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Sphinxa [80]
3 years ago
15

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

Poisson distribution with parameter μ = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†). (Round your answer to three decimal places.) (a) What is the probability that the number of drivers will be at most 15? (b) What is the probability that the number of drivers will exceed 26? (c) What is the probability that the number of drivers will be between 15 and 26, inclusive? What is the probability that the number of drivers will be strictly between 15 and 26? (d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?
Mathematics
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†).

a) P(X\leq 15) = 0.1565=0.157

b) P(X>26) =1-F(26)\\= 1-0.9221\\=0.0779=0.078

c) P(15\leq x\leq 26)\\=F(26)-F(14)\\=0.9221-0.1049\\=0.8172=0.817

d) 2 std dev = 2(20) =40

Hence 2 std deviation means

20-40, 20+40

i.e. (0,60)

P(0

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3(0)^2 - 4 = -4

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2 years ago
Consider the set of differences, denoted with d, between two dependent sets: 84, 85, 83, 63, 61, 100, 98. Find the sample standa
sveta [45]

Answer:

The sample standard deviation is 15.3.

Step-by-step explanation:

Given data items,

84, 85, 83, 63, 61, 100, 98,

Number of data items, N = 7,

Let x represents the data item,

Mean of the data points,

\bar{x}=\frac{84+85+83+63+61+100+98}{7}

=82

Hence, sample standard deviation would be,

\sigma= \sqrt{\frac{1}{N-1}\sum_{i=1}^{N} (x_i-\bar{x})^2}

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=15.2534149182

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2 years ago
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