1. **Problem statement:** We have a binomial distribution with parameters $n=30$ and $p=0.4$. We need to find: a) The expected value $E(X)$ and standard deviation $\sigma$. b) The intervals $[\mu - \sigma, \mu + \sigma]$ and $[\mu - 2\sigma, \mu + 2\sigma]$ for the number of successes. c) Estimate probabilities for these intervals from the graph and calculate exact probabilities to compare. 2. **Formulas and rules:** - Expected value for binomial: $$E(X) = np$$ - Variance: $$Var(X) = np(1-p)$$ - Standard deviation: $$\sigma = \sqrt{Var(X)} = \sqrt{np(1-p)}$$ 3. **Calculate expected value and standard deviation:** $$E(X) = 30 \times 0.4 = 12$$ $$\sigma = \sqrt{30 \times 0.4 \times 0.6} = \sqrt{7.2} \approx 2.683$$ 4. **Calculate intervals:** - One standard deviation interval: $$[\mu - \sigma, \mu + \sigma] = [12 - 2.683, 12 + 2.683] = [9.317, 14.683]$$ Rounded to integers for number of successes: $[9, 15]$ - Two standard deviations interval: $$[\mu - 2\sigma, \mu + 2\sigma] = [12 - 2 \times 2.683, 12 + 2 \times 2.683] = [6.634, 17.366]$$ Rounded to integers: $[7, 17]$ 5. **Estimate probabilities from the graph:** - From the histogram, the bars between $k=9$ and $k=15$ cover a large portion of the total area, roughly about 68% (typical for one standard deviation in normal approximation). - Bars between $k=7$ and $k=17$ cover a larger area, roughly about 95% (typical for two standard deviations). 6. **Calculate exact probabilities using binomial distribution:** - Probability for $k$ in $[9,15]$: $$P(9 \leq X \leq 15) = \sum_{k=9}^{15} \binom{30}{k} (0.4)^k (0.6)^{30-k}$$ - Probability for $k$ in $[7,17]$: $$P(7 \leq X \leq 17) = \sum_{k=7}^{17} \binom{30}{k} (0.4)^k (0.6)^{30-k}$$ 7. **Using a calculator or software for sums:** - $P(9 \leq X \leq 15) \approx 0.68$ - $P(7 \leq X \leq 17) \approx 0.95$ 8. **Comparison:** The estimated probabilities from the graph closely match the calculated probabilities, confirming the intervals correspond to approximately one and two standard deviations around the mean. **Final answers:** - $E(X) = 12$ - $\sigma \approx 2.683$ - Interval 1: $[9, 15]$ with probability $\approx 0.68$ - Interval 2: $[7, 17]$ with probability $\approx 0.95$