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Sagot :
(a) The probability that more than three aircraft arrive within an hour is 0.2970.
(b) The probability that no interval contains more than three arrivals is 2.56.
(c) The length of an interval of time is 0.891 hours.
The time between arrivals of small aircraft at a country airport is exponentially distributed with a mean of one hour.
(a) We will find the probability that more than three aircraft arrive within an hour,
P(X > 3) = 1 - P(X ≤ 3)
= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
= 1 - [ ( [tex]e^{-4/3}[/tex] 1° / 0!) + ( [tex]e^{-4/3}[/tex] 1¹ / 1!) + ( [tex]e^{-4/3}[/tex] 1² / 2!) + [tex]e^{-4/3}[/tex] 1³/ 3!]
= 1 - (0.7029)
P(X > 3) = 0.2970
(b) Now we calculate probability that no interval contains more than three arrivals if 30 separate one-hour intervals are chosen,
(1 - P(X > 3)³°) = (1 - 0.2970)³°
= 2.56
(c) Let the length of an interval of time is t, the calculation for t is given as;
P(X = 0) = 0.41
[tex]e^{-t}[/tex](t)° / 0! = 0.41
[tex]e^{-t}[/tex] = 0.41
- t = ln(0.41)
- t = -0.891
t = 0.891 hours
Therefore the length of an interval of time such that the probability that no arrivals occur during the interval is 0.41 is 0.891 hours.
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