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According to a November 2008 survey completed by the Pew Internet \& American Life Project [http://www. pewinternet.org], about \(74 \%\) of all adult Internet users say they went online for news and information about the 2008 election or to communicate with others about the election race. Assuming the percentage is correct, use the normal approximation to the binomial to find the probability that in a survey of 2000 American adult Internet users, a. at least 1400 used the Internet for information about the 2008 election. b. at least 1565 used the Internet for information about the 2008 election. c. at most 1500 used the Internet for information about the 2008 election. d. at most 1425 used the Internet for information about the 2008 election.

Short Answer

Expert verified
a. The probability that at least 1400 users used the Internet for information about the 2008 election is approximately 1. b. The probability that at least 1565 users used it for the same purpose is nearly 0. c. The probability that at most 1500 users used it for this purpose is 0.834. d. The probability that at most 1425 used it for the same purpose is 0.9963.

Step by step solution

01

Calculate Mean and Standard Deviation

First, calculate the mean and standard deviation for the given problem. The mean \(\mu\) is given by \(np\) and the standard deviation \(\sigma\) is given by \(\sqrt{np(1-p)}\), where \(n = 2000\) (the size of the survey) and \(p = 0.74\) (the probability of an adult Internet user seeking election information online). Thus, we find \(\mu = 2000 * 0.74 = 1480\) and \(\sigma = \sqrt{2000 * 0.74 * (1-0.74)} = 20.49\).
02

Normalize and Apply Z-score

Next, for each part of the problem, we need to normalize the given value and find the corresponding Z-score. The Z-score is calculated as \((x - \mu) / \sigma\) where \(x\) is the value we are examining. Then we'll look up the resulting Z-score in a standard Z-score table or use a calculator to find the area under the normal distribution curve for the Z-score.
03

Calculate Probabilities

a. Calculate the Z-score for 1400: \((1400 - 1480) / 20.49 = -3.91\). Because we want the probability of at least 1400, we look for the area to the right. The Z-table value for -3.91 is negligible, thus the probability is approximately 1. b. For 1565: \((1565 - 1480) / 20.49 = 4.15\). The Z-table value for -4.15 is negligible, so the opposite (i.e., the area to the right): \(1 - 0 = 1\) indicates there's nearly no chance this would occur. c. For 1500: \((1500 - 1480) / 20.49 = 0.97\). Here we want the area to the left, which on the Z-table is 0.834. d. For 1425: \((1425-1480) / 20.49 = -2.68\). The Z-table value for -2.68 is 0.0037, so the opposite (i.e. the area to the right) becomes approximately 0.9963.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is a probability distribution that represents the number of successes in a fixed number of independent trials of a binary scenario. In simpler terms, imagine flipping a coin multiple times: each flip is an independent event with two possible outcomes (heads or tails). Similarly, when we survey American adult Internet users, each individual has two options: they either used the Internet for election information or they did not.

Regarding our exercise, the probability of success on a single trial is approximated to be 74% or 0.74, representing those who used the Internet for election-related information. The total number of trials is the number of users surveyed, which is 2,000. This scenario, when the number of trials is large, and the probability of success is not too close to 0 or 1, lends itself well to the normal approximation, which simplifies calculations significantly.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. In the context of our problem, if we want to find out how unusual or typical a certain value is, such as the number of users who used the Internet for election details, we compare this to what we'd expect — the mean (or average).

A Z-score can be positive or negative, indicating whether the value is above or below the mean, respectively. To calculate it, we subtract the mean from the value we're interested in and then divide by the standard deviation. It is a crucial part of finding probabilities as it allows us to compare different sets of data that may have different means and standard deviations.
Probability Distribution
A probability distribution is essentially a function that provides the probabilities of occurrence of different possible outcomes in an experiment. It quantifies the expected value of each outcome, which is critical in assessing risk and in making decisions under uncertainty.

In the normal approximation to the binomial distribution, we use a normal probability distribution as a stand-in for our binomial distribution because it is easier to work with, especially when dealing with large sample sizes. This normal distribution is characterized by its bell-shaped curve and is determined by two parameters: the mean and the standard deviation. It tells us how likely it is for different results to occur within our binomial experiment—like the number of users engaging with election content online.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most of the numbers are close to the mean (average). A high standard deviation indicates that the values are spread out over a wider range.

Standard deviation plays a key role in the calculation of Z-scores and, consequently, in the application of the normal approximation to binomial problems. To work out the standard deviation of our Internet users survey, for instance, we consider both the chance that someone would seek out election information (0.74) and the chance they wouldn't (0.26). The resulting standard deviation reflects how much the actual number of users could deviate from the expected average.

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Most popular questions from this chapter

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