Chapter 5: Problem 14
Let \(x\) be a continuous random variable with a standard normal distribution. Using Table \(A,\) find each of the following. $$P(-2.01 \leq x \leq 0)$$
Short Answer
Expert verified
0.4778
Step by step solution
01
Identify the Problem
We need to find the probability that a continuous random variable with a standard normal distribution falls between the values ewline ewline ewline-2.01 and 0, denoted as \(P(-2.01 \leq x \leq 0).\)
02
Refer to the Z-Table
Locate the cumulative probability values for \(z=-2.01\) and \(z=0\) in the standard normal distribution table (Table A).
03
Find the Cumulative Probability
From the Z-table, the cumulative probability for \(z=-2.01\) is approximately 0.0222, and the cumulative probability for \(z=0\) is 0.5000.
04
Calculate the Desired Probability
The probability for \(P(-2.01 \leq x \leq 0)\) is the difference between the cumulative probabilities at \(z=0\) and \(z=-2.01\). Therefore, ewlineewline \(P(-2.01 \leq x \leq 0) = \Phi(0) - \Phi(-2.01)\) where \(\Phi(0) = 0.5000\) and \(\Phi(-2.01) = 0.0222\) ewline ewline Hence, \(P(-2.01 \leq x \leq 0) = 0.5000 - 0.0222 = 0.4778.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Continuous Random Variables
A continuous random variable can take on infinitely many values within a given range. Unlike a discrete random variable, which can only assume specific values, a continuous random variable could be any value along a continuum. For instance, think of measuring the exact height of people in a room. You could get values like 5.833 feet, 6.421 feet, and so on. Since these values cover a spectrum, you need a different approach to work with probabilities.
Probabilities for continuous random variables are defined over intervals, not individual points. This is because, in theory, the probability of a continuous random variable taking any exact value is zero. Instead, you find probabilities over specified ranges. For instance, the probability that a height falls between 5.5 and 6.5 feet.
In the context of a standard normal distribution, a continuous random variable often denoted by the letter 'Z', represents standard scores (Z-scores). The probability that this score falls within a specified range can be found using statistical tools like the Z-table.
Probabilities for continuous random variables are defined over intervals, not individual points. This is because, in theory, the probability of a continuous random variable taking any exact value is zero. Instead, you find probabilities over specified ranges. For instance, the probability that a height falls between 5.5 and 6.5 feet.
In the context of a standard normal distribution, a continuous random variable often denoted by the letter 'Z', represents standard scores (Z-scores). The probability that this score falls within a specified range can be found using statistical tools like the Z-table.
How to Use the Z-Table
The Z-table is a tool used to find the cumulative probabilities associated with standard normal distribution values, also known as Z-scores. Z-scores represent how many standard deviations an element is from the mean of the distribution. To use the Z-table effectively, follow these steps:
For example, from the problem, the Z-scores are -2.01 and 0. According to the Z-table, the cumulative probability for \(z = -2.01\) is approximately 0.0222, and for \(z = 0\), it is 0.5000.
These cumulative probabilities are used to determine the likelihood that the random variable falls within a certain range.
- Identify the Z-score value for which you need the cumulative probability. The Z-score is calculated as \(z = \frac{(X - \text{mean})}{\text{standard deviation}}\).
- Locate the Z-score in the Z-table. The table usually lists scores to two decimal places; one digit on the side and the second along the top.
- Read off the cumulative probability corresponding to that Z-score. This value represents the area under the normal curve to the left of the specified Z-score.
For example, from the problem, the Z-scores are -2.01 and 0. According to the Z-table, the cumulative probability for \(z = -2.01\) is approximately 0.0222, and for \(z = 0\), it is 0.5000.
These cumulative probabilities are used to determine the likelihood that the random variable falls within a certain range.
Cumulative Probability Explained
Cumulative probability represents the probability that a random variable takes on a value less than or equal to a specified value. When using the Z-table for a standard normal distribution, the cumulative probability is the area under the curve to the left of a particular Z-score. This gives the probability that Z will be less than or equal to that Z-score.
Let's break down the steps using the exercise example:
To find the probability that a variable falls between \(z = -2.01\) and \(z = 0\), subtract the cumulative probability at \(z = -2.01\) from the cumulative probability at \(z = 0\). Thus, we get:
\(\text{Probability} = \text{Cumulative Probability at } z=0 - \text{Cumulative Probability at } z=-2.01\)
Which is
\(\text{Probability} = 0.5000 - 0.0222 = 0.4778\).
This calculation tells you the likelihood that the random variable falls between the Z-scores given.
Let's break down the steps using the exercise example:
- First, identify the Z-scores: -2.01 and 0.
- Find the cumulative probabilities corresponding to these Z-scores using the Z-table: 0.0222 for \(z = -2.01\) and 0.5000 for \(z = 0\).
- The cumulative probability for any Z-score indicates the area under the normal distribution curve to the left of that Z-score.
To find the probability that a variable falls between \(z = -2.01\) and \(z = 0\), subtract the cumulative probability at \(z = -2.01\) from the cumulative probability at \(z = 0\). Thus, we get:
\(\text{Probability} = \text{Cumulative Probability at } z=0 - \text{Cumulative Probability at } z=-2.01\)
Which is
\(\text{Probability} = 0.5000 - 0.0222 = 0.4778\).
This calculation tells you the likelihood that the random variable falls between the Z-scores given.