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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoLarson 8th EditionCh. 5 - Normal Probability DistributionsProblema 5.R.57c
Capitolo 5, Problema 5.R.57c

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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Step 1: Identify the given information. The population mean (μ) is 20.7, the population standard deviation (σ) is 5.9, the sample size (n) is 36, and we are tasked with finding the probability that the sample mean (x̄) is between 20 and 21.5.
Step 2: Calculate the standard error of the mean (SE). The formula for the standard error is SE = σ / √n. Substitute the given values for σ and n into the formula.
Step 3: Convert the sample mean values (20 and 21.5) into z-scores using the formula z = (x̄ - μ) / SE. Perform this calculation for both 20 and 21.5 to find their respective z-scores.
Step 4: Use the standard normal distribution table (or a statistical software) to find the probabilities corresponding to the z-scores obtained in Step 3. These probabilities represent the cumulative probabilities up to each z-score.
Step 5: Subtract the smaller cumulative probability (corresponding to the z-score for 20) from the larger cumulative probability (corresponding to the z-score for 21.5). This difference gives the probability that the sample mean is between 20 and 21.5.

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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 the population standard deviation is known.
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Calculating the Mean

Standard Error of the Mean

The Standard Error of the Mean (SEM) quantifies the amount of variability in sample means that you would expect if you took multiple samples from the same population. It is calculated as the population standard deviation divided by the square root of the sample size (σ/√n). In this case, with σ = 5.9 and n = 36, the SEM helps determine the range of sample means around the population mean.
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Calculating the Mean

Z-scores

A Z-score indicates how many standard deviations an element is from the mean of a distribution. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. In this context, Z-scores will be used to find the probabilities of the sample mean falling between 20 and 21.5 by converting these values into Z-scores using the SEM.
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Z-Scores From Given Probability - TI-84 (CE) Calculator
Pratica correlata
Domanda del libro di testo

In Exercises 69 and 70, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities.


A survey of U.S. adults found that 72% used a mobile device to manage their bank account at least once in the previous month. You randomly select 70 U.S. adults and ask whether they used a mobile device to manage their bank account at least once in the previous month. Find the probability that the number who have done so is (c) greater than 60.

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Domanda del libro di testo

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 (c) more than 11.1 picomoles per liter. Compare your answers with those in Exercise 34.

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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 (c) greater than 5900 MMT CO2 eq. Compare your answers with those in Exercise 33.

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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

Determine whether any of the events in Exercise 33 are unusual. Explain your reasoning.

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