12. Forty-five percent of adults say they are interested in regularly measurit their carbon footprint. You randomly select 11 adults and ask them if they a interested in regularly measuring their carbon footprint. Find the probabili that the number of adults who say they are interested is (a) exactly eight, ( at least five, and (c) less than two. Are any of these events unusual? Expla your reasoning. (Source: Sacred Heart University Polling)

**Step 1. Define the Situation** Let \( X \) be the number of adults (out of 11) who say they are interested in regularly measuring their carbon footprint. Since each adult independently says “yes” with probability \( p = 0.45 \) (and “no” with probability \( q = 1-p = 0.55 \)), we model \( X \) using the binomial distribution: \[ X \sim \text{Binomial}(n=11,\, p=0.45) \] The probability mass function is given by \[ P(X=k)= \binom{11}{k} (0.45)^k (0.55)^{11-k}. \] --- **Step 2. (a) Probability Exactly Eight Adults are Interested** We need to find \[ P(X=8)= \binom{11}{8}(0.45)^8(0.55)^3. \] Recognizing that \[ \binom{11}{8} = \binom{11}{3} = 165, \] we have \[ P(X=8)= 165\,(0.45)^8\,(0.55)^3. \] *Numerical Approximation:* Calculating approximately, one finds \[ P(X=8) \approx 0.0461 \quad (\text{or } 4.61\%). \] --- **Step 3. (b) Probability that at Least Five Adults are Interested** We need \[ P(X \ge 5)= 1-P(X \le 4)=1-\sum_{k=0}^{4} P(X=k). \] The terms to compute are: 1. \( P(X=0)= \binom{11}{0}(0.45)^0(0.55)^{11}=(0.55)^{11} \). Approximation: \( P(X=0) \approx 0.00139 \). 2. \( P(X=1)= \binom{11}{1}(0.45)^1(0.55)^{10}= 11\,(0.45)(0.55)^{10} \). Approximation: \( P(X=1) \approx 0.01254 \). 3. \( P(X=2)= \binom{11}{2}(0.45)^2(0.55)^9 \). Approximation: \( P(X=2) \approx 0.0513 \). 4. \( P(X=3)= \binom{11}{3}(0.45)^3(0.55)^8 \). Approximation: \( P(X=3) \approx 0.1258 \). 5. \( P(X=4)= \binom{11}{4}(0.45)^4(0.55)^7 \). Approximation: \( P(X=4) \approx 0.2063 \). Now, summing: \[ \sum_{k=0}^{4} P(X=k) \approx 0.00139 + 0.01254 + 0.0513 + 0.1258 + 0.2063 \approx 0.39733. \] Thus, \[ P(X \ge 5) \approx 1 - 0.39733 = 0.60267 \quad (\text{or } 60.27\%). \] --- **Step 4. (c) Probability that Fewer Than Two Adults are Interested** This means \( P(X<2)= P(X=0)+P(X=1) \). Using the approximations from Step 3: \[ P(X<2) \approx 0.00139 + 0.01254 = 0.01393 \quad (\text{or } 1.393\%). \] --- **Step 5. Determining Whether the Events Are Unusual** A common rule of thumb is that an event is considered unusual if its probability is less than \( 5\% \) (i.e. \( 0.05 \)). - For (a): \( P(X=8) \approx 4.61\% \) which is slightly less than \( 5\% \); hence, this event is unusual. - For (b): \( P(X \ge 5) \approx 60.27\% \) which is not unusual. - For (c): \( P(X<2) \approx 1.39\% \) which is well below \( 5\% \); hence, this event is unusual. --- **Final Answers** - (a) \( P(X=8)= 165\,(0.45)^8\,(0.55)^3 \approx 0.0461 \) (4.61%). - (b) \( P(X \ge 5)= 1-\sum_{k=0}^{4} P(X=k) \approx 0.60267 \) (60.27%). - (c) \( P(X<2)= P(X=0)+P(X=1) \approx 0.01393 \) (1.39%). Events (a) and (c) are unusual since their probabilities are less than \( 5\% \); event (b) is not unusual.