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A z-score, often used in the context of a normal distribution, represents the number of standard deviations a particular value is from the mean. It is a useful way to understand how far, and in what direction, a particular value is from the average observation. For the algebra exam example, the z-score helps us determine how extreme or not the time of 30 minutes is compared to the average time of 28 minutes.

To calculate the z-score, use the formula: \[ z = \frac{X - \mu}{\sigma} \]

• Here, \( X \) is the value you're examining, which is 30 minutes in this case.
• \( \mu \) is the mean, 28 minutes.
• \( \sigma \) is the standard deviation, 1.5 minutes.

By applying these values, we find the z-score to be 1.33. This tells us that the 30-minute mark is 1.33 standard deviations above the mean. Understanding this helps us in finding the probability associated with this z-score by looking up the values in a standard normal distribution table.

Standard deviation is a measure that tells us how spread out the values in a dataset are around the mean. In the context of exam times, it helps us understand the variability in the time it takes students to complete the exam. A smaller standard deviation indicates the times are clustered closely around the mean, while a larger one suggests a more spread out distribution. For our algebra exam problem, the standard deviation is 1.5 minutes. This relatively small value suggests that most students finish the exam close to the 28-minute average.

• Helps determine z-score: It serves as the denominator in the z-score formula, giving us standard units to assess how typical or atypical a score is.
• Informs about distribution spread: Assists in understanding how much variation there is.

By helping to convert various times to z-scores, the standard deviation plays a crucial role in calculating probabilities in a normal distribution.

Probability, in this scenario, refers to the likelihood of an event occurring. When dealing with a normal distribution, such as the exam completion times, we can calculate probabilities using z-scores.Once we have a z-score, we can find how probable it is for an event at that z-score to occur using a Z-table or calculator. In the steps provided, we found that a z-score of 1.33 corresponds to a probability of 0.9082, meaning about 90.82% of students will finish within the 30-minute time limit.

Now, if you want to see the probability of a group of independent events happening together, you multiply the probabilities of the individual events. For the six students, we calculate: \[P(\text{all six finish}) = (0.9082)^6 \]

• This gives about 0.6012.
• Interpretation: There's a 60.12% chance that all six students will complete the exam in 30 minutes.

Thus, probability helps in making informed predictions about events and their likelihoods, a critical application in assessing performance scenarios.