Chapter 12: Problem 29
Suppose that \(X\) is normally distributed with mean 2 and standard deviation \(1 .\) Find \(P(0 \leq X \leq 3)\).
Short Answer
Expert verified
The probability that \(0 \leq X \leq 3\) is approximately 0.8185.
Step by step solution
01
Identify the Parameters
The problem states that the random variable \( X \) is normally distributed with mean \( \mu = 2 \) and standard deviation \( \sigma = 1 \). This can be written as \( X \sim N(2, 1^2) \).
02
Standardize the Variable
To find probabilities for a normal distribution, we first standardize the variable. This involves converting \( X \) into a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} \]For the lower limit \( X = 0 \):\[ Z_0 = \frac{0 - 2}{1} = -2 \]For the upper limit \( X = 3 \):\[ Z_3 = \frac{3 - 2}{1} = 1 \]
03
Find Standard Normal Probabilities
To find \( P(0 \leq X \leq 3) \), we need \( P(-2 \leq Z \leq 1) \).Using the standard normal distribution table or calculator, find:- \( P(Z \leq 1) \): approximately 0.8413- \( P(Z \leq -2) \): approximately 0.0228Then calculate the probability as \( P(-2 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -2) \).
04
Calculate the Probability
Using the probabilities found:\[ P(-2 \leq Z \leq 1) = 0.8413 - 0.0228 = 0.8185 \]
05
State the Result
The probability that \( X \) takes a value between 0 and 3 is \( 0.8185 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Deviation
In statistics, the standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. It tells us how much the values in a data set deviate from the mean, giving us insight into the spread of data. When you're dealing with a normal distribution, the standard deviation plays a vital role in determining the shape and spread of the curve.
The larger the standard deviation, the more spread out the data points are around the mean, resulting in a flatter, wider curve. Conversely, a smaller standard deviation indicates that the data points are closer to the mean, leading to a steeper, narrower curve.
The larger the standard deviation, the more spread out the data points are around the mean, resulting in a flatter, wider curve. Conversely, a smaller standard deviation indicates that the data points are closer to the mean, leading to a steeper, narrower curve.
- The formula for standard deviation is: \[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \]
- Here, \( \mu \) is the mean, \( N \) is the number of observations, and \( x_i \) represents each individual observation.
Standard Normal Variable
The standard normal variable is a special type of random variable that has a mean of 0 and a standard deviation of 1. This transformation makes it easier to calculate probabilities because we can use a standard normal distribution table or calculator. The process of converting a normal variable into a standard normal variable is called standardization.
To standardize a variable, we use the formula:\[Z = \frac{X - \mu}{\sigma} \]Where:
To standardize a variable, we use the formula:\[Z = \frac{X - \mu}{\sigma} \]Where:
- \( X \) is the original normal variable.
- \( \mu \) is the mean of \( X \).
- \( \sigma \) is the standard deviation of \( X \).
- For \( X = 0 \): \( Z_0 = \frac{0 - 2}{1} = -2 \)
- For \( X = 3 \): \( Z_3 = \frac{3 - 2}{1} = 1 \)
Probability Calculation
Calculating probability in a normal distribution involves determining the likelihood that a random variable falls within a certain range. Once we have a standard normal variable, we can compute these probabilities more straightforwardly. For the given problem, after standardization, we need to find the probability that the standardized variable \( Z \) falls between -2 and 1.
- Find \( P(Z \leq 1) \), which refers to the cumulative probability up to \( Z = 1 \). This value is approximately 0.8413.
- Find \( P(Z \leq -2) \), which is the cumulative probability up to \( Z = -2 \). This value is approximately 0.0228.