Daily commute for a in average with standard deviation of 3.8 minutes. Assume th

Question

Daily commute for a in average with standard deviation of 3.8 minutes. Assume the distribution of trip times to be normally distributed student from his house to school will take 24 minutes a) What is the probability that a trip to school will take at least 1/2 hour? b) If one of student's class starts at 9:00 A.M. and he leaves his house at 8: 45 AM. daily, what percentage of the time is he late for school? c) If he leaves his house at 8:35 A.M. to meet with his professor from 8:50 A.M. until 9:00 AM, what is the probability that he misses the meeting? d) Find the length of time above which we find the slowest 15% of the trips. e) Find the probability that 2 of the next 3 trips will take at least 1/2 hour

Explanation / Answer

Ans:

a)half an hr=30 min

z=(30-24)/3.8=6/3.8=1.58

P(z>=1.58)=1-P(z<1.58)=1-0.9429=0.0571

b)

He will be late if takes more than 15 min

z=(15-24)/3.8=-2.37

P(z>-2.37)=P(z<2.37)=0.9911

99.11%

c)he will miss meeting if not reach between 8.50 to 9 am

i.e between 15 to 25 min

z(15)=-2.37

z(25)=(25-24)/3.8=0.26

P(-2.37<=z<=0.26)=P(z<=0.26)-P(z<=-2.37)=0.6026-0.0089=0.5937

d)P(Z>=z)=0.15

P(Z<=z)=1-0.15=0.85

z=1.036

x=24+1.036*3.8

x=24+3.94

x=27.94 min

e)

Use binomial distribution with n=3,p=0.0571

P(y=2)=3C2*0.05712*(1-0.0571)1

=0.0092

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