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Chapter 8: Problem 58

In a certain political poll, each person polled has a \(90 \%\) probability of telling his or her real preference. Suppose that 1,000 people are polled and \(51 \%\) say that they prefer candidate Goode, while \(49 \%\) say that they prefer candidate Slick. Find the approximate probability that Goode could do at least this well if, in fact, only \(49 \%\) prefer Goode.

Short Answer

Expert verified
The approximate probability that Goode could do at least this well if, in fact, only 49% prefer him is approximately 0.9394 or 93.94%.

Step by step solution

01

Determine the binomial probability variables

We have the population (n = 1000), the real preference percentage for Goode (p = 0.49), and by default, the real preference percentage for Slick (q = 1 - p = 0.51).
02

Calculate the cumulative probability using the normal approximation (z-score)

Using the normal approximation, we can calculate the z-score which determines how many standard deviations an element is from the mean. This is important for finding the probability of our given situation. The z-score formula is: \(z = \frac{(X - \mu)}{\sigma} \) where X is the number of success (number of people honestly prefer Goode), \(\mu = np\) is the expected mean, and \(\sigma = \sqrt{npq}\) is the standard deviation. From the given information in the question, we have: - n = 1000, - p = 0.49, and - q = 0.51. Since we are to find the probability of at least 515 people honestly preferring Goode, we will use X = 514.5 (we add 0.5 to ensure continuity correction). Now let's find the mean (\(\mu\)) and standard deviation (\(\sigma\)): - \(\mu = np = 1000 * 0.49 = 490\) - \(\sigma = \sqrt{npq} = \sqrt{1000 * 0.49 * 0.51} \approx 15.8\) Lastly, let's find the z-score and use the z-table or calculator (with continuous normal distribution) to find the probability. - \(z = \frac{514.5 - 490}{15.8} \approx 1.55\) Looking up the value 1.55 in a z-table or equivalent calculator, we find that the probability is approximately 0.9394. So, the approximate probability that Goode could do at least this well if, in fact, only 49% prefer him is approximately 0.9394 or 93.94%.

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Key Concepts

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

Normal Approximation
The normal approximation is a robust statistical tool used to estimate the probability of a certain outcome when dealing with binomial distributions. As in our political poll example, when the number of trials (in this case, people polled) is sufficiently large, and the probability of success (people preferring a certain candidate) is known, the binomial distribution of outcomes starts to resemble the normal distribution, which is defined by its mean (average) and standard deviation (spread).

To utilize the normal approximation in the context of our example, we essentially assume that the distribution of the number of people who prefer candidate Goode follows a normal distribution, instead of a binomial one. This technique simplifies the process of finding probabilities for a large number of trials and is particularly helpful when direct calculation of binomial probabilities is cumbersome due to a large sample size.
Z-score Calculation
The z-score essentially tells us how many standard deviations an individual data point is from the mean of the distribution. It represents the relative position of that data point and is crucial in determining the probability of occurrence within a normal distribution.

Calculating a z-score involves subtracting the mean value from the data point (or the opposite if predicting higher values) and then dividing by the standard deviation. The formula is:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In the context of our poll, if the calculated z-score is positive, it indicates that the observed outcome (support for Goode) is above what we would expect by chance alone.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates that the values are spread out over a wider range.

In probability and statistics, the standard deviation plays a pivotal role as it helps us understand the spread and consistency of the data. For a binomial distribution, the standard deviation is calculated using:
\[\sigma = \sqrt{npq}\]
where \(n\) is the number of trials, \(p\) is the probability of success, and \(q = 1 - p\) is the probability of failure. By knowing the standard deviation in our polling example, we can gauge the variability of support for a candidate across the population.
Cumulative Probability
Cumulative probability refers to the likelihood that a random variable takes on a value less than or equal to a particular value. It is the sum of the probabilities of all possible outcomes up to that value. For a normal distribution, it can be represented by the area under the curve to the left of a specific z-score.

In practice, cumulative probabilities are often found using statistical tables or computational tools that integrate the normal distribution curve up to the z-score of interest. In our polling scenario, the cumulative probability tells us the chance that at least a certain percentage of people actually prefer Goode, based on the distribution of the polling data and the properties of the normal distribution.

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Most popular questions from this chapter

Based on data from Nielsen Research, there is a \(15 \%\) chance that any television that is turned on during the time of the evening newscasts will be tuned to \(\mathrm{ABC}\) 's evening news show. \({ }^{58}\) Your company wishes to advertise on a small local station carrying \(\mathrm{ABC}\) that serves a community with 2,500 households that regularly tune in during this time slot. Find the approximate probability that at least 400 households will be tuned in to the show. HINT [See Example 4.]

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In a certain set of scores, the median occurs more often than any other score. It follows that (A) The median and mean are equal. (B) The mean and mode are equal. (C) The mode and median are equal. (D) The mean, mode, and median are all equal.

Y o u r ~ p i c k l e ~ c o m p a n y ~ r a t e s ~ i t s ~ p i c k l e s ~ o n ~ a ~ s c a l e ~ o f spiciness from 1 to \(10 .\) Market research shows that customer preferences for spiciness are normally distributed, with a mean of \(7.5\) and a standard deviation of 1 . Assuming that you sell 100,000 jars of pickles, how many jars with a spiciness of 9 or above do you expect to sell?
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