AP Statistics 4.9 Combining Random Variables- FRQs - Exam Style Questions
▶️ Answer/Explanation
(a)
If the failure rate is 15%, then the probability of success for each igniter is \( 1 – 0.15 = 0.85 \).
Since the igniter failures are independent, the probability that all first 30 igniters succeed is:
\( P(\text{30 successes}) = (0.85)^{30} \approx 0.0076 \)
\( \boxed{0.0076} \)
(b)
Given that the first 30 igniters succeeded, we want the probability that the first failure occurs on the 31st or 32nd trial.
This is equivalent to finding the probability of first failure on trial 1 or 2 in a new sequence.
• Probability first failure on trial 31: \( 0.15 \)
• Probability first failure on trial 32: \( (0.85)(0.15) = 0.1275 \)
Total probability: \( 0.15 + 0.1275 = 0.2775 \)
\( \boxed{0.2775} \)
(c)
\( \boxed{\text{Yes}} \), it is reasonable to believe that the failure rate is less than 15%.
The probability of getting 30 consecutive successful launches if the failure rate is 15% is only 0.0076, which is less than 1%. This very small probability suggests that such an outcome would be extremely unlikely if the true failure rate were 15%. Since this probability is smaller than conventional significance levels (such as \( \alpha = 0.05 \) or \( \alpha = 0.01 \)), we have convincing evidence that the super igniters have a lower failure rate than 15%.