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A normal distribution is a continuous probability distribution characterized by its bell-shaped curve. This curve is symmetric about its mean (average) value. In the example exercise, the mean is given as \( \mu = 50 \) and the standard deviation \( \sigma = 7 \). When a random variable like X is normally distributed, its probability distribution follows this bell-shaped curve.
Some important properties of a normal distribution include:

• The mean, median, and mode are all equal.
• It is defined by two parameters: its mean \( \mu \) and standard deviation \( \sigma \).
• Approximately 68% of the data falls within one standard deviation from the mean.
• About 95% of the data lies within two standard deviations.
• Nearly all (99.7%) of the data are within three standard deviations.

In our case, to find the 81st percentile, we need to understand how these properties help us use the normal distribution table (or Z-score table). The percentile shows how the values of X are distributed along this curve.

The Z-score, also called the standard score, helps us understand how far a particular value is from the mean in terms of standard deviations. The Z-score formula is given by:
\[ Z = \frac{x - \mu}{\sigma} \]

• Where Z is the Z-score.
• X is the value whose percentile we need to determine.
• \( \mu \) is the mean of the distribution.
• \( \sigma \) is the standard deviation.

By converting our value into a Z-score, we can use standard Z-score tables to find the percentile or other statistics. For the 81st percentile, we look up the Z-score that corresponds to 0.81 in the Z-table. The Z value for the 81st percentile is approximately 0.88. This Z value tells us that 81% of the data falls below this value.

The mean and standard deviation are critical parameters for any normal distribution. The mean, represented by \( \mu \), indicates the central value or the 'average' of the distribution. Standard deviation, denoted as \( \sigma \), measures the dispersion or spread of the data points.
In the exercise:

• The mean is \( \mu = 50 \), which is the center of our normal distribution.
• The standard deviation is \( \sigma = 7 \), showing how much the values deviate from the mean on average.

Together, these help us understand the location and spread of data in a normal distribution. They are foundational in calculating Z-scores and subsequently finding percentiles in normal distributions.
For example, using the Z-score formula:
\[ x = Z \sigma + \mu \]
Helps us convert the Z value back to a specific data point in our distribution. In this context, it's vital for computing the 81st percentile:
\[ x = 0.88 \times 7 + 50 = 56.16 \]

Percentiles are used in statistics to describe how a value compares to the rest of the data. A percentile represents the value below which a given percentage of observations fall. For example, the 81st percentile is the value below which 81% of the data lies.
In a normal distribution, these percentiles help us understand the spread and distribution of the data. To find the percentile, we use the Z-score, which converts a normal distribution value into a standard normal distribution (mean of 0 and standard deviation of 1).
In the exercise, we calculated the Z value corresponding to the 81st percentile, which is about 0.88. We then used this Z-score to find the specific data value (X) in our original distribution:
\[ x = Z \sigma + \mu = 0.88 \times 7 + 50 = 56.16 \]
So, 56.16 is the value below which 81% of the data in our normal distribution lies. Percentiles are especially useful in comparing scores and understanding relative standings, such as test scores in education or measurements in health-related fields.