Percentile Cutoff for Normal Distribution Calculator

Understanding the Percentile Cutoff Calculation

This calculator finds the cutoff value for a specified percentile in a normally distributed population. Given a population mean (μ) and standard deviation (σ), it uses the inverse cumulative distribution function to find the value that corresponds to the desired percentile.

Real-Life Example: Exam Score Percentile

Suppose a standardized test has scores that are normally distributed with a mean of 85 and a standard deviation of 4. You want to know the minimum score required to be in the top 5% of all test-takers, which corresponds to the 95th percentile.

Here’s the step-by-step calculation:

• Step 1: Set up the values. In this example, the population mean (μ) is 85, the standard deviation (σ) is 4, and the target percentile is 95, meaning we want to find the score that is higher than 95% of test-takers.
• Step 2: Convert the percentile to a decimal. Since the calculator uses decimals for percentile input, divide 95 by 100 to get 0.95.
• Step 3: Find the z-score for the 95th percentile. Using the normal distribution’s properties, we find that the z-score for the 95th percentile is approximately 1.645. This z-score means that a score 1.645 standard deviations above the mean would be at or above the 95th percentile.
• Step 4: Apply the z-score formula to find the cutoff value. The formula for finding a specific score (X) given the z-score, mean, and standard deviation is:
  \( X = \mu + (Z \times \sigma) \)
  where:
  • μ = 85 (population mean)
  • Z = 1.645 (z-score for the 95th percentile)
  • σ = 4 (standard deviation)
• Step 5: Calculate the result. Plugging in the values, we get:
  \( X = 85 + (1.645 \times 4) \)

  Which simplifies to:

  \( X = 85 + 6.58 = 91.58 \)
• Interpretation: To score higher than 95% of test-takers, a student would need to score approximately 91.58 or higher on this exam.