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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoTriola 14th EditionCh. 6 - Normal Probability DistributionsProblema 6.R.1e
Capitolo 6, Problema 6.R.1e

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.
e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.23.

Guida verificata passo dopo passo
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Step 1: Identify the given information. The population of z-scores is normally distributed with a mean (μ) of 0 and a standard deviation (σ) of 1. The sample size (n) is 9, and we are tasked with finding the probability that the sample mean (x̄) is greater than 0.23.
Step 2: Calculate the standard error of the mean (SE). The formula for the standard error is SE = σ / √n. Substitute the given values: σ = 1 and n = 9, so SE = 1 / √9.
Step 3: Standardize the sample mean to find the z-score corresponding to x̄ = 0.23. Use the formula z = (x̄ - μ) / SE. Substitute the values: x̄ = 0.23, μ = 0, and SE (calculated in Step 2).
Step 4: Use the z-score obtained in Step 3 to find the cumulative probability from the standard normal distribution table or a statistical software. This gives the probability that the sample mean is less than 0.23.
Step 5: Subtract the cumulative probability from 1 to find the probability that the sample mean is greater than 0.23. This is because the total probability under the normal curve is 1, and we are interested in the upper tail.

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Z Score

A z score indicates how many standard deviations an element is from the mean of a distribution. In the context of a bone density test, a z score of 0 means the score is exactly at the mean, while a z score of 0.23 indicates the score is 0.23 standard deviations above the mean. This standardization allows for comparison across different populations and tests.
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Z-Scores From Given Probability - TI-84 (CE) Calculator

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. In this case, the z scores of bone density tests are normally distributed, which means that statistical methods based on this distribution can be applied to calculate probabilities and make inferences about the population.
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Finding Standard Normal Probabilities using z-Table

Central Limit Theorem

The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution, provided the sample size is sufficiently large. In this scenario, with a sample size of 9, the theorem allows us to use the normal distribution to find the probability that the mean bone density score exceeds 0.23.
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Calculating the Mean
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Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.


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