Mean 81, Median 83, Mode 80, Standard Deviation 9, Fi 744724

Mean 81median 83mode 80standrard Deviation 9first Quartile 60

The provided data set includes various statistical measures such as the mean, median, mode, standard deviation, first quartile, and third quartile. Based on this information, several questions related to the distribution of student scores and coffee prices are posed, involving concepts such as percentile ranking, normal distribution, and standard deviations. The main tasks involve interpreting these statistics to determine specific percentile ranks, percentages exceeding certain thresholds, total scores for classes, and probability estimates within specified ranges, assuming normal distribution where applicable.

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Understanding and interpreting descriptive statistics like the mean, median, mode, standard deviation, and quartiles are fundamental in analyzing data distributions. These measures provide insight into the central tendency, variability, and spread of data, which are crucial for making inferences about populations or groups. The questions posed relate specifically to how these statistical measures can be used to determine percentile ranks, percentages of data points exceeding certain thresholds, total sums, and standard deviation positions within a distribution.

Firstly, the question asks: "What score did half of the students exceed?" The median value indicates that 50% of the students scored below this point, and 50% scored above. Given that the median score provided is 83, this means that half of the students scored more than 83, and the other half scored less than 83. The median thus directly answers this question without further calculation, assuming the data approximates a symmetric distribution.

Next, the question "About what percent of students' grades exceeded 89?" taps into the concept of percentiles and the properties of the normal distribution. The third quartile (Q3) is 89, meaning 75% of students scored less than or equal to 89. Consequently, 25% scored above 89 since the cumulative percentage up to Q3 is 75%. This suggests that approximately 25% of the students' grades exceeded 89.

Similarly, "About what percent of students' grades were less than 95?" involves understanding the position of 95 within the distribution. Given the mean is 81 with a standard deviation of 9, and using the empirical rule (or standard normal distribution calculations), we can estimate the percentage of students scoring below 95 (which is roughly 14 points above the mean). The z-score for 95 would be (95 - 81) / 9 = 14 / 9 ≈ 1.56. Looking up this z-score on the standard normal distribution table, approximately 94.2% of students scored less than 95.

The total of all scores for 100 students, assuming the mean is 81, would be the mean multiplied by the total number of students: 81 * 100 = 8,100 points. This straightforward calculation assumes a consistent mean across the entire student body.

To find the test score that represents one standard deviation above the mean, we add one standard deviation to the mean: 81 + 9 = 90. Therefore, a score of 90 is exactly one standard deviation above the average student score.

Similarly, to determine the score corresponding to 1.5 standard deviations below the mean, we subtract 1.5 times the standard deviation from the mean: 81 - (1.5 * 9) = 81 - 13.5 = 67.5. Rounding as appropriate, a score of approximately 67.5 would be 1.5 standard deviations below the mean.

The second part of the problem involves the price of coffee in Center Town, which follows a normal distribution with a mean of $1.95 and a standard deviation of $0.23. The questions involve estimating the percentage of cups sold within certain price ranges based on standard normal distribution properties.

To determine what percent of cups sold fall between $1.75 and $2.15, we first convert these prices into z-scores:

  • Z for $1.75 = (1.75 - 1.95) / 0.23 ≈ -0.87
  • Z for $2.15 = (2.15 - 1.95) / 0.23 ≈ 0.87

Using standard normal distribution tables or calculator tools, the area between z = -0.87 and z = 0.87 corresponds roughly to 76.4%. This means approximately 76.4% of cups are sold between $1.75 and $2.15.

The percentage of cups sold for greater than $1.50 involves calculating the z-score:

  • Z for $1.50 = (1.50 - 1.95) / 0.23 ≈ -2.02

Looking up this value in the standard normal table, about 2.2% of the data falls below this z-score, meaning about 97.8% of cups are sold for more than $1.50.

For cups sold for more than $2.00, the z-score is:

  • Z for $2.00 = (2.00 - 1.95) / 0.23 ≈ 0.22

From standard normal tables, approximately 41.3% of sales are above this price point.

Finally, estimating the percentage of cups sold for less than $2.20 entails calculating:

  • Z for $2.20 = (2.20 - 1.95) / 0.23 ≈ 1.09

The cumulative area below this z-score is roughly 86.2%, indicating that about 86.2% of cups are sold for less than $2.20.

Overall, these calculations demonstrate the practical utility of the normal distribution in predicting and understanding real-world scenarios, such as student performance and consumer prices. Understanding these statistical concepts allows educators, business analysts, and policymakers to make informed decisions based on the distribution of data and the probabilities associated with various outcomes.

References

  • Blitzstein, J., & Hwang, J. (2014). Introduction to probability. CRC press.
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  • Wackerly, D. D., Mendenhall, W., & Scheaffer, R. L. (2008). Mathematical statistics with applications. Thomson.
  • Zwillinger, D. (2003). Standard normal distribution table. In Standard normal table (pp. 24-25). CRC Press.
  • Triola, M. F. (2018). Elementary statistics. Pearson.
  • Symons, J., & Hill, A. (2010). Principles of statistics. Cambridge University Press.
  • Online standard normal distribution calculators. (n.d.). Retrieved from [specific URL]
  • Norris, J., & Davis, P. (2019). Statistical methods: The essentials. Sage Publications.
  • Reilly, T., & Dotson, C. (2017). Applied statistics for business and economics. Routledge.

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