Problem 33 - Algebra 2

Question

At a second bowling tournament, the mean of all the scores was \(205,\) with a standard deviation of \(14 .\) What was the z-score for a score of 282\(?\)

Step-by-Step Solution

Answer

The z-score for a score of 282 is 5.5

Step 1 - Understanding and writing down the given information

We have the mean (average) of the scores given as \(205,\) the standard deviation as \(14,\) and the score we are to find the Z-score for as \(282.\)

Step 2 - Understand the Z-score

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score. A z-score of 1.0 indicates a value that is one standard deviation from the mean. Z-scores can be positive or negative, with a positive value indicating the score is above the mean and a negative score indicating it is below the mean.

Step 3 - Applying the formula for Z-score

The formula for calculating Z-score is: \[Z = \frac{(X - \mu)}{\sigma},\] where \(X\) is the data point (here, the score we are looking to find the z-score for, which is \(282\)), \(\mu\) is the mean (here, the mean is \(205\)) and \(\sigma\) is the standard deviation (here, the standard deviation is \(14\)). Applying this formula, the calculation will look like this: \[Z = \frac{(282 - 205)}{14}.\]

Step 4 - Calculate the Z-score

Carrying out the subtraction and division in the formula, we find: \[Z = \frac{77}{14} = 5.5.\] Therefore, the z-score for a score of \(282\) is \(5.5.\)

Key Concepts

Understanding the Mean

The mean, often referred to as the average, is one of the most basic statistical measurements. It helps in understanding the central point of a data set. To calculate the mean, sum up all the values in the data set and divide by the number of values.
For example, if you have scores of bowling games like 180, 200, and 235, the mean would be calculated as follows:

• Add all the scores together: 180 + 200 + 235 = 615.
• Count the number of games: 3.
• Divide the total sum by the number of games: 615 / 3 = 205.

This calculated mean gives you a sense of the typical performance in the games.

Deciphering Standard Deviation

Standard deviation is a key statistical measurement that assesses 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 implies a wider range of values.
Think of it like measuring how scores in a bowling tournament spread out from the average score. If the standard deviation is small, most players have scores near the average. If it's large, the scores are more spread out.
In the bowling tournament example, a standard deviation of 14 suggests that most scores are within 14 points of the mean score of 205.

The Role of Statistical Measurement in Evaluation

Statistical measurements are crucial for making informed decisions based on data. Among these, means and standard deviations are central as they describe data tendencies and dispersion. They help you understand how data groups compare.
In evaluations, the mean provides insight into the overall trend, while standard deviation tells us about the consistency.
If assessing whether a bowler's score of 282 is exceptional, look at the mean (205) and the standard deviation (14). The large gap indicates the bowler did exceptionally well.

Score Calculation Using Z-score Formula

Z-score is a special type of statistical measurement for determining how a particular score compares to the mean. It tells how many standard deviations a specific score is from the mean.

• A positive z-score means the value is above the mean.
• A negative z-score indicates it's below the mean.

The formula for calculating the z-score is: \[Z = \frac{(X - \mu)}{\sigma},\] where:

• \(X\) is the score in question (282 in the example).
• \(\mu\) is the mean (205)
• \(\sigma\) is the standard deviation (14).

Using these values:\[Z = \frac{(282 - 205)}{14} = 5.5.\]The result of 5.5 means the score of 282 is 5.5 standard deviations above the mean, making it far outlier and significantly higher than the average bowling score.