Chapter 8: Problem 13
Find the indicated probabilities. $$ \mu=100, \sigma=15, \text { find } P(110 \leq X \leq 130) $$
Short Answer
Expert verified
The probability that \(X\) falls between 110 and 130 is 0.2286 or 22.86%.
Step by step solution
01
Identify the given values
Mean (\(\mu\)) = 100
Standard deviation (\(\sigma\)) = 15
We want to find the probability \(P(110 \leq X \leq 130)\).
02
Calculate the z-scores for the given values of \(X\)
To find the z-scores, apply the z-score formula:
\[z = \frac{X - \mu}{\sigma}\]
Calculate the z-score for \(X = 110\):
\[z_1 = \frac{110 - 100}{15} = \frac{10}{15} = 0.67\]
Calculate the z-score for \(X = 130\):
\[z_2 = \frac{130 - 100}{15} = \frac{30}{15} = 2\]
Now we have the z-scores, \(z_1 = 0.67\) and \(z_2 = 2\).
03
Find the probabilities under the standard normal curve
To find the probability, use a standard normal distribution table (z-table) or appropriate software to find the area under the curve to the left of the z-scores:
\[P(Z \leq z_1) = P(Z \leq 0.67)\]
\[P(Z \leq z_2) = P(Z \leq 2)\]
Using a z-table or software, we find:
\[P(Z \leq 0.67) = 0.7486\]
\[P(Z \leq 2) = 0.9772\]
04
Calculate the probability of the given range
In order to find the probability of the given range, \(P(110 \leq X \leq 130)\), subtract the smaller probability found above from the larger one:
\[P(110 \leq X \leq 130) = P(Z \leq z_2) - P(Z \leq z_1)\]
\[P(110 \leq X \leq 130) = 0.9772 - 0.7486\]
\[P(110 \leq X \leq 130) = 0.2286\]
The probability that \(X\) falls between 110 and 130 is 0.2286 or 22.86%.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Z-scores
When working with normal distributions, z-scores play an essential role. A z-score is a measure of how many standard deviations an element is away from the mean of the distribution. You can think of it as a measure of distance within the distribution.
Calculating z-scores involves using the formula:
\[z = \frac{X - \mu}{\sigma}\]
\[z = \frac{110 - 100}{15} = 0.67\]
This z-score tells us that 110 is 0.67 standard deviations above the mean.
Calculating z-scores involves using the formula:
\[z = \frac{X - \mu}{\sigma}\]
- \(X\) is the value in question from the data set.
- \(\mu\) is the mean of the distribution.
- \(\sigma\) is the standard deviation.
\[z = \frac{110 - 100}{15} = 0.67\]
This z-score tells us that 110 is 0.67 standard deviations above the mean.
Probability in Normal Distribution
Once z-scores are determined, they help in finding probabilities within a normal distribution—a critical concept in statistics. Probability reflects how likely an event is to occur, and for continuous data like a normal distribution, we look at areas under the curve.
Utilizing a z-score, you can use a z-table or software to find the probability of a score being less than the given z-score, which represents the area to the left of that z-score in the standard normal distribution. By assessing these areas, we derive probabilities for ranges.
For instance, if \(z = 0.67\), a z-table indicates that \(P(Z \leq 0.67) = 0.7486\). This number shows there's a 74.86% chance a randomly chosen score from the distribution will be 110 or lower.
Utilizing a z-score, you can use a z-table or software to find the probability of a score being less than the given z-score, which represents the area to the left of that z-score in the standard normal distribution. By assessing these areas, we derive probabilities for ranges.
For instance, if \(z = 0.67\), a z-table indicates that \(P(Z \leq 0.67) = 0.7486\). This number shows there's a 74.86% chance a randomly chosen score from the distribution will be 110 or lower.
Role of Standard Deviation
Standard deviation is a vital statistic measuring how spread out numbers are in a data set. In a normal distribution, it directly impacts the shape of the curve.
Here's precisely what it represents:
For this exercise, \(\sigma = 15\) reflects how much individual test scores vary around the average score of 100. Calculating z-scores with this standard deviation helps in quantifying these variances within the dataset.
Here's precisely what it represents:
- A small standard deviation means data points are close to the mean, creating a narrow and steep curve.
- A larger standard deviation indicates a wider spread of data points, resulting in a flatter and wider curve.
For this exercise, \(\sigma = 15\) reflects how much individual test scores vary around the average score of 100. Calculating z-scores with this standard deviation helps in quantifying these variances within the dataset.