Bone Density Test. In Exercises 1–4, assume that scores on a bone...

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1. Intro to Stats and Collecting Data55m
2. Describing Data with Tables and Graphs1h 55m
3. Describing Data Numerically1h 45m
4. Probability2h 16m
5. Binomial Distribution & Discrete Random Variables2h 33m
6. Normal Distribution and Continuous Random Variables1h 38m
7. Sampling Distributions & Confidence Intervals: Mean1h 3m
- Introduction to Confidence Intervals
  15m
- Confidence Intervals for Population Mean
  48m
8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- Sampling Distribution of Sample Proportion
  29m
- Confidence Intervals for Population Proportion
  42m
9. Hypothesis Testing for One Sample1h 1m
- Steps in Hypothesis Testing
  1h 1m
10. Hypothesis Testing for Two Samples2h 8m
11. Correlation48m
- Scatterplots & Intro to Correlation
  26m
- Correlation Coefficient
  21m
12. Regression1h 4m
13. Chi-Square Tests & Goodness of Fit1h 20m
14. ANOVA1h 0m

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

Statistics

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

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1:36 minutes

Problem 6.q.2

Textbook Question

Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.

Bone Density Find the bone density score that is the 90th percentile, which is the score separating the lowest 90% from the top 10%.

Verified step by step guidance

Step 1: Recognize that the problem involves a normal distribution with a mean (μ) of 0 and a standard deviation (σ) of 1. This is a standard normal distribution (Z-distribution).

Step 2: Understand that the 90th percentile corresponds to the Z-score where 90% of the data lies below it. This means finding the Z-score such that the cumulative probability (area under the curve to the left of the Z-score) is 0.90.

Step 3: Use a Z-table, statistical software, or a calculator with inverse cumulative distribution function capabilities (often denoted as invNorm or similar) to find the Z-score corresponding to a cumulative probability of 0.90.

Step 4: Interpret the Z-score obtained. This Z-score represents the bone density score that separates the lowest 90% of the population from the top 10%.

Step 5: Verify the result by checking the cumulative probability for the obtained Z-score using a Z-table or statistical software to ensure it is approximately 0.90.

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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 mineral density scores are assumed to follow this distribution, which is crucial for determining percentiles.

Percentiles

A percentile is a measure used in statistics to indicate the value below which a given percentage of observations fall. For example, the 90th percentile is the score below which 90% of the data points lie. Understanding percentiles is essential for interpreting the results of the bone density test, as it helps identify how an individual's score compares to the population.

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 corresponds to the mean, while positive and negative values indicate how many standard deviations a score is above or below the mean. To find the 90th percentile score in this context, one would typically look up the corresponding Z-score in a standard normal distribution table.

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Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.Bone Density Find the bone density score that is the 90th percentile, which is the score separating the lowest 90% from the top 10%.