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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
All textbooksLarson 8th EditionCh. 5 - Normal Probability DistributionsProblem 5.R.55b
Chapter 5, Problem 5.R.55b

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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1
Step 1: Identify the key components of the problem. You are tasked with finding the probability that the mean amount of greenhouse gases for a random sample of 2 years is between 6000 and 6500 MMT CO2 eq. This involves using the sampling distribution of the sample mean.
Step 2: Recall the formula for the sampling distribution of the sample mean. The mean of the sampling distribution is the population mean (μ), and the standard deviation of the sampling distribution (σₓ̄) is given by: σx¯=σn, where σ is the population standard deviation and n is the sample size.
Step 3: Standardize the values 6000 and 6500 using the z-score formula: z=(x-μ)σx¯. This will convert the raw values into z-scores, which can then be used to find probabilities from the standard normal distribution.
Step 4: Use the z-scores obtained in Step 3 to find the cumulative probabilities from the standard normal distribution table or a statistical software. Subtract the cumulative probability corresponding to the lower z-score (6000) from the cumulative probability corresponding to the upper z-score (6500) to find the probability that the sample mean is between these two values.
Step 5: Compare the calculated probability with the results from Exercise 33. Interpret the findings by discussing how the sample size and the sampling distribution affect the probability, and whether the results align with expectations based on Exercise 33.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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 means vary and is crucial for understanding how likely it is to obtain a sample mean within a certain range, such as between 6000 and 6500 MMT CO2 eq.
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Sampling Distribution of Sample Proportion

Central Limit Theorem (CLT)

The Central Limit Theorem states that, regardless of the population's distribution, the distribution of the sample means will approach a normal distribution as the sample size increases. This theorem is essential for calculating probabilities related to sample means, especially when determining the likelihood of the mean amount of greenhouse gases falling within a specified range.
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Probability Calculation

Probability calculation involves determining the likelihood of a specific event occurring, often using statistical methods. In this context, it requires using the properties of the normal distribution to find the probability that the sample mean of greenhouse gases lies between 6000 and 6500 MMT CO2 eq, which can be computed using z-scores and standard normal distribution tables.
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Related Practice
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

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.


Refer to Exercise 34. A random sample of six days is selected. Find the probability that the mean surface concentration of carbonyl sulfide for the sample is (a) between 5.1 and 15.7 picomoles per liter. Compare your answers with those in Exercise 34.

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 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)

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.