Domanda del libro di testo
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
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Ch. 5 - Normal Probability Distributions
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470
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TOC.tableOfContents
- 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: 9780137493470Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.R.55c
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.
Guida verificata passo dopo passo1
Identify the key information from the problem: the population mean (μ), the population standard deviation (σ), the sample size (n), and the value of interest (5900 MMT CO2 eq). These values should be referenced from Exercise 33.
Determine the sampling distribution of the sample mean. The mean of the sampling distribution is the same as the population mean (μ), and the standard error (SE) is calculated as SE = σ / √n, where σ is the population standard deviation and n is the sample size.
Standardize the value of interest (5900 MMT CO2 eq) to a z-score using the formula: z = (X̄ - μ) / SE, where X̄ is the sample mean (5900), μ is the population mean, and SE is the standard error.
Use the z-score to find the probability that the sample mean is greater than 5900. This can be done by looking up the z-score in a standard normal distribution table or using statistical software to find the area to the right of the z-score.
Compare the calculated probability with the results from Exercise 33 to interpret how the probability changes when considering the sample mean instead of individual values.
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Sampling Distribution
The sampling distribution is the probability distribution of a statistic (like the sample mean) obtained from a large number of samples drawn from a specific population. It describes how the sample mean varies from sample to sample and is crucial for understanding how to calculate probabilities related to sample means.
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05:11
Sampling Distribution of Sample Proportion
Central Limit Theorem (CLT)
The Central Limit Theorem states that, for a sufficiently large sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the population's distribution. This theorem allows statisticians to make inferences about population parameters using sample statistics, particularly when calculating probabilities.
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04:52Calculating the Mean
Z-Score
A Z-score measures how many standard deviations an element is from the mean of a distribution. In the context of probability, Z-scores are used to determine the probability of a sample mean being above or below a certain value by converting the sample mean into a standard normal variable, facilitating the use of standard normal distribution tables.
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06:31Z-Scores From Given Probability - TI-84 (CE) Calculator
Pratica correlata
Domanda del libro di testo
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 (b) more than 502? Assume sigma=10.6.
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Domanda del libro di testo
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.
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Domanda del libro di testo
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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Domanda del libro di testo
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.
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Domanda del libro di testo
Determine whether any of the events in Exercise 33 are unusual. Explain your reasoning.
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