Chapter 7: Problem 9
A variable is normally distributed with a mean of 120 and a standard deviation of \(5 .\) One score is randomly sampled. What is the probability it is above \(127 ?\)
Short Answer
Expert verified
The probability is approximately 8.08\%.
Step by step solution
01
Understand the Problem
We are given a normally distributed variable with mean \(\mu = 120\) and standard deviation \(\sigma = 5\). We need to find the probability that a randomly sampled score is greater than \(127\).
02
Calculate the Z-Score
The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For \(X = 127\), the Z-score is \( Z = \frac{127 - 120}{5} = 1.4 \).
03
Find the Probability Using the Z-Table
Using a standard normal distribution table (Z-table), we look up the probability corresponding to a Z-score of 1.4. The table provides the probability that a score is less than 1.4. This is approximately 0.9192.
04
Calculate the Probability Greater Than the Z-Score
Since we need the probability of the score being greater than \(127\), we calculate this as \(1 - P(Z < 1.4)\). Thus, \(1 - 0.9192 = 0.0808\).
05
Final Answer
The probability that a randomly sampled score from this distribution is greater than \(127\) is \(0.0808\), or approximately 8.08\%.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-score calculation
Calculating the Z-score is an essential skill when working with a normal distribution. A Z-score tells us how many standard deviations an individual data point is from the mean of the distribution. To calculate a Z-score, use the formula:
For instance, consider if you want to determine how a score of 127 compares to a normal distribution with a mean (\(\mu)\) of 120 and a standard deviation (\(\sigma)\) of 5. The calculation goes as follows:
- \( Z = \frac{X - \mu}{\sigma} \)
For instance, consider if you want to determine how a score of 127 compares to a normal distribution with a mean (\(\mu)\) of 120 and a standard deviation (\(\sigma)\) of 5. The calculation goes as follows:
- \( Z = \frac{127 - 120}{5} = 1.4 \)
probability calculation
Once you have calculated the Z-score, the next step is finding the probability associated with it. This process helps you understand the likelihood of a random sample score being above or below a certain value in a normal distribution.
For a Z-score of 1.4, you need to determine the probability that corresponds to this score in the context of the standard normal distribution.
For a Z-score of 1.4, you need to determine the probability that corresponds to this score in the context of the standard normal distribution.
- First, look up the Z-score in a Z-table to find the probability \( P(Z < 1.4) \).
- The table typically gives the cumulative probability of a score being less than your Z-score.
- In our case, \( P(Z < 1.4) \) is approximately 0.9192, indicating a 91.92% chance that a score is less than 127.
- \( P(Z > 1.4) = 1 - P(Z < 1.4) = 0.0808 \)
Z-table usage
A Z-table, also known as a standard normal distribution table, is a valuable tool for finding the probability associated with a specific Z-score. Here’s how you can effectively use it:
Using the Z-table allows straightforward interpretation of normal distribution probabilities, making it a critical resource for anyone dealing with statistical data analysis.
- First, identify the Z-score you need. For example, a Z-score of 1.4.
- Look for the value of 1.4 in the Z-table. The table usually displays cumulative probabilities from the left side of the distribution to your Z-score.
- Find the corresponding probability that shows the likelihood of a score being below the Z-score. For a Z-score of 1.4, the cumulative probability is 0.9192.
Using the Z-table allows straightforward interpretation of normal distribution probabilities, making it a critical resource for anyone dealing with statistical data analysis.