We wil assume a variable x to be the total number of calls received by the switchboard.
The question also says to assume that the calls were made independently.
Given:
Calls are independent.
Calls are uniformly distributed over a 1 hour period.
Ans (a). The calls are distributed uniformly over 1 hour, hence: (0, 1).
So we have,
f1(x1) = 1
f2(x2) = 1
X1 and X2 are considered to be independent of each other. Hence,
f(x1,x2) = f1 (x1) f2 (x2)
f(x1,x2) = 1 (1)
f(x1,x2) = 1
Thus,
P(X1 <= 0.5; X2 <= 0.5) = ∫0.50 ∫0.50 f(x1,x2) dx2 dx1
= ∫^0.5 0 ∫^0.5 0 (1) dx2 dx1
= ∫^0.5 0 (x2^0.5 0) dx1
= ∫^0.5 0 (0.5 - 0) dx1
= 0.5 ∫^0.5 0 dx1
= 0.5 (x1^0.5 0)
= 0.5 (0.5 - 0)
= 0.25
Therefore, the probability that the calls were received within the first 30 minutes or first half hour is 0.25.
Ans (b). Steps 1 and 2 are the same as the above answer.
Probability = [∫^11/12 0 ∫^x1 + 1/12 x1 1dx2 dx1] + [∫^1 1/12 ∫^x1 x1-1/12 1dx2 dx1
= [∫^11/12 0 (x2 ^x1+1/12 x1 dx1] + [∫^1 1/12 (x2 ^x1 x1-1/12 dx1)]
= [∫^11/12 0 (x1 + 1/12 -x1) dx1] + [∫^1 1/12 (x1 - x1 + 1/12) dx1]
= [(1 + x1/12 - 1) ^11/12 0] + [( 1 - 1 + x1/12) ^1 1/12]
= 11/144 + 11/144
= 0.1528
Therefore, the probability that the calls were received within five minutes of each other is 0.15.
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