Question #98919
6. The time it takes for Norman to travel from his house to the 10:00 AM MATH 1F92 lecture is
approximately normal with a mean of 25 minutes and standard deviation of 3 minutes.
(Note: Labelled diagrams and proper notation are required for all parts.)
a) One morning, Norman leaves his house at 9:40 AM. The professor is going to give hints about
the test during the first 7 minutes of lecture. What is the probability that he will miss all the hints?
b) On the morning of the test, Norman wants to arrive by 9:30 AM. What time must Norman leave
his home to ensure a 97% probability that he will arrive before 9:30 AM?
1
Expert's answer
2019-11-18T12:10:42-0500

a)

P(X>27)=P(Z>27253)=P(Z>0.67)=1P(Z<0.67)===0.2514.P(X>27)=P(Z>\frac{27−25}{3})=P(Z>0.67)=1−P(Z<0.67)==\\=0.2514.



b)

P(Z<z)=0.97z=1.881.88=x253x=31min.P(Z<z)=0.97→z=1.88→1.88=x−253→x=31 min.

Norman must leave his home at 9:30-31=8:59.


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Comments

Assignment Expert
25.11.19, 00:58

Part a) asks about missing professor's hints. It passed 20 minutes since 9:40 AM till 10:00 AM before the lecture plus the first 7 minutes of lecture when the professor was expected to give hints about the test. Thus, 20+7=27 (minutes).

Al
24.11.19, 19:25

In question 98919, can you please explain how you got the value of 27 in (a)

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