Please explain in the problem below how a) λ was obtained and b) what went wrong with the Poisson approximation.Thank you Example 5.8. It is known that of 313.6 million of US population, about 200 million have high-speedInternet access at home. Thus the population proportion of high-speed Internet users is 0.6378. Suppose1500 people are randomly selected. What is the probability that at least 1000 of those responding to thesurvey have high-speed Internet access?Solution. We interpret 0.6378 proportion as p and a sample of 1500 people selected for the survey as asample of 1500 independent Bernoulli trials, so that X = X₁ + ... + X1500 € [B(1500, 0.6378)] givesthe number of those in the surveyed sample who have high-speed Internet access. Thus, we areinterested to find 1200) 0.6378 0.36221500-kNow the above is a computational challenge. The normal approximation, due to the Central LimitTheorem (to be explored in Chapter V, section 2), would be a remedy. Alternatively, we can use Rlanguage to compute the above probability precisely. The procedure will be as follows.P{X > 1000} = 1− P{X < 1000} = 1 − P{X ≤ 999}P{X ≥ 1000} = 150implyingk=1000> 1-pbinom (999,1500,.6378)[1] 0.01042895which readsP{X > 1000} = 0.01042895.Here pbinom (999,1500,.6378) calculates999Σ (¹500) 0.6378* 0.3622¹500-kNow if we use the Poisson approximation with λ = 956.7 (why?) we arrive at> 1-ppois(999,956.7)[1] 0.08395288The result significantly differs from that of direct binomial. The student needs to explain thediscrepancy. (See Problem 5.6.)

Answered: Please explain in the problem below how…

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Please explain in the problem below how a) λ was obtained and b) what went wrong with the Poisson approximation.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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Question

Please explain in the problem below how a) λ was obtained and b) what went wrong with the Poisson approximation.

Thank you

Example 5.8. It is known that of 313.6 million of US population, about 200 million have high-speed
Internet access at home. Thus the population proportion of high-speed Internet users is 0.6378. Suppose
1500 people are randomly selected. What is the probability that at least 1000 of those responding to the
survey have high-speed Internet access?
Solution. We interpret 0.6378 proportion as p and a sample of 1500 people selected for the survey as a
sample of 1500 independent Bernoulli trials, so that X = X₁ + ... + X1500 € [B(1500, 0.6378)] gives
the number of those in the surveyed sample who have high-speed Internet access. Thus, we are
interested to find

1200) 0.6378 0.36221500-k
Now the above is a computational challenge. The normal approximation, due to the Central Limit
Theorem (to be explored in Chapter V, section 2), would be a remedy. Alternatively, we can use R
language to compute the above probability precisely. The procedure will be as follows.
P{X > 1000} = 1− P{X < 1000} = 1 − P{X ≤ 999}
P{X ≥ 1000} = 150
implying
k=1000
> 1-pbinom (999,1500,.6378)
[1] 0.01042895
which reads
P{X > 1000} = 0.01042895.
Here pbinom (999,1500,.6378) calculates
999
Σ (¹500) 0.6378* 0.3622¹500-k
Now if we use the Poisson approximation with λ = 956.7 (why?) we arrive at
> 1-ppois(999,956.7)
[1] 0.08395288
The result significantly differs from that of direct binomial. The student needs to explain the
discrepancy. (See Problem 5.6.)

Expert Solution

Step 1

Given

The population proportion of high speed internet users $p = 0.6378$

Sample size $n = 1500$

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