A bus is scheduled to leave at 5.00 pm. A person who wants to catch the bus takes a rickshaw at 4.35 pm from his home. If there is not much traffic on the road he can reach the bus stand in 10 minutes, while in heavy traffic, it may take him 35 minutes. Further it takes him 3 minutes to locate his bus in the stand. Let X denote the time taken for the rickshaw to reach the stand from his home and suppose X has uniform distribution on the interval (10, 35). Obtain the probability that the person catches the bus.
2007-10-04 03:31:35 · 3 answers · asked by mark_fraser 1 in Science & Mathematics ➔ Mathematics
The bus leaves at 5:00 and it takes three minutes just to find the bus, so the person must be at the stand before 5:57 to catch the bus.
if the rickshaw takes less than 22 minutes the person will be able to catch the bus.
X ~ Uniform(10,35)
P(X < 22) =
22
∫ (1/(35-10) dx
10
= 22/25 - 10/25
= 12/25
the probability the person catches the bus is 12/25
2007-10-04 13:23:43 · answer #1 · answered by Merlyn 7 · 0⤊ 0⤋
he must actually arrive at the station at 4:57 ...
given the distribution X, which is a uniform distribution on (10 , 35)
this means he will arrive between 4:45 and 5:10
(a span of 25 minutes)
from 4:45 to 4:57 (a span of 12 minutes) he can catch his bus...
therefore the probability is 12/25 = 48%.
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2007-10-04 11:06:34 · answer #2 · answered by Alam Ko Iyan 7 · 0⤊ 0⤋
the probability to catch the bus is = 49 . 44 %
5.00pm -4.35 pm =25 mint .
25 mint - 3mint ( looking for the bus ) = 22 mint
his Average time = ( 35 + 10) / 2 = 45 / 2 = 22.5 mint .
so the chance ( % ) to catch the bus =100% ( 22 mint + 22.5 mint ) x 22 =(100 / 44.5 ) x 22 =2. 2472 x 22 = 49. 44 %
2007-10-04 10:52:01 · answer #3 · answered by Anonymous · 0⤊ 0⤋