Textbook Question
In Exercises 7–18, find the indicated area under the standard normal curve. If convenient, use technology to find the area.
To the left of z = -2.825
Ch. 5 - Normal Probability Distributions
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470
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Table of contents
- Ch. 1 - Introduction to Statistics87
- Ch. 2 - Descriptive Statistics176
- Ch. 3 - Probability184
- Ch. 4 - Discrete Probability Distributions102
- Ch. 5 - Normal Probability Distributions183
- Ch. 6 - Confidence Intervals154
- Ch. 7 - Hypothesis Testing with One Sample165
- Ch. 8 - Hypothesis Testing with Two Samples134
- Ch. 9 - Correlation and Regression87
- Ch. 10 - Chi-Square Tests and the F-Distribution87
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 5, Problem 5.RE.17
In Exercises 7–18, find the indicated area under the standard normal curve. If convenient, use technology to find the area.
To the left of z = -1.5 and to the right of z = 1.5
Verified step by step guidance1
Step 1: Understand the problem. You are tasked with finding the area under the standard normal curve in two regions: (1) to the left of z = -1.5 and (2) to the right of z = 1.5. The standard normal curve is symmetric, with a mean of 0 and a standard deviation of 1.
Step 2: Recall that the total area under the standard normal curve is 1. The area to the left of a z-score can be found using a z-table, statistical software, or a calculator with normal distribution functions.
Step 3: Use a z-table or technology to find the cumulative probability (area) to the left of z = -1.5. This value represents the proportion of the distribution that lies to the left of z = -1.5.
Step 4: Similarly, use a z-table or technology to find the cumulative probability to the left of z = 1.5. To find the area to the right of z = 1.5, subtract the cumulative probability to the left of z = 1.5 from 1. This is because the total area under the curve is 1.
Step 5: Add the two areas together: the area to the left of z = -1.5 and the area to the right of z = 1.5. This sum represents the total area under the curve in the specified regions.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Standard Normal Distribution
The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It is represented by the z-score, which indicates how many standard deviations an element is from the mean. This distribution is crucial for calculating probabilities and areas under the curve, as it allows for the standardization of different normal distributions.
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Z-scores
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. Z-scores are essential for determining the area under the standard normal curve, as they allow us to find probabilities associated with specific values in the distribution.
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Area Under the Curve
The area under the curve (AUC) in a probability distribution represents the likelihood of a random variable falling within a particular range. For the standard normal distribution, this area can be found using z-scores and standard normal distribution tables or technology. In the context of the question, finding the area to the left of z = -1.5 and to the right of z = 1.5 involves calculating the cumulative probabilities for these z-scores.
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Related Practice
Textbook Question
In Exercises 21–26, find the indicated probability using the standard normal distribution. If convenient, use technology to find the probability.
P(0.42 < z < 3.15)
Textbook Question
In Exercises 33 and 34, find the indicated probabilities. If convenient, use technology to find the probabilities.
The daily surface concentration of carbonyl sulfide on the Indian Ocean is normally distributed, with a mean of 9.1 picomoles per liter and a standard deviation of 3.5 picomoles per liter. Find the probability that on a randomly selected day, the surface concentration of carbonyl sulfide on the Indian Ocean is
a. between 5.1 and 15.7 picomoles per liter.
Textbook Question
In Exercises 7–18, find the indicated area under the standard normal curve. If convenient, use technology to find the area.
Between z = -1.55 and z = 1.04
Textbook Question
In Exercises 27–32, the random variable x is normally distributed with mean mu=74 and standard deviation sigma=8. Find the indicated probability.
P(72 < x < 82)
Textbook Question
In Exercises 21–26, find the indicated probability using the standard normal distribution. If convenient, use technology to find the probability.
P(z < 1.28)