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Based on data from ACT in \(2008,\) the average science reasoning test score was \(20.8,\) with a standard deviation of \(4.6 .\) Assuming that the scores are normally distributed, a. find the probability that a randomly selected student has a science reasoning ACT score of least 25. b. find the probability that a randomly selected student has a science reasoning ACT score between 20 and 26. c. find the probability that a randomly selected student has a science reasoning ACT score less than 16.

Short Answer

Expert verified
a) There's approximately 18% probability for a student to score at least 25. b) There's approximately 44% probability for a student to score between 20 and 26. c. There's approximately 15% probability for a student to score less than 16.

Step by step solution

01

- Calculate the Z-score for each scenario

For scenario (a), a score of 25 is given. The formula for calculating the z-score is \( z = (X - μ) / σ \) where X is the value from the data, μ is the mean and σ is the standard deviation. So, \( z = (25 - 20.8) / 4.6 = 0.913 \). For scenario (b), we are given two scores, 20 and 26. We calculate two z-scores: for 20, \( z1 = (20 - 20.8) / 4.6 = -0.174 \); for 26, \( z2 = (26 - 20.8) / 4.6 = 1.130 \). For scenario (c), with a score of 16, \( z = (16 - 20.8) / 4.6 = -1.043 \)
02

- Refer to Z-tables for corresponding probabilities

Once we've obtained the z-scores, we refer to Z-tables to find the corresponding probabilities. Because Z-tables typically give the probability from the mean up to a given score, for scenarios (a) and (c) where we want the probability for scores 'at least' and 'less than' respectively, we need to subtract the obtained value from 1. So, for (a) the probability \( P(X ≥ 25) = 1 - P(Z < 0.913) = 1 - 0.8191 = 0.1809. In scenario (b), since we're looking for a range between two scores, we find the probabilities for both and subtract them: \( P(20 ≤ X ≤ 26) = P(Z < 1.130) - P(Z < -0.174 ) = 0.8708 - 0.4306 = 0.4402. In scenario (c), \( P(X < 16) = P(Z < -1.043) = 0.1482
03

- Interpret the Results

In each case, we have found the probabilities for a student's score to be in a certain range. For (a), there's an 18% chance that a student scores at least 25. For (b), there's a 44% chance that a student scores between 20 and 26. And for (c), there's almost a 15% probability for a student's score to be less than 16.

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Key Concepts

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

Z-score
A Z-score helps to understand how far away a data point is from the mean, measured in standard deviations. It is a way to standardize different data points.
In simpler terms, if the Z-score is positive, the data point is above the mean. If it is negative, the data point is below the mean. This information is key in assessing the likelihood of a score occurring in a normal distribution.
For example, a Z-score of 0.913 means the score is 0.913 standard deviations above the mean. Similarly, a Z-score of -1.043 indicates the data point is 1.043 standard deviations below the mean.
Here's how the Z-score formula works:
  • \( z = \frac{X - \mu}{\sigma} \)
  • X is the data point
  • \( \mu \) is the mean of the dataset
  • \( \sigma \) is the standard deviation
Understanding Z-scores helps find probabilities related to the data point, as seen in our example with ACT scores.
Probability
Probability refers to the likelihood of an event happening. In the context of a normal distribution, it is often used to determine how often a score occurs within certain boundaries.
When dealing with ACT scores, probabilities can answer questions like "What is the chance of scoring at least 25?" or "What is the likelihood of scoring between 20 and 26?"
To find these probabilities, Z-scores are converted using a Z-table. The Z-table provides the probability of a score occurring below a specific Z-score.
For example, if you want the probability of a score being at least 25:
  • Calculate the Z-score (0.913 in this case).
  • Find the probability from the Z-table (0.8191 for 0.913).
  • Subtract from 1 because we want at least 25: \(1 - 0.8191 = 0.1809\).
Understanding probability helps in determining how common or rare any given test score is.
ACT Scores
ACT scores are standardized test results used in college admissions in the United States. They cover several subjects, one being science reasoning.
These scores are normally distributed, which means most students will score around the average, and fewer students will have extremely high or low scores.
In 2008, the average science reasoning score was 20.8 with a standard deviation of 4.6. Understanding the normal distribution of ACT scores allows us to calculate how many students score above or below certain scores.
For example, knowing that scoring above 25 (a Z-score of 0.913) has an 18% probability makes it clear that scoring this high is relatively uncommon compared to the majority who score around the mean. ACT scores illustrate practical applications of statistics in the educational landscape.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In a normal distribution, it indicates how spread out the values are around the mean.
If the standard deviation is high, the data points are spread out over a larger range of values. Conversely, a low standard deviation indicates that the data points are closer to the mean.
In the case of the science reasoning ACT scores, the standard deviation is 4.6. This tells us that each student's score will typically vary by 4.6 points from the average score of 20.8.
Understanding standard deviation is crucial because it provides context for interpreting different scores. A smaller standard deviation might indicate that most scores are close to the average, whereas a larger one would suggest the scores are more variable. This measurement makes it easier to analyze and explain the distribution of students' test results.

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