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mart [117]
3 years ago
13

How might an organization like Pew Research obtain an accurate estimate of the number of cell-phone only households in the first

place? What kind of sample would you need to get this estimate? How would you contact this sample?
Mathematics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:

a) Sample of households using cell-phone only in a particular region or in a country.

b) It has to be Systematic sampling techniques

Step-by-step explanation:

a) Since the goal of the organization is to estimate the number of cell phone only households. Only households that uses cell-phone only is the appropriate population to sample from. The first step is to identify the households that fall into the population of interest. Then number them systematically and sample from the frame.

b) The moment an household is selected into the sample from the frame, the researcher can visit to conduct the survey. Or, he or she can use other method of data collection such as phoning the household before physical visit, etc.

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Lubov Fominskaja [6]

Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

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And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

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\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

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