Textbook Question
In Exercises 37–42, use the Standard Normal Table or technology to find the z-score that corresponds to the cumulative area or percentile.
P85
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.R.67
In Exercises 63–68, write the binomial probability in words. Then, use a continuity correction to convert the binomial probability to a normal distribution probability.
P(x < 60)
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Step 1: Understand the problem. The binomial probability P(x < 60) represents the probability of observing fewer than 60 successes in a binomial experiment. The goal is to express this probability in words and then apply a continuity correction to approximate it using a normal distribution.
Step 2: Write the binomial probability in words. The probability P(x < 60) can be described as 'the probability that the number of successes is less than 60 in a binomial distribution.'
Step 3: Recall the concept of continuity correction. When approximating a binomial distribution using a normal distribution, a continuity correction is applied to account for the discrete nature of the binomial distribution. For P(x < 60), the continuity correction adjusts the inequality to P(x ≤ 59.5) in the normal distribution.
Step 4: Convert the binomial probability to a normal distribution probability. To do this, you need the mean (μ) and standard deviation (σ) of the binomial distribution. The mean is calculated as μ = n * p, and the standard deviation is calculated as σ = √(n * p * (1 - p)), where n is the number of trials and p is the probability of success.
Step 5: Standardize the normal distribution probability. Use the z-score formula z = (x - μ) / σ to convert the value x = 59.5 into a z-score. Then, use the standard normal distribution table or software to find the probability corresponding to this z-score.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Binomial Probability
Binomial probability refers to the likelihood of obtaining a fixed number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is calculated using the binomial formula, which incorporates the number of trials, the number of successes, and the probability of success on each trial. This concept is essential for understanding scenarios where outcomes are binary, such as success/failure or yes/no.
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Normal Distribution
The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, defined by its mean and standard deviation. It is significant in statistics because many phenomena tend to follow this distribution due to the Central Limit Theorem, which states that the sum of a large number of independent random variables will approximate a normal distribution, regardless of the original distribution of the variables.
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Continuity Correction
Continuity correction is a technique used when approximating a discrete probability distribution, like the binomial distribution, with a continuous distribution, such as the normal distribution. This correction involves adjusting the discrete value by 0.5 to account for the fact that the normal distribution is continuous. For example, to find P(X < 60) in a binomial context, one would calculate P(X < 60.5) in the normal approximation.
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Related Practice
Textbook Question
In Exercises 55–60, find the indicated probabilities and interpret the results.
Refer to Exercise 33. A random sample of 2 years is selected. Find the probability that the mean amount of greenhouse gases for the sample is (c) greater than 5900 MMT CO2 eq. Compare your answers with those in Exercise 33.
Textbook Question
In Exercises 55–60, find the indicated probabilities and interpret the results.
The mean MCAT total score in a recent year is 500.9. A random sample of 32 MCAT total scores is selected. What is the probability that the mean score for the sample is (a) less than 503? Assume sigma=10.6.
Textbook Question
In Exercises 55–60, find the indicated probabilities and interpret the results.
Refer to Exercise 33. A random sample of 2 years is selected. Find the probability that the mean amount of greenhouse gases for the sample is (b) between 6000 and 6500 MMT CO2 eq. Compare your answers with those in Exercise 33.
Textbook Question
In Exercises 55–60, find the indicated probabilities and interpret the results.
The mean ACT composite score in a recent year is 20.7. A random sample of 36 ACT composite scores is selected. What is the probability that the mean score for the sample is (c) between 20 and 21.5? Assume σ=5.9.
Textbook Question
Determine whether any of the events in Exercise 33 are unusual. Explain your reasoning.