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.58a
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.
Guida verificata passo dopo passo1
Step 1: Identify the given values in the problem. The population mean (μ) is 500.9, the population standard deviation (σ) is 10.6, the sample size (n) is 32, and we are tasked with finding the probability that the sample mean (x̄) is less than 503.
Step 2: Recall that the sampling distribution of the sample mean follows a normal distribution if the population is normal or the sample size is sufficiently large (n ≥ 30). The mean of the sampling distribution is μ, and the standard error (SE) is calculated as SE = σ / √n.
Step 3: Calculate the standard error (SE) using the formula SE = σ / √n. Substitute the given values: σ = 10.6 and n = 32. This will give you the standard deviation of the sampling distribution.
Step 4: Convert the sample mean (503) to a z-score using the formula z = (x̄ - μ) / SE. Substitute the values x̄ = 503, μ = 500.9, and the SE calculated in the previous step.
Step 5: Use the z-score obtained in Step 4 to find the cumulative probability from the standard normal distribution table or a statistical software. This cumulative probability represents the probability that the sample mean is less than 503.
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Central Limit Theorem
The Central Limit Theorem states that the distribution of the sample mean will approach a normal distribution as the sample size increases, regardless of the population's distribution, provided the sample size is sufficiently large (typically n > 30). This theorem is crucial for calculating probabilities related to sample means, especially when dealing with a known population mean and standard deviation.
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04:52
Calculating the Mean
Standard Error of the Mean
The Standard Error of the Mean (SEM) quantifies how much the sample mean is expected to vary from the true population mean. It is calculated by dividing the population standard deviation (sigma) by the square root of the sample size (n). In this case, SEM helps determine the probability of the sample mean being less than a specific value, such as 503.
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04:52Calculating the Mean
Z-Score
A Z-score measures how many standard deviations an element is from the mean. In the context of this problem, the Z-score is used to standardize the sample mean to find the corresponding probability in the standard normal distribution. By calculating the Z-score for the sample mean of 503, we can determine the likelihood of obtaining a sample mean less than this value.
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06:31Z-Scores From Given Probability - TI-84 (CE) Calculator
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In Exercises 55–60, find the indicated probabilities and interpret the results.
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