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Chapter 6: Problem 20

Assume that a randomly selected subject is given a bone density test. Those 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 probability of the given bone density test scores. If using technology instead of Table \(A-2,\) round answers to four decimal places. Less than 2.56

Short Answer

Expert verified
The probability is 0.9948.

Step by step solution

01

- Understanding the Problem

The problem states that bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. We need to find the probability that a randomly selected test score is less than 2.56.
02

- Drawing the Normal Distribution

Draw a normal distribution curve with the mean at the center (which is 0). Shade the area to the left of 2.56 since we are interested in the probability of scores less than 2.56.
03

- Standardizing the Value

Since the distribution is already standard with a mean (\textmu) of 0 and standard deviation (\textsigma) of 1, the z-score for 2.56 remains 2.56. The z-score formula is given by: \[ z = \frac{x - \textmu}{\textsigma} \]In this case, \[ z = \frac{2.56 - 0}{1} = 2.56. \]
04

- Using the Z-Table or Technology

To find the probability corresponding to a z-score of 2.56, use the standard normal distribution table (Z-table) or technology (such as a calculator or statistical software). According to the Z-table, the probability for a z-score of 2.56 is 0.9948.
05

- Concluding the Result

The probability that a randomly selected subject has a bone density test score less than 2.56 is 0.9948.

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Key Concepts

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

Z-Score
The z-score tells us how many standard deviations a data point is from the mean. It provides a way to compare individual scores to the overall distribution. The formula for calculating the z-score is given by: \( z = \frac{x - \mu}{\sigma} \)Here,
  • \(x\) is the data point,
  • \(\mu\) is the mean of the distribution,
  • and \(\sigma\) is the standard deviation.
A positive z-score means the data point is above the mean, while a negative z-score means it is below the mean. In our exercise, the z-score for 2.56 is:\[ z = \frac{2.56 - 0}{1} = 2.56 \].
Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. It shows how much the individual data points differ from the mean. A smaller standard deviation means the values are closer to the mean, while a larger standard deviation indicates more spread out values. For a normal distribution, the standard deviation helps define the 'spread' of the data. In our problem, the standard deviation is 1, meaning that on average, the bone density test scores differ from the mean by 1 unit. The standard deviation is a critical factor when standardizing values to obtain the z-score.
Probability Calculation
Probability calculation in a normal distribution involves finding the area under the curve. This area represents the likelihood of a value occurring. To find the probability for a z-score in our problem, we use the Z-table or statistical software. The Z-table displays the cumulative probability from the mean up to a z-score. For the z-score of 2.56, we lookup in the table or use technology to find that:
  • The probability is 0.9948, or 99.48%.
This means there's a 99.48% chance that a randomly selected bone density test score is less than 2.56.
Mean
The mean is the average value of a dataset. It is calculated by summing all the values and dividing by the number of values. In the context of our exercise, the mean is 0. For a normal distribution, the mean is the center point around which the data is symmetrically distributed. Knowing the mean helps us understand where the 'middle' of our data lies. In this problem, the mean of 0 indicates that the average bone density test score is 0. Combined with the standard deviation, the mean helps standardize our values to determine the z-score.

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