Find the model of number of successful contacts from 1,000 sampled households.

ddaeeric
2021-09-21
Answered

Find the model of number of successful contacts from 1,000 sampled households.

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Mitchel Aguirre

Answered 2021-09-22
Author has **94** answers

According to the pew research centre report, only with 76% of the randomly selected households can be contactable. There are 1,000 households were selected.

Properties of Bernoulli trials:

- In each trial there will be only two possible outcomes, success and failure.

- The probability of success for each trial is same.

- Each trial is independent to each other.

Binomial probability for Bernoulli trials:

The number of successes in a specified number of Bernoulli trials:

The number of successes in a specified number of Bernoulli trials is described by Binomial model.

If the number of trials is n, with the probability of success is p and the probability of failure is$q=1-p$ , and X be the event of the number of success in n independent trials, then the binomial probability model is defined as,

$P(X=x){=}_{n}{C}_{x}{p}^{x}{q}^{n-x}{,}_{n}{C}_{x}=\frac{n!}{x!(n-x)!}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x=0,1\cdots n$

With mean is$\mu =np$ and standard deviation is $\sigma =\sqrt{npq}$ . It is denoted by $X\sim B\in om(n,p)$ .

Normal approximation from Binomial distribution:

The success and failure condition:

If$np\ge 10\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}nq\ge 10$ , then the binomial distribution can be approximated as Normal distribution with mean $\mu =np$ and variance ${\sigma}^{2}=npq$ . That is, random variable follows $N(np,\sqrt{npq})$ .

Let X be the number contactable households.

Checking the Bernoulli properties:

- In this problem there are only two possible outcome that the randomly selected household is contactable or not contactable. Hence 1st condition is satisfied.

- Here the probability of getting contact is 0.76 for each trial. Hence, 2nd condition is also satisfied.

- Each household is selected randomly. Hence, the independence condition is also satisfied.

Hence, the phone calls can be considered as Bernoulli trials.

Hence, the number of phone calls can be approximated as Binomial distribution, that is$X\sim B\in om(1,000,0.76)$ .

Checking success and failure condition:

Here,$np=1,000\times 0.76$

$=760\ge 10$

and$nq=1,000\times (1-0.76)$

$=240\ge 10$

Hence, the success and failure condition is satisfied.

Hence approximately,

$X\sim N(1,000\times 0.76,\sqrt{1,000\times 0.76\times (1-0.76)})$

$X\sim N(760,13.5)$

Hence, the number of successful contacts can be modelled as normal distribution.

Properties of Bernoulli trials:

- In each trial there will be only two possible outcomes, success and failure.

- The probability of success for each trial is same.

- Each trial is independent to each other.

Binomial probability for Bernoulli trials:

The number of successes in a specified number of Bernoulli trials:

The number of successes in a specified number of Bernoulli trials is described by Binomial model.

If the number of trials is n, with the probability of success is p and the probability of failure is

With mean is

Normal approximation from Binomial distribution:

The success and failure condition:

If

Let X be the number contactable households.

Checking the Bernoulli properties:

- In this problem there are only two possible outcome that the randomly selected household is contactable or not contactable. Hence 1st condition is satisfied.

- Here the probability of getting contact is 0.76 for each trial. Hence, 2nd condition is also satisfied.

- Each household is selected randomly. Hence, the independence condition is also satisfied.

Hence, the phone calls can be considered as Bernoulli trials.

Hence, the number of phone calls can be approximated as Binomial distribution, that is

Checking success and failure condition:

Here,

and

Hence, the success and failure condition is satisfied.

Hence approximately,

Hence, the number of successful contacts can be modelled as normal distribution.

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