A bird at the top of a tree looks down at a field mouse with an angle of depression of 65Á. If the field mouse is 30 meters from
the base of the tree, find the distance from the field mouse to the birdÍs eyes. Round the answer to the nearest tenth.
a.
29.6 m
b.
50.0 m
c.
64.3 m
d.
71.0 m
a.
29.6 m
b.
50.0 m
c.
64.3 m
d.
71.0 m
1 answer:
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Answer:
(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.
(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.
(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.
Step-by-step explanation:
Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.
The arrival rate is, <em>λ</em>t = 8 per hour.
(a)
For <em>t</em> = 1 the average number of aircraft arrival is:
The probability distribution of a Poisson distribution is:
Compute the value of P (X = 6) as follows:
Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.
Compute the value of P (X ≥ 6) as follows:
Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.
Compute the value of P (X ≥ 10) as follows:
Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.
(b)
For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:
The expected value of the number of small aircraft that arrive during a 90-min period is 12.
The standard deviation is:
The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.
(c)
For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:
Compute the value of P (X ≥ 20) as follows:
Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.
Compute the value of P (X ≤ 10) as follows:
Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.