IB Mathematics SL 4.9 The normal distribution -AI HL Paper 1- Exam Style Questions- New Syllabus
Question
In a school year, Thomas attends school for 180 days. Every day, he leaves home at exactly 7:45 am and takes public transport to school. His travel times are normally distributed with a mean of 35.5 minutes and a standard deviation of 5.2 minutes. The travel times are independent of each other. He is considered late if he arrives at school after 8:30 am.
(a) Find the probability that Thomas is late on at least 10 days during the year.
Mrs. Beale records student lateness. However, she does not always mark it accurately. On days when Thomas is actually late, there is a 35% chance Mrs. Beale will still mark him as on time. Her accuracy on any day is independent of her accuracy on other days. Thomas receives a detention if he is marked late at least 4 times during the year.
(b) Find the probability that Thomas receives a detention in the year.
Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL):
• SL 4.9: Normal distribution — probability calculations — part (a)
• SL 4.8: Binomial distribution — situations and modelling — parts (a), (b)
• SL 4.6: Combined events — probability calculations — part (b)
• SL 4.8: Binomial distribution — situations and modelling — parts (a), (b)
• SL 4.6: Combined events — probability calculations — part (b)
▶️ Answer/Explanation
(a)
Travel time available: from 7:45 to 8:30 is 45 minutes. Probability of being late on a single day: \( P(T > 45) \) where \( T \sim N(35.5, 5.2^2) \). Using normalcdf: \( p \approx 0.0338556 \).
Let \( X \) = number of late days in 180 days. \( X \sim B(180, 0.0338556) \).
We need \( P(X \geq 10) = 1 – P(X \leq 9) \approx 0.0869 \).
\( \boxed{0.0869} \)
(b)
Probability Thomas is late and correctly marked late = \( 0.0338556 \times (1 – 0.35) = 0.0220061 \).
Let \( Y \) = number of days marked late. \( Y \sim B(180, 0.0220061) \).
Probability of detention = \( P(Y \geq 4) = 1 – P(Y \leq 3) \approx 0.561 \).
\( \boxed{0.561} \)