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alexgriva [62]
3 years ago
9

An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results wer

e: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. For the Poisson goodness-of-fit test, what is the expected frequency of the data value X
Mathematics
1 answer:
liraira [26]3 years ago
8 0

Step-by-step explanation:

since Poisson distribution parameter is not given so we have to estimate it from the sample data. The average number of of arrivals per minute at an ATM is

\hat{\lambda}=\bar{x}=\frac{\sum x}{n}=\frac{30}{30}=1

So probabaility for \mathrm{X}=1 is

P(X=1)=\frac{e^{-\lambda} \lambda^{x}}{x !}=\frac{e^{-1} \cdot 1^{1}}{1 !}=0.3679

So expected frequency for X=1 is 0.3679^{*} 30=11.037 (or \left.11.04\right) .

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Ivahew [28]

Let C be the set of all students in the <u>c</u>lassroom.

Let P and M be the sets of students that pass <u>p</u>hysics and <u>m</u>ath, respectively.

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ii. 9 students fails both subjects, so we find

n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51

By the inclusion/exclusion principle,

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Using the result from part (i), we have

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\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}

7 0
2 years ago
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