1. **State the problem:** We are given a discrete probability distribution of the number of modules John can finish in a day, with values of $x$ and their probabilities $P(x)$. We need to find the mean ($\mu$), variance ($\sigma^2$), and standard deviation ($\sigma$) of this distribution. Then, we will assess if John's claim of finishing 5 modules in a day is believable. 2. **Recall formulas:** - Mean (expected value): $$\mu = \sum x P(x)$$ - Variance: $$\sigma^2 = \sum (x - \mu)^2 P(x)$$ - Standard deviation: $$\sigma = \sqrt{\sigma^2}$$ 3. **Calculate the mean:** Calculate $x \cdot P(x)$ for each $x$: - $1 \times 0.20 = 0.20$ - $2 \times 0.35 = 0.70$ - $3 \times 0.10 = 0.30$ - $4 \times 0.15 = 0.60$ - $5 \times 0.10 = 0.50$ - $6 \times 0.10 = 0.60$ Sum these values: $$\mu = 0.20 + 0.70 + 0.30 + 0.60 + 0.50 + 0.60 = 2.90$$ 4. **Calculate variance:** First, find $x - \mu$ for each $x$: - $1 - 2.90 = -1.90$ - $2 - 2.90 = -0.90$ - $3 - 2.90 = 0.10$ - $4 - 2.90 = 1.10$ - $5 - 2.90 = 2.10$ - $6 - 2.90 = 3.10$ Square these differences: - $(-1.90)^2 = 3.61$ - $(-0.90)^2 = 0.81$ - $(0.10)^2 = 0.01$ - $(1.10)^2 = 1.21$ - $(2.10)^2 = 4.41$ - $(3.10)^2 = 9.61$ Multiply each squared difference by its probability: - $3.61 \times 0.20 = 0.722$ - $0.81 \times 0.35 = 0.2835$ - $0.01 \times 0.10 = 0.001$ - $1.21 \times 0.15 = 0.1815$ - $4.41 \times 0.10 = 0.441$ - $9.61 \times 0.10 = 0.961$ Sum these values to get variance: $$\sigma^2 = 0.722 + 0.2835 + 0.001 + 0.1815 + 0.441 + 0.961 = 2.59$$ 5. **Calculate standard deviation:** $$\sigma = \sqrt{2.59} \approx 1.61$$ 6. **Interpretation:** The mean number of modules John can finish in a day is approximately 2.90, with a standard deviation of about 1.61. John's claim of finishing 5 modules in a day is above the mean by about $5 - 2.90 = 2.10$ modules, which is roughly $\frac{2.10}{1.61} \approx 1.3$ standard deviations above the mean. This is possible but somewhat above average, so it may not be typical but not impossible. **Final answers:** - Mean $\mu = 2.90$ - Variance $\sigma^2 = 2.59$ - Standard deviation $\sigma = 1.61$