A Professor Knows That Her Statistics Students' Final Exam Scores Have A Mean Of 78 And A Standard Deviation Of 11.3. In Her Class, An A Is Any Exam Score Of 90 Or Higher. This Quarter She Has 25 Students In Her Class. What Is The Probability That 6

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Introduction

As a professor of statistics, it is essential to understand the concepts and principles of probability and statistics to effectively teach and guide students. In this article, we will delve into a real-world scenario where a professor wants to calculate the probability of her students achieving high scores on a final exam. The professor has 25 students in her class, and she wants to know the probability that 6 or more students will score 90 or higher, which is the criteria for an "A" grade.

Understanding the Problem

The professor knows that the mean score of her students is 78, with a standard deviation of 11.3. This information is crucial in understanding the distribution of scores in the class. The standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

Calculating the Probability

To calculate the probability that 6 or more students will score 90 or higher, we need to use the concept of the normal distribution. The 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. In this case, we can use the z-score formula to calculate the probability.

The z-Score Formula

The z-score formula is:

z = (X - μ) / σ

where:

  • z is the z-score
  • X is the value of the element
  • μ is the mean of the dataset
  • σ is the standard deviation of the dataset

In this case, we want to find the probability that a student will score 90 or higher. We can use the z-score formula to calculate the z-score for a score of 90.

Calculating the z-Score

Using the z-score formula, we get:

z = (90 - 78) / 11.3 z = 12 / 11.3 z = 1.07

Interpreting the z-Score

The z-score of 1.07 indicates that a score of 90 is 1.07 standard deviations above the mean. This means that about 85.1% of the students in the class will score below 90, and about 14.9% will score 90 or higher.

Calculating the Probability of 6 or More Students Scoring 90 or Higher

To calculate the probability that 6 or more students will score 90 or higher, we need to use the binomial distribution. The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, where each trial has a constant probability of success.

The Binomial Distribution Formula

The binomial distribution formula is:

P(X = k) = (nCk) * (p^k) * (q^(n-k))

where:

  • P(X = k) is the probability of k successes
  • n is the number of trials
  • k is the number of successes
  • p is the probability of success
  • q is the probability of failure
  • nCk is the number of combinations of n items taken k at a time

In this case, we want to find the probability that 6 or more students will score 90 or higher. We can use the binomial distribution formula to calculate the probability.

Calculating the Probability

Using the binomial distribution formula, we get:

P(X ≥ 6) = 1 - P(X < 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))

We can calculate the probability of each of these events using the binomial distribution formula.

Calculating the Probability of 0, 1, 2, 3, 4, and 5 Students Scoring 90 or Higher

Using the binomial distribution formula, we get:

P(X = 0) = (25C0) * (0.149^0) * (0.851^25) = 0.0003

P(X = 1) = (25C1) * (0.149^1) * (0.851^24) = 0.0063

P(X = 2) = (25C2) * (0.149^2) * (0.851^23) = 0.0613

P(X = 3) = (25C3) * (0.149^3) * (0.851^22) = 0.1739

P(X = 4) = (25C4) * (0.149^4) * (0.851^21) = 0.3125

P(X = 5) = (25C5) * (0.149^5) * (0.851^20) = 0.3846

Calculating the Probability of 6 or More Students Scoring 90 or Higher

Using the binomial distribution formula, we get:

P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)) = 1 - (0.0003 + 0.0063 + 0.0613 + 0.1739 + 0.3125 + 0.3846) = 0.9689

Conclusion

In conclusion, the probability that 6 or more students will score 90 or higher in the professor's class is approximately 0.9689. This means that there is a high probability that 6 or more students will achieve an "A" grade in the class.

References

  • Moore, D. S., & McCabe, G. P. (2013). Introduction to the practice of statistics. W.H. Freeman and Company.
  • Johnson, R. A., & Wichern, D. W. (2007). Applied multivariate statistical analysis. Prentice Hall.
  • DeGroot, M. H., & Schervish, M. J. (2012). Probability and statistics. Addison-Wesley.
    A Professor's Dilemma: Calculating the Probability of High-Scoring Students - Q&A ====================================================================================

Introduction

In our previous article, we explored a real-world scenario where a professor wanted to calculate the probability of her students achieving high scores on a final exam. We used the concept of the normal distribution and the binomial distribution to calculate the probability that 6 or more students will score 90 or higher. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the normal distribution?

A: The 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.

Q: How is the z-score formula used in this scenario?

A: The z-score formula is used to calculate the z-score for a given score. In this scenario, we used the z-score formula to calculate the z-score for a score of 90.

Q: What is the significance of the z-score in this scenario?

A: The z-score indicates how many standard deviations a score is away from the mean. In this scenario, the z-score of 1.07 indicates that a score of 90 is 1.07 standard deviations above the mean.

Q: How is the binomial distribution used in this scenario?

A: The binomial distribution is used to calculate the probability of 6 or more students scoring 90 or higher. We used the binomial distribution formula to calculate the probability.

Q: What is the significance of the binomial distribution in this scenario?

A: The binomial distribution is used to model the number of successes in a fixed number of independent trials, where each trial has a constant probability of success. In this scenario, we used the binomial distribution to calculate the probability of 6 or more students scoring 90 or higher.

Q: How is the probability of 6 or more students scoring 90 or higher calculated?

A: The probability of 6 or more students scoring 90 or higher is calculated using the binomial distribution formula. We calculated the probability of 0, 1, 2, 3, 4, and 5 students scoring 90 or higher and then subtracted the sum of these probabilities from 1 to get the probability of 6 or more students scoring 90 or higher.

Q: What is the significance of the probability of 6 or more students scoring 90 or higher?

A: The probability of 6 or more students scoring 90 or higher indicates the likelihood of 6 or more students achieving an "A" grade in the class. In this scenario, the probability is approximately 0.9689, indicating a high likelihood of 6 or more students achieving an "A" grade.

Q: What are some common applications of the normal distribution and binomial distribution?

A: The normal distribution and binomial distribution have many common applications in statistics and data analysis. Some common applications include:

  • Modeling the distribution of exam scores
  • Modeling the number of successes in a fixed number of independent trials
  • Calculating probabilities and percentiles
  • Analyzing the relationship between variables

Conclusion

In conclusion, the normal distribution and binomial distribution are powerful tools for modeling and analyzing data. In this article, we answered some frequently asked questions related to the scenario of a professor calculating the probability of her students achieving high scores on a final exam. We hope that this article has provided a better understanding of the normal distribution and binomial distribution and their applications in statistics and data analysis.

References

  • Moore, D. S., & McCabe, G. P. (2013). Introduction to the practice of statistics. W.H. Freeman and Company.
  • Johnson, R. A., & Wichern, D. W. (2007). Applied multivariate statistical analysis. Prentice Hall.
  • DeGroot, M. H., & Schervish, M. J. (2012). Probability and statistics. Addison-Wesley.