1. **Problem statement:** Connor Hellebuyck has a save percentage of 92.6%, meaning the probability of making a save on any shot is $p=0.926$. There are $n=20$ shots. We want to find: - Probability of making saves on more than 18 shots. - Probability of allowing at least 2 shots (not stopping at least 2). - Expected number of saves. - Standard deviation of the number of saves. 2. **Model and formulas:** This is a binomial distribution problem where the number of successes (saves) $X$ follows $X \sim \text{Binomial}(n=20, p=0.926)$. - Probability mass function: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - Expected value: $$E(X) = np$$ - Variance: $$\text{Var}(X) = np(1-p)$$ - Standard deviation: $$\sigma = \sqrt{np(1-p)}$$ 3. **Calculate probability of more than 18 saves:** More than 18 means $X=19$ or $X=20$. $$P(X>18) = P(X=19) + P(X=20)$$ Calculate each term: $$P(X=19) = \binom{20}{19} (0.926)^{19} (0.074)^1 = 20 \times (0.926)^{19} \times 0.074$$ $$P(X=20) = \binom{20}{20} (0.926)^{20} (0.074)^0 = 1 \times (0.926)^{20}$$ Calculate powers: $(0.926)^{19} \approx 0.1887$ $(0.926)^{20} = (0.926)^{19} \times 0.926 \approx 0.1887 \times 0.926 = 0.1748$ Calculate probabilities: $$P(X=19) = 20 \times 0.1887 \times 0.074 = 20 \times 0.01396 = 0.2792$$ $$P(X=20) = 0.1748$$ Sum: $$P(X>18) = 0.2792 + 0.1748 = 0.4540$$ 4. **Calculate probability of allowing at least 2 shots:** Allowing at least 2 means number of saves $\leq 18$ because 20 shots minus at least 2 allowed means at most 18 saves. So, $$P(\text{allow} \geq 2) = P(X \leq 18) = 1 - P(X=19) - P(X=20)$$ From above, $$P(X=19) + P(X=20) = 0.4540$$ Therefore, $$P(\text{allow} \geq 2) = 1 - 0.4540 = 0.5460$$ 5. **Expected number of saves:** $$E(X) = np = 20 \times 0.926 = 18.52$$ 6. **Standard deviation:** $$\sigma = \sqrt{np(1-p)} = \sqrt{20 \times 0.926 \times 0.074} = \sqrt{1.37048} \approx 1.170$$