Problem 12 - Algebra 2 and Trigonometry

Question

In \(11-14,\) select the numeral that precedes the choice that best completes the statement or answers the question. The scores of a test are normally distributed. If the mean is 50 and the standard deviation is \(8,\) then a student who scored 38 had a z-score of $$ \begin{array}{llll}{\text { (1) } 1.5} & {\text { (2) }-1.5} & {\text { (3) } 12} & {\text { (4) }-12}\end{array} $$

Step-by-Step Solution

Answer

The correct choice is (2) -1.5.

Step 1: Understand the z-score Formula

The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is \( z = \frac{x - \mu}{\sigma} \), where \( x \) is the student's score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Step 2: Identify the Given Values

In the problem, the mean \( \mu \) is 50, the standard deviation \( \sigma \) is 8, and the student's score \( x \) is 38.

Step 3: Calculate the z-score

Substitute the given values into the z-score formula: \( z = \frac{38 - 50}{8} \). This simplifies to \( z = \frac{-12}{8} \).

Step 4: Simplify the Expression

Calculate \( z = \frac{-12}{8} \), which simplifies to \( z = -1.5 \).

Step 5: Identify the Correct Choice

Based on the calculated z-score of \(-1.5\), the correct option from the given choices is (2) -1.5.

Key Concepts

normal distribution

Normal distribution is a key concept in statistics. It is useful to understand when dealing with data sets or analyzing patterns. This type of distribution has a characteristic bell-shaped curve. Many natural phenomena fall into this pattern. For instance, test scores often resemble a normal distribution. This happens when most score near an average, with fewer at the extremes.

Let's break down some characteristics of normal distribution:

• Symmetry: The curve is symmetrical around the mean. This means that the left side of the curve is a mirror image of the right side.
• Mean, median, and mode coincidence: In a perfectly normal distribution, these three measures of central tendency are all the same.
• 68-95-99.7 Rule: About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

This makes it easier to make predictions about how data behave.

standard deviation

Standard deviation is a statistical tool used to measure the amount of variability or dispersion in a set of data. In the context of a normal distribution, it tells us how much the data points deviate from the mean score. A smaller standard deviation indicates that the data points are closer to the mean.

Consider the following points about standard deviation:

• Consistency: A smaller standard deviation implies that the scores are more consistent and closely clustered around the mean.
• Data spread: A larger standard deviation suggests a wider spread of scores, with more variation.
• Use in z-score calculation: It influences the z-score, as you divide by the standard deviation in the calculation.

In the exercise, the standard deviation is 8, revealing how scores are spread around the mean of 50.

mean

The mean, often known as the average, is a measure of central tendency in data. It shows the typical value in a data set and is a foundation for other statistical measures. Calculating the mean involves adding up all the values and then dividing by the number of values.

Here are key points about the mean:

• Simplicity: Being straightforward to calculate makes it a popular and easily understandable measure.
• Influence by outliers: The mean is sensitive to extremely high or low values, known as outliers, which can skew the result.
• Conceptual foundation: The mean is vital in concepts like z-score calculation and normal distributions.

In our problem, the mean test score is 50, which acts as a benchmark to compare other scores.

z-score calculation

The z-score is a statistical measure that indicates how many standard deviations a data point is from the mean. It's useful for understanding a score's position within the data set. Calculating the z-score helps identify how typical or unusual a given value is.

To calculate the z-score, use this formula:

• \[ z = \frac{x - \mu}{\sigma} \]
• x: The value you are evaluating (e.g., a student's score).
• \(\mu\): The mean of the data set.
• \(\sigma\): The standard deviation.

In the exercise, for a score of 38, the z-score is calculated as \[ z = \frac{38-50}{8} \]Which simplifies to \[ z = -1.5 \]This means the score is 1.5 standard deviations below the mean.