Problem 3 - Algebra 2

Question

For Exercises \(2-4,\) use the following information. Mr. Bash gave a quiz in his social studies class. The scores were normally distributed with a mean of 21 and a standard deviation of \(2 .\) What percent would you expect to score between 23 and 25\(?\)

Step-by-Step Solution

Answer

Approximately 13.59% would score between 23 and 25.

Step 1: Understand the Normal Distribution

The quiz scores are normally distributed with a mean, \( \mu = 21 \), and a standard deviation, \( \sigma = 2 \). We want to find the percentage of scores that fall between 23 and 25.

Step 2: Convert to Z-scores

We need to convert the raw scores of 23 and 25 to Z-scores using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \]For 23: \[ Z = \frac{23 - 21}{2} = 1 \]For 25:\[ Z = \frac{25 - 21}{2} = 2 \]

Step 3: Use Z-table to Find Probabilities

Look up the Z-scores in the standard normal distribution (Z) table. The Z-table gives the probability that a score is less than a given value.- For \( Z = 1 \), the cumulative probability is approximately 0.8413.- For \( Z = 2 \), the cumulative probability is approximately 0.9772.

Step 4: Calculate the Probability for the Range

The probability of a score being between 23 and 25 is the difference between the probabilities of Z = 2 and Z = 1:\[ P(23 < X < 25) = P(Z < 2) - P(Z < 1) \]\[ P(23 < X < 25) = 0.9772 - 0.8413 = 0.1359 \]

Step 5: Convert Probability to Percentage

To convert the probability to a percentage, multiply by 100:\[ 0.1359 \times 100 \approx 13.59\% \]

Key Concepts

Z-Score

The Z-score is an important concept in statistics that helps us understand how far away a data point is from the mean of a dataset. This measure is particularly helpful when dealing with a normally distributed set of data, like test scores.

• A positive Z-score indicates the score is above the mean.
• A negative Z-score indicates the score is below the mean.
• A Z-score of zero means the score is exactly at the mean.

Calculating a Z-score involves using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the raw score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For example, in Mr. Bash's quiz, a score of 23 results in a Z-score of 1. This means 23 is one standard deviation above the mean. Meanwhile, a score of 25 corresponds to a Z-score of 2, indicating it is two standard deviations above the mean.

Mean and Standard Deviation

The mean and standard deviation are two fundamental concepts in statistics that describe data distribution. The mean is the average of all data points, while the standard deviation measures the spread of the data around the mean.

• The mean is calculated by adding all data scores and dividing by the number of scores.
• The standard deviation is the square root of the variance, providing insight into the dispersion of scores.

In the context of Mr. Bash's quiz:- The mean score \( \mu \) is 21, indicating the average score attained by the students.- The standard deviation \( \sigma \) is 2, which tells us how much the scores typically deviate from the mean, with smaller values indicating scores close to the mean, and larger values showing more spread.

Probability and Statistics

Probability and statistics are essential areas of mathematics that deal with analyzing and interpreting data. They help to predict the likelihood of various outcomes.

• Probability refers to the measure of the likelihood that an event will occur.
• Statistics involves collecting, analyzing, and interpreting data to make informed decisions.

In the exercise involving Mr. Bash's quiz scores: - We use probability to find the percentage of students scoring between 23 and 25. - This involves calculating the probability of scores within a Z-score range.
Using the Z-table, the cumulative probability for different Z-scores is found: For a Z-score of 1, the cumulative probability is 0.8413, and for a Z-score of 2, it’s 0.9772. The probability for the scores falling between these Z-scores is the difference, which results in approximately 13.59%. By converting this probability into a percentage, we understand that about 13.59% of students scored within this range.