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{ "quality": "blurry", "question_count": 2, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Normal Distribution Applications", "Z-score Calculation", "Statistical Interpretation" ], "new_concepts": [], "current_concepts": [ "Normal Distribution Applications", "Z-score Calculation", "Statistical Interpretation", "Percentile Calculations" ] } ## Calculate the Z-score for 306 days $$ \begin{aligned} \mu &= 269 \ \sigma &= 15 \ x &= 306 \ z &= \frac{x - \mu}{\sigma} = \frac{306 - 269}{15} = \frac{37}{15} \approx 2.4667 \end{aligned} $$ ## Find the probability of a pregnancy lasting 306 days or longer $$ \begin{aligned} P(X \ge 306) &= P(Z \ge 2.47) \ &= 1 - P(Z < 2.47) \ &= 1 - 0.9932 \ &= 0.0068 \end{aligned} $$ ## Interpret the statistical significance of the result $$ \begin{aligned} P(X \ge 306) &= 0.0068 \le 0.05 \end{aligned} $$ Since the probability is extremely low, this suggests that either a very rare event occurred or the husband is not the father. ## Question 1 The probability that a pregnancy will last 306 days or longer is 0.0068. ## Question 2 (A) The result suggests an uncommon but not significant event occurred.

(B) The result suggests the event did not occur.

(C) The result suggests that the husband is the father.

(D) The result suggests that either a very rare event occurred or the husband is not the father.

{ "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Statistics", "Normal Distribution Applications" ] }

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Explanation:

Calculate the Z-score for 306 days

Find the probability of a pregnancy lasting 306 days or longer

Interpret the statistical significance of the result

Answer:

Question 1

Question 2