A student commutes daily from his home to school. The average timefor a one way
ID: 2917313 • Letter: A
Question
A student commutes daily from his home to school. The average timefor a one way trip is 25 minutes, with standard deviation of 5minutes. Assume the distribution of trip times to be normallydistributed.a) what is the probability that the trip will take at least 35minutes?
b) if his early class is at 9:00 AM and he leaves his home at 8:35AM daily, what percentage of the time does he show up late forclass?
c) if he leaves the house at 8:25 and free lunch coupons aredistributed from 8:50 until 9:00, what is the probability he willmiss the opportunity of securing a free meal for the day?
d)find the probability of 2 of the next 3 trips taking at least 35minutes.
e) find the length of time above which we find the slowest 15% ofthe trip.
Explanation / Answer
given that the average time for a one way trip is 25min, = 25min and the standard deviation is , =5min 1)The probability that the trip will take at least 35 minutes,P(X 35) = P[(X-/)(35-25)/5] =P(Z 2) =1 - P(Z 2) = 1- 0.97725 = 0.02275 2)if his early class is at 9:00 AM and he leaves his home at8:35 AM daily, the percentage of the time does he show up latefor class ,P(X>25) = P[(X-)/ > (25-25)/5) =P(Z > 0) =0.5 3) if he leaves the house at 8:25 andfree lunch coupons are distributed from 8:50 until 9:00, what isthe probability he will miss the opportunity of securing a freemeal for the day if he reaches within 9.00 that is within 35 min, he can securethe free lunch coupon , the probability is P(X < 35) = P[(X-/) (35-25)/5] = P(Z 2) =0.97725 = P(Z 2) =0.97725Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.