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Standard deviation is a key concept in statistics that helps us understand how data points differ from the mean. In a normal distribution, most data points cluster around the mean, but some are spread out. The standard deviation, denoted as \(\sigma\), measures this spread.
For example, if a set of test scores has a standard deviation of \(1.6\), like in our ITBS vocabulary scores example, it tells us how much the students' scores vary from the average score of \(6.8\).

• A small standard deviation means data points are close to the mean.
• A large standard deviation indicates more spread out data.
• In our problem, a standard deviation of \(1.6\) suggests moderate variability around the average score.

Understanding the standard deviation gives insight into the consistency of students' test results, showing if most scores are around the average or if there are many that deviate significantly.

To work with normally distributed data, it's often useful to convert raw scores into standard scores, known as z-scores. This process is called standardization. By calculating a z-score, we transform a raw score into a value that represents how many standard deviations the data point is from the mean. The z-score formula is \[ z = \frac{X - \mu}{\sigma} \].
In the context of our ITBS example, we calculated the z-score for a score of \(9\):

• This involved subtracting the mean \(\mu = 6.8\) from the score \(9\).
• Then, dividing the result \((2.2)\) by the standard deviation \(\sigma = 1.6\).
• This gives us a z-score of about \(1.375\).

A z-score tells us where a score stands in relation to the mean and allows us to use statistical tables or technology to find probabilities related to the normal distribution.

Probability in the context of normal distribution is all about finding the likelihood of a certain score or range of scores. Once we've calculated the z-score, we can use a standard normal distribution table to find related probabilities.
In our problem, we wanted to find the probability that a student scores \(9\) or higher on the test. After converting this to a z-score of \(1.375\), we looked up the probability of a score being below \(1.375\), which was approximately \(0.9154\). Thus, the probability of scoring higher than \(9\) is \[ P(X \geq 9) = 1 - P(Z < 1.375) = 1 - 0.9154 \approx 0.0846 \].
This outcome means there's an \(8.46\)% chance that a student will score \(9\) or more. In simpler terms, among 100 students, we can expect around \(8 ext{-}9\) to achieve a score of \(9\) or higher. Understanding these probabilities helps educators assess student performance and make critical decisions regarding instructional needs.