$ | |-----|--------| | 1 | 0.2 | | 4 | 0.4 | | 5 | 0.2 | | 7 | 0.2 | 4. **Calculate $X \times P(X)$:** - $1 \times 0.2 = 0.2$ - $4 \times 0.4 = 1.6$ - $5 \times 0.2 = 1.0$ - $7 \times 0.2 = 1.4$ Sum these to find the mean: $$\mu = 0.2 + 1.6 + 1.0 + 1.4 = 4.2$$ 5. **Calculate $X - \mu$ for each $x$:** - $1 - 4.2 = -3.2$ - $4 - 4.2 = -0.2$ - $5 - 4.2 = 0.8$ - $7 - 4.2 = 2.8$ 6. **Calculate $(X - \mu)^2$ for each $x$:** - $(-3.2)^2 = 10.24$ - $(-0.2)^2 = 0.04$ - $(0.8)^2 = 0.64$ - $(2.8)^2 = 7.84$ 7. **Calculate $(X - \mu)^2 \times P(X)$:** - $10.24 \times 0.2 = 2.048$ - $0.04 \times 0.4 = 0.016$ - $0.64 \times 0.2 = 0.128$ - $7.84 \times 0.2 = 1.568$ Sum these to find variance: $$\sigma^2 = 2.048 + 0.016 + 0.128 + 1.568 = 3.76$$ 8. **Calculate standard deviation:** $$\sigma = \sqrt{3.76} \approx 1.94$$ 9. **Interpretation:** - The mean $\mu = 4.2$ represents the expected value of $X$. - The variance $\sigma^2 = 3.76$ measures the spread of the distribution. - The standard deviation $\sigma = 1.94$ shows the average distance of values from the mean. - Since the standard deviation is less than half the mean, the data points are moderately spread around the mean. **Final completed table:** | $x$ | $P(X)$ | $X \times P(X)$ | $X - \mu$ | $(X - \mu)^2$ | $(X - \mu)^2 \times P(X)$ | |-----|--------|-----------------|-----------|---------------|----------------------------| | 1 | 0.2 | 0.2 | -3.2 | 10.24 | 2.048 | | 4 | 0.4 | 1.6 | -0.2 | 0.04 | 0.016 | | 5 | 0.2 | 1.0 | 0.8 | 0.64 | 0.128 | | 7 | 0.2 | 1.4 | 2.8 | 7.84 | 1.568 | $$\mu = 4.2, \quad \sigma^2 = 3.76, \quad \sigma = 1.94$$