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vesna_86 [32]
3 years ago
8

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number o

f arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t.
Required:
a. What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6? At least 10?
b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period?
c. What is the probability that at least 20 small aircraft arrive during a 2.5-hour period? That at most 10 arrive during this period?
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

             = 8(1)

             = 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =  

              =  

              =  

              = 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

                = 1 - {+++++}

                = 1 - (){+++++}

                = 1 - (0.000335){+++++}

                = 1 - (0.000335){1+8+32+85.34+170.67+273.07}

                = 1 - (0.000335){570.08}

                = 1 - 0.1909

                = 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

                 = 1 - {+++++

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A) In a group of 60 students, 15 liked maths only, 20 liked science only and 5 did not like
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Part (i)

We have 60 students total, and 5 didn't like any of the two subjects, so that must mean 60-5 = 55 students liked at least one subject.

<h3>Answer: 55</h3>

=========================================================

Part (ii)

We have 15 who like math only, 20 who like science only, and 55 who like either (or both). Let x be the number of people who like both classes.

We can then say

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<h3>Answer: 20</h3>

=========================================================

Part (iii)

There are 15 people who like math only, and 20 who like both. Therefore, there are 15+20 = 35 people who like math (and some of these people also like science)

<h3>Answer: 35</h3>

=========================================================

Part (iv)

We'll follow the same idea as the previous part. There are 20 people who like science only and 20 who like both subjects. That yields 40 people total who like science (and some of these people also like math).

<h3>Answer: 40</h3>

=========================================================

Part (v)

We'll draw a rectangle to represent the entire group of 60 students. This is considered the universal set. Inside the rectangle will be two overlapping circles to represent math (M) and science (S).

We'll have 15 go in circle M, but outside circle S to represent the 15 people who like math only. Then we have 20 go in circle S but outside circle M to show the 20 people who like science only. We have another copy of 20 go in the overlapped region between the circles. This is the 20 people who like both classes. And finally, we have 5 go outside both circles, but inside the rectangle. These are the 5 people who don't like either subject.

Note how all of the values in the diagram add up to 60

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This helps confirm we have the correct values.

<h3>Answer: See the venn diagram below</h3>

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