### Understanding Percentiles in a Normal DistributionIn this exercise, we consider a normally distributed set of values characterized by a mean (average) of 187.9 and a standard deviation of 71.2.To solve the question, we are asked to identify **P₈₀**, which is the value below which 80% of the observations fall (also known as the 80th percentile). This score will effectively separate the bottom 80% of the data from the top 20%.Given:- Mean (μ) = 187.9- Standard Deviation (σ) = 71.2To find **P₈₀**, follow these steps:1. **Determine the Z-score for the 80th percentile**: The Z-score indicates how many standard deviations a value is from the mean. For the 80th percentile (P₈₀), the Z-score can be found in a Z-table or using statistical software or a calculator. The Z-score for the 80th percentile is approximately **0.8416**.2. **Convert the Z-score to the actual value**: Use the formula for converting a Z-score to an X value: \[ X = μ + (Z * σ) \] where: - $ μ $ is the mean - $ Z $ is the Z-score - $ σ $ is the standard deviation Plugging in the values: \[ P₈₀ = 187.9 + (0.8416 * 71.2) \]3. **Calculate**: \[ P₈₀ = 187.9 + 59.87 ≈ 247.7700 \]### AnswerHence, **P₈₀** ≈ 247.7700Enter this value accurate to 4 decimal places.---This example illustrates the process of finding a specific percentile in a normally distributed dataset. It's a key concept in the study of statistics, particularly useful for understanding data distribution and probability.

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Answered: A distribution of values is normal with…

MATLAB: An Introduction with Applications

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A distribution of values is normal with a mean of 187.9 and a standard deviation of 71.2. Find P80, which is the score separating the bottom 80% from the top 20%. P80 = Enter your answer as a number accurate to 4 decimal places.