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Scores on a national exam are normally distributed with mean 382 and standard deviation \(26 .\) a. Find the score that is the 50 th percentile. b. Find the score that is the 90 th percentile.

Short Answer

Expert verified
50th percentile: 382; 90th percentile: 415.28.

Step by step solution

01

Understanding the Problem

We need to find the 50th and 90th percentiles for scores on a national exam. Scores are normally distributed with a mean (\(\mu\)) of 382 and a standard deviation (\(\sigma\)) of 26.
02

Find the z-score for the 50th percentile

The 50th percentile is the median of a normal distribution. For a standard normal distribution, the z-score for the 50th percentile is 0, because the median is the same as the mean.
03

Calculate the 50th Percentile Score

Use the formula \(X = \mu + z\sigma\) to convert the z-score to the actual score. Here, \(X\) is the score, \(\mu = 382\), \(z = 0\), and \(\sigma = 26\). Thus, \(X = 382 + 0 \times 26 = 382\).
04

Find the z-score for the 90th percentile

Use a standard normal distribution table to find the z-score corresponding to the 90th percentile. This z-score is approximately 1.28.
05

Calculate the 90th Percentile Score

Use the formula \(X = \mu + z\sigma\) to convert the z-score to the actual score. Here, \(\mu = 382\), \(z = 1.28\), and \(\sigma = 26\). Thus, \(X = 382 + 1.28 \times 26 = 415.28\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
Normal distribution is a commonly used statistical concept that describes how a set of data points is spread around a central value, known as the mean. In a normal, or bell-shaped, distribution:
  • Most of the data points are clustered around the mean.
  • The probabilities taper off symmetrically on either side, creating a bell curve shape.
  • The width of this curve is determined by the standard deviation.
The mean (\(\mu\)) and standard deviation (\(\sigma\)) are key parameters that define this distribution. A larger standard deviation results in a wider curve, while a smaller standard deviation makes the curve steeper. Understanding normal distribution helps in calculating probabilities and percentiles, like determining exam scores at specific intervals from the mean.
Z-Score
The z-score is a measure that tells us how many standard deviations a data point is from the mean. It is a useful concept when dealing with normal distributions, as it allows us to standardize scores and compare them across different distributions. To calculate the z-score:
  • Subtract the mean from the data point.
  • Divide by the standard deviation.
Symbolically, the formula is:\[z = \frac{X - \mu}{\sigma}\]Where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. A z-score of 0 indicates the data point is exactly the mean, positive z-scores indicate values above the mean, and negative scores indicate values below the mean. Understanding how to convert z-scores to actual values is key to solving percentile problems.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual data points differ from the mean. In simpler terms, it helps to quantify the spread of data points in a dataset. A low standard deviation means that the data points are close to the mean, while a high standard deviation means that the data points are spread out over a wider range. The formula for standard deviation is:\[\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2}\]Where:
  • \(N\) is the number of data points.
  • \(X_i\) are the individual data points.
  • \(\mu\) is the mean of the data points.
This concept is crucial in interpreting the normal distribution, as it affects the shape of the distribution curve. Understanding standard deviation is essential when calculating percentiles, such as finding the 50th or 90th percentile of exam scores.

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