Solution:
We are given that arrival times follows uniform distribution with a = 4 hours = 4*60 = 240 minutes and b = 4 hours 30 minutes = 4*60 + 30 = 270 minutes.
a = 240, b = 270
Part a
The required graph for the above uniform distribution is given as below:
Part b
Here, we have to find P(X<4 hours 10 minutes) = P(X < 4*60 + 10 ) = P(X<250)
P(X<m) = (m – a)/(b – a)
P(X<250) = (250 – 240) / (270 – 240) = 10/30 = 1/3 = 0.3333
Required probability = 0.3333
Part c
Here, we have to find P(X>4 hours 20 minutes) = P(X>4*60+20) = P(X>260)
P(X>m) = (b – m)/(b – a)
P(X>260) = (270 – 260)/(270 – 240) = 10/30 = 1/3 = 0.3333
Required probability = 0.3333
Part d
We might expect such an objection because usually arrival times do not follows uniform distribution most of the times. In most of the cases, arrival times follow Poisson distribution.
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