Question

David can take on three possible routes from his home to his office in the morning.

The percentage of days on routes A, B, and C to his office were 45%, 30%, and 25%, respectively.

The traveling time on each of these routes is normally distributed with mean and standard deviation given in the following table:

	Route A	Route B	RouteC
Mean	30 mins	35 mins	28 mins
Standard Deviation	10 mins	8 mins	12 mins

David left his home at 10:28am and arrived at his office before 11:00am.

Calculate the probability that he took Route B to work? (10 marks)

Answer 1

Sol: Given that, David from his home to can take on three possible toutes his office in the morning. The percentage of days on toutes A = 45%. ; P(A) = 45/100 = 0.45. B = 30:1 ; P(B) = 30/100 = 0.30. C = 254. ; PCC) = 25/100 = 0.25. Route c. Route A Route B | 16 = 35 mins Mean th 30 mins HC-38 mins standard deviation * = 40 mins B-8 mins = 12 mins Now, David left his home at 10:28 am and arrived at his office before 11:00 am. therefore, the time between 10:28 am to 11:00 am is 32 mins. that means that he arrived at office in less than 32 mins. Let X be the event that he teach his office in less than 32 mins. i X= 32minute. Since, the travelling time on each foule is noomally distributed, we can find the following probability using the no omal distribution.

Now, the probability on soule A that he teach his office in less than 32 minule is - НА We know that, P(x10 32- PIZ- = P(Z < 3250) = f(z< = P(z< 02) i plxA) = 0.5793 (: By using Z-Scope Calculator) Now, the poobability on foute B, that he reach his office in less than 32 minute is, We know that, PCx18) X - MB) c B 2 - 35 -Plz< 33-35) =plz< 3) =P(Z< -0.375) :: P(x16) = 0.352 (-:- By using 2-Gcode Calculator)

teach his office in less than me poobability on foule C, that he 32 minule is We know that, Pexle) = P(Z < ANCE) = Plze 32-02) = P(z< 0.4) :: P(xlc) = 0.6554 Now, the probability that he took Route B given he arrived his office in less than 32 minutes is conditional probability We know that, the formula for conditional poobability is given 1 PCBnx) by PC B1x) = pl BUT PCA) P(x/a) + P(B) P(x16) + PCC) P(xic) = _ P(B) P(xIB) PLA) PCX16) + P(8) plx 8) + P(c) pcxlc) 0.30 x 0.359 0.45X0.5793 +0.300.352 +0.25 x 0.6554

P(81x) - - 0.1056 0.260685 + 0.1056 + 0.16385 0.1056 : 0.530135 0.19919 : PC6lx) = 0.20 (approximately)