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Trip time normal distribution probabilities

Short answer

Convert to standard normal z-scores. (a) For at least 35 min, z = 1.00, probability = 0.1587 (15.9%). (b) With only 20 min available, z = –2.00, probability of being late = 0.9772 (97.7%).

Average trip time for office is 30 minutes with standard deviation of 5 min. Assuming normal distribution of trip time, calculate the followings: a) Probability of trip time being at least 35 minutes b) If the working hour starts at 10:00 AM and trip starts at 9:40 AM what is the probability of being late?

Trip time TT is normally distributed with mean μ=30\mu = 30 min and standard deviation σ=5\sigma = 5 min. Use the standard normal variable z=(T−μ)/σz = (T - \mu)/\sigma.

(a) Probability that the trip time is at least 35 minutes

z=35−305=1.00,P(T≥35)=1−Φ(1.00)=1−0.8413=0.1587z = \frac{35 - 30}{5} = 1.00,\quad P(T \ge 35) = 1 - \Phi(1.00) = 1 - 0.8413 = 0.1587

Answer (a): 0.1587 (about 15.9 %).

(b) Probability of being late

Work starts at 10:00 AM and the trip starts at 9:40 AM, so only 20 minutes are available. The person is late if the trip takes more than 20 minutes:

z=20−305=−2.00,P(T>20)=1−Φ(−2.00)=Φ(2.00)=0.9772z = \frac{20 - 30}{5} = -2.00,\quad P(T > 20) = 1 - \Phi(-2.00) = \Phi(2.00) = 0.9772

Answer (b): 0.9772 (about 97.7 %). Leaving at 9:40 makes the person late almost every day. To be on time with 95 percent confidence, the trip must allow 30+1.645×5=38.230 + 1.645 \times 5 = 38.2 min, so it must start by about 9:21 AM (10:00 minus 38.2 min).

✓ Answer checked by Claude Opus 5.5 (high effort)

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