Chapter 6: Problem 8
Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(7 \leq x \leq 9) ; \mu=5 ; \sigma=1.2 $$
Short Answer
Expert verified
The probability that \( x \) is between 7 and 9 is approximately 0.0471.
Step by step solution
01
Define the Standard Normal Variable
To find the probability of a normally distributed random variable between two points, we need to standardize it using the standard normal variable, also known as the Z-score. The Z-score is given by the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the random variable, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We'll do this for the bounds 7 and 9.
02
Compute the Z-scores for the Bounds
Using the Z-score formula, compute the Z-scores for \( x = 7 \) and \( x = 9 \). For \( x = 7 \), \[ Z_1 = \frac{7 - 5}{1.2} = \frac{2}{1.2} \approx 1.67 \] For \( x = 9 \), \[ Z_2 = \frac{9 - 5}{1.2} = \frac{4}{1.2} \approx 3.33 \] These Z-scores will help us find the corresponding probabilities.
03
Look Up Z-Scores in the Standard Normal Table
Using the standard normal distribution table, find the probabilities corresponding to the Z-scores 1.67 and 3.33. From the table, \( P(Z < 1.67) \approx 0.9525 \) and \( P(Z < 3.33) \approx 0.9996 \).
04
Calculate the Desired Probability
The probability of the variable \( X \) being between 7 and 9 is the difference between these two probabilities: \[ P(7 \leq x \leq 9) = P(Z < 3.33) - P(Z < 1.67) \approx 0.9996 - 0.9525 = 0.0471 \] This is the probability that \( x \) lies between 7 and 9.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Normal Variable
The standard normal variable, often represented as the Z-score, is crucial in converting any normal distribution into a standard normal distribution. This helps us simplify calculations by allowing us to use a universal reference, the standard normal distribution table.
The concept of the standard normal variable centers on the idea of standardizing a normally distributed random variable. A normal distribution can vary significantly with different means (\( \mu \)) and standard deviations (\( \sigma \)). By turning it into a standard normal variable, we set the mean to 0 and the standard deviation to 1. This is where the Z-score comes in, because it represents how many standard deviations an element is from the mean.
The concept of the standard normal variable centers on the idea of standardizing a normally distributed random variable. A normal distribution can vary significantly with different means (\( \mu \)) and standard deviations (\( \sigma \)). By turning it into a standard normal variable, we set the mean to 0 and the standard deviation to 1. This is where the Z-score comes in, because it represents how many standard deviations an element is from the mean.
- A normal distribution is standardized using the formula: \[ Z = \frac{X - \mu}{\sigma} \]
- \( X \) stands for the random variable of interest.
- \( \mu \) represents the mean of the distribution.
- \( \sigma \) is the standard deviation.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean.
When you calculate a Z-score, you're essentially converting a raw score into a form that you can easily analyze using standard normal distribution tables.
The formula for calculating the Z-score is:
\[ Z = \frac{X - \mu}{\sigma} \]
Here, Z represents the Z-score, \( X \) is the raw score you are investigating, \( \mu \) is the mean of the population, and \( \sigma \) is the standard deviation.
For our problem:
When you calculate a Z-score, you're essentially converting a raw score into a form that you can easily analyze using standard normal distribution tables.
The formula for calculating the Z-score is:
\[ Z = \frac{X - \mu}{\sigma} \]
Here, Z represents the Z-score, \( X \) is the raw score you are investigating, \( \mu \) is the mean of the population, and \( \sigma \) is the standard deviation.
- A positive Z-score means the data point is above the mean.
- A negative Z-score indicates it is below the mean.
For our problem:
- Finding the Z-scores for 7 and 9 gives us a sense of where these values lie within the entire distribution.
- This enables the use of the next tool in our kit: the standard normal distribution table.
Standard Normal Distribution Table
The standard normal distribution table, or Z-table, is a critical resource in statistics for finding probabilities associated with standard normal variables. This table compiles the cumulative probabilities of a standard normal distribution (mean = 0, standard deviation = 1) for various Z-score values.
The Z-table can tell you the probability of a value being less than a certain Z-score. For instance, a Z-score of 1.67 corresponds to a cumulative probability of approximately 0.9525 according to the Z-table, meaning that there's a 95.25% chance a randomly selected value from the distribution will be lower than 1.67.
The Z-table can tell you the probability of a value being less than a certain Z-score. For instance, a Z-score of 1.67 corresponds to a cumulative probability of approximately 0.9525 according to the Z-table, meaning that there's a 95.25% chance a randomly selected value from the distribution will be lower than 1.67.
- Locate the Z-score on the left-hand column and top row of the table.
- The intersection of these row and column entries gives the cumulative probability.
- If you were looking at Z-scores of 1.67 and 3.33, you use the Z-table to find probabilities 0.9525 and 0.9996, respectively.
- The probability of a variable being between these two Z-scores is the difference in probabilities: 0.9996 - 0.9525 = 0.0471.