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College campuses are graying! According to a recent article, one in four college students is aged 30 or older. Assume that the \(25 \%\) figure is accurate, that your college is representative of colleges at large, and that you sample \(n=200\) students, recording \(x\), the number of students age 30 or older. a. What are the mean and standard deviation of \(x ?\) b. If there are 35 students in your sample who are age 30 or older, would you be willing to assume that the \(25 \%\) figure is representative of your campus? Explain.

Short Answer

Expert verified
Is the 25% figure representative of the campus if 35 students in the 200 student sample are aged 30 or older? Answer: The mean number of college students aged 30 or older in a sample of 200 students is 50, and the standard deviation is √37.5. Based on the z-score calculation (-2.44), the 25% figure may not be representative of the campus as the sample of 35 students older than 30 is less likely when the true proportion is 25%.

Step by step solution

01

Determine the values for n and p

We are given that the proportion of college students aged 30 or older is 25%, which means that the probability of a student being older than 30 is \(p = 0.25\). We are also given that we need to sample 200 students, which means that there are 200 trials, or \(n=200\).
02

Calculate the mean

We will use the formula \(\mu = np\) to calculate the mean. Plugging in the values of n and p, we get: \(\mu = (200)(0.25) = 50\)
03

Calculate the standard deviation

We will use the formula \(\sigma = \sqrt{np(1-p)}\) to calculate the standard deviation. Plugging in the values of n and p, we get: \(\sigma = \sqrt{(200)(0.25)(1-0.25)} = \sqrt{(200)(0.25)(0.75)} = \sqrt{37.5}\) So, the mean of x is 50 and the standard deviation of x is \(\sqrt{37.5}\). #b. Determine if 35 students in the sample would be representative#
04

Calculate the z-score

To determine if the sample of 35 students older than 30 is representative of the campus, we need to calculate the z-score. The z-score is calculated as: \(z = \frac{x-\mu}{\sigma}\) In this case, we have x = 35, \(\mu = 50\), and \(\sigma = \sqrt{37.5}\). So the z-score is: \(z = \frac{35-50}{\sqrt{37.5}} \approx -2.44\)
05

Interpret the z-score

A z-score of -2.44 indicates that the sample result of 35 students older than 30 is approximately 2.44 standard deviations below the mean. Typically, a z-score between -2 and 2 is considered within the range of normal variation (approximately 95% of the data). Since the z-score is outside the range of -2 to 2, it would be unusual to find a sample with 35 students older than 30 in a population where the true proportion is 25%. Therefore, based on this sample, you may not assume that the 25% figure is representative of your campus, as having a sample with 35 students older than 30 is less likely when the true proportion is 25%.

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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 specific type of probability distribution used when there are exactly two mutually exclusive outcomes of an experiment. For example, when looking at college students who are 30 years old or older, there are only two possibilities: a student is either 30 or older, or they are not.

In such scenarios, the binomial distribution helps us model the probability of a certain number of successes (students aged 30 or above) in a fixed number of trials (total students sampled).

Key characteristics of a binomial distribution include:
  • A fixed number of trials, denoted as \(n\). In this case, \(n = 200\).
  • Two possible outcomes in each trial, success or failure.
  • A constant probability of success, \(p\). Here, \(p = 0.25\).
To determine probabilities or calculate statistics like the mean or variance, we rely on the binomial distribution formulas for mean, \(\mu = np\), and standard deviation, \(\sigma = \sqrt{np(1-p)}\).

Understanding the binomial distribution helps in assessing the expected and typical outcomes, guiding further statistical analysis like calculation of Z-scores.
Z-score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. When dealing with binomial distributions, it's used for determining how "unusual" a particular finding is.

The formula for calculating a z-score is:\[ z = \frac{x - \mu}{\sigma} \]where:\
  • \(x\) is a single observation or the sample mean we're analyzing (in the exercise, this is 35 students).
  • \(\mu\) is the population mean. For our example, \(\mu = 50\).
  • \(\sigma\) is the standard deviation. Here, it is \(\sqrt{37.5}\).
The result of the z-score informs how many standard deviations away \(x\) is from the mean of the distribution.

A high z-score, whether positive or negative, indicates a result that is deviant from the expected mean. In our situation, a z-score of approximately -2.44 suggests that the fraction of students 30 and older is much less common than we would predict under the assumed conditions, thus questioning the representativeness of the 25% statistic.
Statistical Variation
Statistical variation speaks to how data points in a study differ from each other and from the mean. This concept is essential when examining outcomes like mean and standard deviation, especially in probability distribution models like the binomial distribution.

In the context of our example, understanding variation can help explain why a z-score of -2.44 indicates something may be amiss. When variation is low, most of the data points cluster closely around the mean. Conversely, high variation suggests data points are spread out over a wider range of values.

By calculating the standard deviation (a measure of statistical variation), we understand the dispersion or variability in the sample of students aged 30 and above. The key takeaway here is that significant deviations from the expected mean (in this case, drastically fewer or more students than anticipated) may automatically trigger deeper investigation.

Learning about statistical variation helps us appreciate why certain anomalies warrant further scrutiny and how they can impact the validity of our assumptions about a population.

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