Chapter 29: Problem 1
A random variable, \(x\), has a standard normal distribution. Calculate the probability that \(x\) lies in the following intervals: (a) \((0.25,0.75)\) (b) \((-0.3,0.1)\) (c) within \(1.5\) standard deviations of the mean (d) more than two standard deviations from the mean (e) \((-1.7,-0.2)\)
Short Answer
Expert verified
(a) 0.1747, (b) 0.1577, (c) 0.8664, (d) 0.0456, (e) 0.3761
Step by step solution
01
Understanding the Standard Normal Distribution
A standard normal distribution has a mean of 0 and a standard deviation of 1. The probability of a random variable, \(x\), lying between two points is calculated using the standard normal distribution table (Z-table) or a calculator with cumulative distribution function capabilities.
02
Finding Probability for Interval (0.25, 0.75)
Find the Z-scores for 0.25 and 0.75. Use the Z-table or a calculator to get the cumulative probability for these Z-scores. Then, subtract the cumulative probability at 0.25 from that at 0.75: \[ P(0.25 < x < 0.75) = P(x < 0.75) - P(x < 0.25) \] Using the Z-table, find: \( P(x < 0.75) \approx 0.7734 \), \( P(x < 0.25) \approx 0.5987 \). Thus, the probability is approximately \( 0.7734 - 0.5987 = 0.1747 \).
03
Finding Probability for Interval (-0.3, 0.1)
Find the Z-scores for -0.3 and 0.1 and use these to get their cumulative probabilities from the Z-table or a calculator. Subtract the cumulative probability at -0.3 from that at 0.1: \[ P(-0.3 < x < 0.1) = P(x < 0.1) - P(x < -0.3) \] Using the Z-table, find: \( P(x < 0.1) \approx 0.5398 \), \( P(x < -0.3) \approx 0.3821 \). Thus, the probability is approximately \( 0.5398 - 0.3821 = 0.1577 \).
04
Within 1.5 Standard Deviations of the Mean
For a standard normal distribution, \(1.5\) standard deviations above and below the mean are between -1.5 and 1.5. Use the Z-table to find probabilities: \[ P(-1.5 < x < 1.5) = P(x < 1.5) - P(x < -1.5) \] \( P(x < 1.5) \approx 0.9332 \), \( P(x < -1.5) \approx 0.0668 \). Thus, the probability is approximately \( 0.9332 - 0.0668 = 0.8664 \).
05
More Than Two Standard Deviations from the Mean
This is the union of the probabilities less than -2 and greater than 2: \[ P(x < -2) + P(x > 2) \] Find using the Z-table: \( P(x < -2) \approx 0.0228 \), \( P(x > 2) = 1 - P(x < 2) = 1 - 0.9772 = 0.0228 \). Thus, the total probability is approximately \( 0.0228 + 0.0228 = 0.0456 \).
06
Finding Probability for Interval (-1.7, -0.2)
Find Z-scores for -1.7 and -0.2, then use a Z-table to find their cumulative probabilities: \[ P(-1.7 < x < -0.2) = P(x < -0.2) - P(x < -1.7) \] Using the Z-table, find: \( P(x < -0.2) \approx 0.4207 \), \( P(x < -1.7) \approx 0.0446 \). Thus, the probability is approximately \( 0.4207 - 0.0446 = 0.3761 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cumulative Probability
In the context of a standard normal distribution, cumulative probability refers to the probability that a random variable will take on a value less than or equal to a specific value. This cumulative probability is often expressed as the area under the curve to the left of a given value on a distribution graph.
To determine this probability, we utilize the Z-table or a calculator. The Z-table provides the cumulative probabilities for standard normal distribution values, allowing one to quickly see what percentage of the data is expected to fall below a given Z-score.
For example, to find the probability that a standard normal random variable is below 0.75, we look up the Z-score of 0.75 in the Z-table, which gives us a cumulative probability of approximately 0.7734. This indicates that about 77.34% of the data lies to the left of 0.75 on the standard normal distribution curve.
To determine this probability, we utilize the Z-table or a calculator. The Z-table provides the cumulative probabilities for standard normal distribution values, allowing one to quickly see what percentage of the data is expected to fall below a given Z-score.
For example, to find the probability that a standard normal random variable is below 0.75, we look up the Z-score of 0.75 in the Z-table, which gives us a cumulative probability of approximately 0.7734. This indicates that about 77.34% of the data lies to the left of 0.75 on the standard normal distribution curve.
Z-score
A Z-score is a statistical measurement that describes a value's position relative to the mean of a set of values, measured in terms of standard deviations.
In a standard normal distribution, where the mean is 0 and the standard deviation is 1, a Z-score quantifies how many standard deviations a data point is from the mean. Calculating Z-scores is an essential part of working with normal distributions, as it allows for the standardization of different datasets to make comparisons easier.
If you have a score for a random variable and want to find its Z-score, you can use the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \( X \) is your data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, let's say we have a value of 0.25 in a standard normal distribution. The Z-score would be 0.25 as the distribution's mean is 0 and its standard deviation is 1.
In a standard normal distribution, where the mean is 0 and the standard deviation is 1, a Z-score quantifies how many standard deviations a data point is from the mean. Calculating Z-scores is an essential part of working with normal distributions, as it allows for the standardization of different datasets to make comparisons easier.
If you have a score for a random variable and want to find its Z-score, you can use the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \( X \) is your data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, let's say we have a value of 0.25 in a standard normal distribution. The Z-score would be 0.25 as the distribution's mean is 0 and its standard deviation is 1.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells you how much the values in a dataset deviate from the mean on average.
In a standard normal distribution, the standard deviation is set to 1. A smaller standard deviation indicates that values are close to the mean, whereas a larger one suggests more spread out data.
An essential feature of the normal distribution is that a consistent percentage of data falls within a certain number of standard deviations from the mean. For instance:
In a standard normal distribution, the standard deviation is set to 1. A smaller standard deviation indicates that values are close to the mean, whereas a larger one suggests more spread out data.
An essential feature of the normal distribution is that a consistent percentage of data falls within a certain number of standard deviations from the mean. For instance:
- Approximately 68% of the data falls within one standard deviation.
- Approximately 95% within two standard deviations.
- About 99.7% within three standard deviations.
Probability Intervals
Probability intervals, within the context of a normal distribution, refer to the probability of a random variable falling between two points. These intervals are calculated using Z-scores and their corresponding cumulative probabilities.
When you have two Z-scores, finding the probability interval means subtracting the cumulative probability of the lower Z-score from that of the higher Z-score. This process gives the likelihood that a random variable falls within this specific interval.
For instance, if we want to find the probability that a random variable is between -0.3 and 0.1, we would find the cumulative probabilities for these Z-scores and calculate the difference:
When you have two Z-scores, finding the probability interval means subtracting the cumulative probability of the lower Z-score from that of the higher Z-score. This process gives the likelihood that a random variable falls within this specific interval.
For instance, if we want to find the probability that a random variable is between -0.3 and 0.1, we would find the cumulative probabilities for these Z-scores and calculate the difference:
- Find \( P(x < 0.1) \) and \( P(x < -0.3) \).
- Calculate \( P(-0.3 < x < 0.1) = P(x < 0.1) - P(x < -0.3) \).