Question

# If a phone call is received once every 10 minutes on average, during a one-hour interval,...

If a phone call is received once every 10 minutes on average, during a one-hour interval,

1) The probability that exactly one phone call is received during the first 30 minutes and exactly two calls in the last 30 minutes

2) The expected value of the distribution of the time of the arrival of the first call

3) The probability that the second phone call is received between 30-40 minutes

4) The probability that the third phone call is received during the last 20 minutes

5) The probability that two or more phone calls are received during the 60-minute interval

events/time = 1/10 calls/min = 0.1 calls/min

1.

P(1 call in first 30 min , 2 call in last 30 min)

= P(1 call in first 30 min)*P(2 call in last 30 min)

= [e^(-0.1*30)*(0.1*30)^1 / (1!)] * [ e^(-0.1*30) * (0.1*30)^2 / (2!)]

= 0.0335

2.

E(ariival if first call) = 1 / rate = 1 / 0.1 = 10 min

3.

P(second call between 30-40 min) = P(1st call in 30 min)*P(1 call in next 10 min{i.e. between 30-40min})

= [e^(-0.1*30)*(0.1*30)^1 / (1!)] * [e^(-0.1*10)*(0.1*10)^1 / (1!)]

= 0.0550

4.

P(3 rd phone call in last 20 min) = P(2 call in 40 min)*P(1 call in 20 min)

= [e^(-0.1*40)*(0.1*40)^2 / (2!)] * [e^(-0.1*20)*(0.1*20)^1 / (1!)]

= 0.0397

5.

P(2 or more call in 60 min)

= 1 - P(0) - P(1)

= 1 - [e^(-0.1*60)*(0.1*60)^0 / (0!)] - [e^(-0.1*60)*(0.1*60)^1 / (1!)]

= 0.9826

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