Question

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is 89. Did you receive an A? Explain.

Short answer

Yes, with a score of 89, an 'A' was received because the Z-score of this score (1.57) is greater than the Z-score corresponding to the top 15% of scores (1.04).

Step-by-step solution

Calculate the Z-score

To calculate the z-score, use the formula: z = (X - μ) / σ. Here, X is the given score of 89, μ is the mean (78), and σ is the standard deviation (7). So, z = (89 - 78) / 7 = 1.57.

Compare the Z-score to standard Z-values

Next, you need to compare this Z-score to the standard Z-values to find out if it falls in the top 15%. From the standard normal tables or using calculation (such as in a statistical software), a Z-value of 1.04 approximately corresponds to the top 15%. This means that only the scores with a Z-value greater than 1.04 are given 'A's.

Draw Conclusion

Finally, we compare the calculated Z-value of 1.57 to the Z-value for top 15% (1.04). Since 1.57 > 1.04, we conclude that the score of 89 did fall into the top 15% of the class, and thus an 'A' was received.

Key concepts

Z-score Calculation

Understanding the concept of a Z-score is fundamental in statistics, particularly when working with the normal distribution. A Z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean of a data set. To calculate a Z-score, the formula \( z = \frac{X - \mu}{\sigma} \) is employed, where \( X \) is a value in the dataset, \( \mu \) is the mean of the dataset, and \( \sigma \) the standard deviation.

For example, if you score 89 on a test, and the test scores are normally distributed with a mean (average) of 78 and a standard deviation of 7, your Z-score would be \( z= \frac{89 - 78}{7} = 1.57 \) which indicates that your score is 1.57 standard deviations above the mean score of the class. A high Z-score means the data point is very unusual compared to the average result, making it a critical tool for determining how remarkable a particular score is within the distribution.

Standard Deviation

The standard deviation is a crucial concept in statistics as it measures 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, while a high standard deviation indicates that the values are spread out over a wider range.

To calculate standard deviation, one must execute a series of steps including squaring the difference between each data point and the mean, averaging these values, and then taking the square root of this average. Mathematically, it is typically represented with the symbol \( \sigma \) for a population and \( s \) for a sample. In our exercise, the standard deviation of the midterm exam scores was given as 7, meaning that most students' scores were, on average, 7 points away from the mean in either direction. This measure helps determine the distribution of scores and is essential when calculating Z-scores.

Normal Probability Tables

Normal probability tables, often known as Z-tables, are tools that allow users to find the probability of a score falling within a particular section of a normal distribution. These tables represent the areas under the curve for a standard normal distribution as it relates to different Z-scores. Since it's impractical to calculate these areas by hand for every Z-score, the tables provide a quick reference.

The tables typically show the probability that a statistic is less than a given Z-score. In our example, after calculating a Z-score of 1.57, we'd refer to the normal probability table to see what percentile that score represents. The table would tell us the proportion of scores falling below a Z-score of 1.57. To determine if this score is within the top 15%, we would look for the Z-score associated with 85% (100% - 15%), which typically is around 1.04. Since 1.57 is greater than 1.04, it indicates that the score of 89 would indeed be in the top 15% of scores, confirming an 'A' grade.