6. Normal Distribution and Continuous Random Variables
Standard Normal Distribution
2:39 minutesProblem 6.1.37
Textbook Question
Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places.
Find P99, the 99th percentile. This is the bone density score separating the bottom 99% from the top 1%.
Verified step by step guidance1
Step 1: Understand the problem. The goal is to find the bone density test score corresponding to the 99th percentile (P99) in a standard normal distribution. This means we are looking for the z-score that separates the bottom 99% of the distribution from the top 1%.
Step 2: Recall that the standard normal distribution has a mean (μ) of 0 and a standard deviation (σ) of 1. The z-score corresponding to a given percentile can be found using a z-table, statistical software, or a calculator with inverse cumulative distribution function capabilities.
Step 3: Use the cumulative probability value of 0.99 (since 99% of the data lies below this point) to find the z-score. This involves finding the z-score such that the cumulative area under the standard normal curve to the left of this z-score is 0.99.
Step 4: Once the z-score is identified, interpret it as the bone density test score corresponding to the 99th percentile. This z-score represents the value separating the bottom 99% of the distribution from the top 1%.
Step 5: Round the z-score to two decimal places as instructed in the problem. This rounded value is the final answer for the bone density test score at the 99th percentile.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Normal Distribution
Normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its bell-shaped curve, defined by its mean and standard deviation. In this context, the bone density scores are normally distributed with a mean of 0 and a standard deviation of 1, which allows for the application of statistical methods to find percentiles.
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Percentiles
A percentile is a measure used in statistics indicating the value below which a given percentage of observations fall. For example, the 99th percentile (P99) is the score below which 99% of the data points lie. Understanding percentiles is crucial for interpreting the results of the bone density test, as it helps to identify how a specific score compares to the overall distribution of scores.
Z-scores
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. In a standard normal distribution, a Z-score of 0 indicates the mean, while positive and negative values indicate how many standard deviations a score is above or below the mean. To find P99 in this scenario, one would typically look up the Z-score that corresponds to the 99th percentile in the standard normal distribution table.
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