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taurus [48]
4 years ago
5

Internet sites often vanish or move so that references to them cannot be followed. In fact, 13% of Internet sites referenced in

major scientific journals are lost within two years after publication. If a paper contains nine Internet references, what is the probability that all nine are still good two years later? Round your answer to three decimal places. What specific assumption must be made in order to calculate the probability? The paper containing the references must be obtained by random sampling. The occurrence of the site references in the paper are independent events. The occurrence of the site references in the paper are disjoint events, One does not need to make any assumptions; this is just a straightforward calculation. P(all nine are still good)=
Mathematics
1 answer:
Dmitriy789 [7]4 years ago
4 0

Answer:

The probability that all nine references are still good two years later is P(all nine are still good)=0.29.

The assumption we have to make is that the occurrence of the site references in the paper are independent events.

Step-by-step explanation:

The assumptions we have to make about the papers is that their probabilities of survival are independent of each other. This is the same as saying: The occurrence of the site references in the paper are independent events.

The probabilities of survival of any reference is:

P(survive)=1-P(lost)=1-0.13=0.87

The probability of the nine references are still good later can be calculated as:

P=\prod_{k=1}^{9}P(x_k=S)=P(x=S)^9=0.87^9=0.29

The probability that all nine references are still good two years later is P=0.29.

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Answer:

n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337

So the answer for this case would be n=1337 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=1.1 represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got z_{\alpha/2}=2.326, replacing into formula (b) we got:

n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337

So the answer for this case would be n=1337 rounded up to the nearest integer

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⇒ 180-90-50

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