- 7.1 The Central Limit Theorem for Sample Means (Averages)
- Introduction
- 1.1 Definitions of Statistics, Probability, and Key Terms
- 1.2 Data, Sampling, and Variation in Data and Sampling
- 1.3 Frequency, Frequency Tables, and Levels of Measurement
- 1.4 Experimental Design and Ethics
- 1.5 Data Collection Experiment
- 1.6 Sampling Experiment
- Chapter Review
- Bringing It Together: Homework
- 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
- 2.2 Histograms, Frequency Polygons, and Time Series Graphs
- 2.3 Measures of the Location of the Data
- 2.4 Box Plots
- 2.5 Measures of the Center of the Data
- 2.6 Skewness and the Mean, Median, and Mode
- 2.7 Measures of the Spread of the Data
- 2.8 Descriptive Statistics
- Formula Review
- 3.1 Terminology
- 3.2 Independent and Mutually Exclusive Events
- 3.3 Two Basic Rules of Probability
- 3.4 Contingency Tables
- 3.5 Tree and Venn Diagrams
- 3.6 Probability Topics
- Bringing It Together: Practice
- 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
- 4.2 Mean or Expected Value and Standard Deviation
- 4.3 Binomial Distribution (Optional)
- 4.4 Geometric Distribution (Optional)
- 4.5 Hypergeometric Distribution (Optional)
- 4.6 Poisson Distribution (Optional)
- 4.7 Discrete Distribution (Playing Card Experiment)
- 4.8 Discrete Distribution (Lucky Dice Experiment)
- 5.1 Continuous Probability Functions
- 5.2 The Uniform Distribution
- 5.3 The Exponential Distribution (Optional)
- 5.4 Continuous Distribution
- 6.1 The Standard Normal Distribution
- 6.2 Using the Normal Distribution
- 6.3 Normal Distribution—Lap Times
- 6.4 Normal Distribution—Pinkie Length
- 7.2 The Central Limit Theorem for Sums (Optional)
- 7.3 Using the Central Limit Theorem
- 7.4 Central Limit Theorem (Pocket Change)
- 7.5 Central Limit Theorem (Cookie Recipes)
- 8.1 A Single Population Mean Using the Normal Distribution
- 8.2 A Single Population Mean Using the Student's t-Distribution
- 8.3 A Population Proportion
- 8.4 Confidence Interval (Home Costs)
- 8.5 Confidence Interval (Place of Birth)
- 8.6 Confidence Interval (Women's Heights)
- 9.1 Null and Alternative Hypotheses
- 9.2 Outcomes and the Type I and Type II Errors
- 9.3 Distribution Needed for Hypothesis Testing
- 9.4 Rare Events, the Sample, and the Decision and Conclusion
- 9.5 Additional Information and Full Hypothesis Test Examples
- 9.6 Hypothesis Testing of a Single Mean and Single Proportion
- 10.1 Two Population Means with Unknown Standard Deviations
- 10.2 Two Population Means with Known Standard Deviations
- 10.3 Comparing Two Independent Population Proportions
- 10.4 Matched or Paired Samples (Optional)
- 10.5 Hypothesis Testing for Two Means and Two Proportions
- 11.1 Facts About the Chi-Square Distribution
- 11.2 Goodness-of-Fit Test
- 11.3 Test of Independence
- 11.4 Test for Homogeneity
- 11.5 Comparison of the Chi-Square Tests
- 11.6 Test of a Single Variance
- 11.7 Lab 1: Chi-Square Goodness-of-Fit
- 11.8 Lab 2: Chi-Square Test of Independence
- 12.1 Linear Equations
- 12.2 The Regression Equation
- 12.3 Testing the Significance of the Correlation Coefficient (Optional)
- 12.4 Prediction (Optional)
- 12.5 Outliers
- 12.6 Regression (Distance from School) (Optional)
- 12.7 Regression (Textbook Cost) (Optional)
- 12.8 Regression (Fuel Efficiency) (Optional)
- 13.1 One-Way ANOVA
- 13.2 The F Distribution and the F Ratio
- 13.3 Facts About the F Distribution
- 13.4 Test of Two Variances
- 13.5 Lab: One-Way ANOVA
- A | Appendix A Review Exercises (Ch 3–13)
- B | Appendix B Practice Tests (1–4) and Final Exams
- C | Data Sets
- D | Group and Partner Projects
- E | Solution Sheets
- F | Mathematical Phrases, Symbols, and Formulas
- G | Notes for the TI-83, 83+, 84, 84+ Calculators

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose

- μ x μ x = the mean of X
- σ x σ x = the standard deviation of X

If you draw random samples of size n , then as n increases, the random variable X ¯ X ¯ , which consists of sample means, tends to be normally distributed and

The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means , the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done.

To put it more formally, if you draw random samples of size n , the distribution of the random variable X ¯ X ¯ , which consists of sample means, is called the sampling distribution of the mean . The sampling distribution of the mean approaches a normal distribution as n , the sample size , increases.

The random variable X ¯ X ¯ has a different z -score associated with it from that of the random variable X . The mean x ¯ x ¯ is the value of X ¯ X ¯ in one sample.

μ X is the average of both X and X ¯ X ¯ .

σ x ¯ = σ x n σ x ¯ = σ x n = standard deviation of X ¯ X ¯ and is called the standard error of the mean .

## Using the TI-83, 83+, 84, 84+ Calculator

To find probabilities for means on the calculator, follow these steps.

2nd DISTR 2:normalcdf

n o r m a l c d f ( l o w e r v a l u e o f t h e a r e a , u p p e r v a l u e o f t h e a r e a , m e a n , s t a n d a r d d e v i a t i o n s a m p l e s i z e ) n o r m a l c d f ( l o w e r v a l u e o f t h e a r e a , u p p e r v a l u e o f t h e a r e a , m e a n , s t a n d a r d d e v i a t i o n s a m p l e s i z e )

- mean is the mean of the original distribution
- standard deviation is the standard deviation of the original distribution
- sample size = n

## Example 7.1

A distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population.

a. Find the probability that the sample mean is between 85 and 92.

a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean .

Let X ¯ X ¯ = the mean of a sample of size 25. Because μ x = 90, σ x = 15, and n = 25,

Find P (85 < x ¯ x ¯ < 92). Draw a graph.

P (85 < x ¯ x ¯ < 92) = 0.6997

The probability that the sample mean is between 85 and 92 is 0.6997.

normalcdf (lower value, upper value, mean, standard error of the mean)

The parameter list is abbreviated (lower value, upper value, μ , σ n σ n ).

normalcdf (85,92,90, 15 25 ) = 0.6997 15 25 ) = 0.6997

b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.

b. To find the value that is two standard deviations above the expected value 90, use the following formula

The value that is two standard deviations above the expected value is 96.

The standard error of the mean is σ x n σ x n = 15 25 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n .

An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.

## Example 7.2

The length of time, in hours, it takes a group of people, 40 years old and older, to play one soccer match is normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours . A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.

Let X = the time, in hours, it takes to play one soccer match.

The probability question asks you to find a probability for the sample mean time, in hours , it takes to play one soccer match.

Let X ¯ X ¯ = the mean time, in hours, it takes to play one soccer match.

If μ X = _________, σ X = __________, and n = ___________, then X ~ N (______, ______) by the central limit theorem for means .

μ X = 2, σ X = 0.5, n = 50, and X ~ N ( 2, 0.5 50 ) ( 2, 0.5 50 )

Find P (1.8 < x ¯ x ¯ < 2.3). Draw a graph.

The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.

The length of time taken on the SAT exam for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.

To find percentiles for means on the calculator, follow these steps.

2 nd DIStR 3:invNorm

- k = the k th percentile

## Example 7.3

In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.

- What are the mean and standard deviation for the sample mean ages of tablet users?
- What does the distribution look like?
- Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
- Find the 95 th percentile for the sample mean age (to one decimal place).
- Because the sample mean tends to target the population mean, we have μ χ = μ = 34. The sample standard deviation is given by σ χ σ χ = σ n σ n = 15 100 15 100 = 15 10 15 10 = 1.5.
- The central limit theorem states that for large sample sizes ( n ), the sampling distribution will be approximately normal.
- The probability that the sample mean age is more than 30 is given by P ( Χ > 30) = normalcdf (30,E99,34,1.5) = 0.9962.
- Let k = the 95 th percentile. k = invNorm ( 0. 95,34, 15 100 ) ( 0. 95,34, 15 100 ) = 36.5

A gaming marketing gap for men between the ages of 30 to 40 has been identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy?

## Example 7.4

The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.

- What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user?
- What is the standard error of the mean?
- Find the 90 th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
- Find the probability that the sample mean is between eight minutes and 8.5 minutes.
- μ x ¯ = μ = 8.2 σ x ¯ = σ n = 1 60 = 0.13 μ x ¯ = μ = 8.2 σ x ¯ = σ n = 1 60 = 0.13
- This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
- Let k = the 90 th percentile. k = invNorm ( 0. 90,8 .2, 1 60 ) ( 0. 90,8 .2, 1 60 ) = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
- P (8 < x ¯ x ¯ < 8.5) = normalcdf ( 8,8 .5,8 .2, 1 60 ) ( 8,8 .5,8 .2, 1 60 ) = 0.9293

Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x ¯ x ¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?

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## Central Limit Theorem

- Statistics Homework Help

One of the most important parts of statistics and probability theory is the Central Limit Theorem.

The central limit theorem states that if some certain conditions are satisfied, then the distribution of the arithmetic mean of a number of independent random variables approaches a normal distribution as the number of variables approaches infinity.

In simple words, If a random sample of n observations is selected from a population ( any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal.

(The larger the sample size, the better will be the normal approximation to the sampling distribution of x.)

Here are two main things you should remember and use when proving agreement with the Central Limit Theorem:

1) If the sample size is greater than 30, the Distribution of Sample Means will approximately follow a normal distribution Regardless of the underlying population distribution.

If the underlying population distribution is Normally Distributed, the Distribution of Sample Means will be normally distributed Regardless of the sample sizes used.

2) The mean of the Distribution of Sample Means will be the same as the underlying populations mean i.e. The mean of the sample means will be the population mean µ.

The standard deviation of the Distribution of Sample Means, however, will be the underlying population’s standard deviation divided by the square root of the sample size i.e. the standard deviation of the sample means will approach σ/√n.

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- Introduction
- 1.1 Definitions of Statistics, Probability, and Key Terms
- 1.2 Data, Sampling, and Variation in Data and Sampling
- 1.3 Frequency, Frequency Tables, and Levels of Measurement
- 1.4 Experimental Design and Ethics
- 1.5 Data Collection Experiment
- 1.6 Sampling Experiment
- Chapter Review
- Bringing It Together: Homework
- 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
- 2.2 Histograms, Frequency Polygons, and Time Series Graphs
- 2.3 Measures of the Location of the Data
- 2.4 Box Plots
- 2.5 Measures of the Center of the Data
- 2.6 Skewness and the Mean, Median, and Mode
- 2.7 Measures of the Spread of the Data
- 2.8 Descriptive Statistics
- Formula Review
- 3.1 Terminology
- 3.2 Independent and Mutually Exclusive Events
- 3.3 Two Basic Rules of Probability
- 3.4 Contingency Tables
- 3.5 Tree and Venn Diagrams
- 3.6 Probability Topics
- Bringing It Together: Practice
- 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
- 4.2 Mean or Expected Value and Standard Deviation
- 4.3 Binomial Distribution (Optional)
- 4.4 Geometric Distribution (Optional)
- 4.5 Hypergeometric Distribution (Optional)
- 4.6 Poisson Distribution (Optional)
- 4.7 Discrete Distribution (Playing Card Experiment)
- 4.8 Discrete Distribution (Lucky Dice Experiment)
- 5.1 Continuous Probability Functions
- 5.2 The Uniform Distribution
- 5.3 The Exponential Distribution (Optional)
- 5.4 Continuous Distribution
- 6.1 The Standard Normal Distribution
- 6.2 Using the Normal Distribution
- 6.3 Normal Distribution—Lap Times
- 6.4 Normal Distribution—Pinkie Length

## 7.1 The Central Limit Theorem for Sample Means (Averages)

7.2 the central limit theorem for sums (optional), 7.3 using the central limit theorem.

- 7.4 Central Limit Theorem (Pocket Change)
- 7.5 Central Limit Theorem (Cookie Recipes)
- 8.1 A Single Population Mean Using the Normal Distribution
- 8.2 A Single Population Mean Using the Student’s t-Distribution
- 8.3 A Population Proportion
- 8.4 Confidence Interval (Home Costs)
- 8.5 Confidence Interval (Place of Birth)
- 8.6 Confidence Interval (Women's Heights)
- 9.1 Null and Alternative Hypotheses
- 9.2 Outcomes and the Type I and Type II Errors
- 9.3 Distribution Needed for Hypothesis Testing
- 9.4 Rare Events, the Sample, and the Decision and Conclusion
- 9.5 Additional Information and Full Hypothesis Test Examples
- 9.6 Hypothesis Testing of a Single Mean and Single Proportion
- 10.1 Two Population Means with Unknown Standard Deviations
- 10.2 Two Population Means with Known Standard Deviations
- 10.3 Comparing Two Independent Population Proportions
- 10.4 Matched or Paired Samples (Optional)
- 10.5 Hypothesis Testing for Two Means and Two Proportions
- 11.1 Facts About the Chi-Square Distribution
- 11.2 Goodness-of-Fit Test
- 11.3 Test of Independence
- 11.4 Test for Homogeneity
- 11.5 Comparison of the Chi-Square Tests
- 11.6 Test of a Single Variance
- 11.7 Lab 1: Chi-Square Goodness-of-Fit
- 11.8 Lab 2: Chi-Square Test of Independence
- 12.1 Linear Equations
- 12.2 Scatter Plots
- 12.3 The Regression Equation
- 12.4 Testing the Significance of the Correlation Coefficient (Optional)
- 12.5 Prediction (Optional)
- 12.6 Outliers
- 12.7 Regression (Distance from School) (Optional)
- 12.8 Regression (Textbook Cost) (Optional)
- 12.9 Regression (Fuel Efficiency) (Optional)
- 13.1 One-Way Anova
- 13.2 The F Distribution and the F Ratio
- 13.3 Facts About the F Distribution
- 13.4 Test of Two Variances
- 13.5 Lab: One-Way ANOVA
- A | Review Exercises (Ch 3–13)
- B | Practice Tests (1–4) and Final Exams
- C | Data Sets
- D | Group and Partner Projects
- E | Solution Sheets
- F | Mathematical Phrases, Symbols, and Formulas
- G | Notes for the TI-83, 83+, 84, 84+ Calculators
- TEA Statistics PowerPoint Slides
- TEA Statistics Textbook PDF

Previously, De Anza's statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

- In words, Χ = ____________.

Χ ~ _____(_____, _____)

- In words, X ¯ X ¯ = ____________.
- X ¯ X ¯ ~ ______ (______, ______)
- Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
- Find the probability that the average amount of change of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
- Explain why there is a difference in part (e) and part (f).

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

- If X ¯ X ¯ = average distance in feet for 49 fly balls, then X ¯ X ¯ ~ _______(_______, _______).
- What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X ¯ X ¯ . Shade the region corresponding to the probability. Find the probability.
- Find the 80 th percentile of the distribution of the average of 49 fly balls.

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

- In words, Χ = _____________.
- In words, X ¯ X ¯ = _____________.
- X ¯ X ¯ ~ _____(_____, _____)
- Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
- Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let X ¯ X ¯ be the average of the 49 races.

- Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
- Find the 80 th percentile for the average of these 49 marathons.
- Find the median of the average running times.

The length of songs in a collector’s online album collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.

- In words, Χ = _________.
- Χ ~ _____________
- Find the first quartile for the average song length.
- The IQR for the average song length is _______–_______.

In 1940, the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

- The IQR for X ¯ X ¯ is from _______ acres to _______ acres.

Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

- When the sample size is large, the mean of X ¯ X ¯ is approximately equal to the mean of Χ .
- When the sample size is large, X ¯ X ¯ is approximately normally distributed.
- When the sample size is large, the standard deviation of X ¯ X ¯ is approximately the same as the standard deviation of Χ .

The percentage of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X ¯ X ¯ = average percentage of fat calories.

- X ¯ X ¯ ~ ______(______, ______)
- For the group of 16, find the probability that the average percentage of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.
- Find the first quartile for the average percentage of fat calories.

The distribution of income in some economically developing countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge-shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.

- How is it possible for the standard deviation to be greater than the average?
- Why is it more likely that the average salary of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?

Which of the following is NOT true about the distribution for averages?

- The mean, median, and mode are equal.
- The area under the curve is 1.
- The curve never touches the x -axis.
- The curve is skewed to the right.

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:

- X ¯ X ¯ ~ N (4.59, 0.10)
- X ¯ X ¯ ~ N ( 4 .59, 0.10 16 ) ( 4 .59, 0.10 16 )
- X ¯ X ¯ ~ N ( 4 .59, 16 0.10 ) ( 4 .59, 16 0.10 )

Which of the following is NOT true about the theoretical distribution of sums?

- The area under the curve is one.

Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.

- In words, ΣX = ______________.
- ΣX ~ _____(_____, _____)
- Find the probability that the total length of the nine trials is at least 225 days.
- Ninety percent of the total of nine of these types of trials will last at least how long?

Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.

- In words, X = _____________.
- The distribution is _______.
- In words, ΣX = _______________.
- Find the probability that the total weight of the open boxes is less than 250 pounds.
- Find the 35 th percentile for the total weight of open boxes of cereal.

Salaries for entry-level managers at a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from these restaurants.

- In words, X = ______________.
- X ~ _____(_____, _____)
- In words, ΣX = _____________.
- Find the probability that the managers earn a total of over $400,000.
- Find the 90 th percentile for an individual manager's salary.
- Find the 90 th percentile for the sum of ten managers' salary.
- If we surveyed 70 managers instead of ten, graphically, how would that change the distribution in part (d)?
- If each of the 70 managers received a $3,000 raise, graphically, how would that change the distribution in part (b)?

The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.

- In words, Χ = _______.
- The probability that an individual attention span is less than 10 minutes.
- The probability that the average attention span for the 60 children is less than 10 minutes.
- Calculate the probabilities in part (e).
- Explain why the distribution for X ¯ X ¯ is not exponential.

The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows:

- In words, Χ = ______________.
- x ¯ x ¯ = _____
- s x = _____
- Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of 10.
- In words, describe the distribution of the stock prices.
- Randomly average five stock prices together. (Use a random number generator.) Continue averaging five prices together until you have 10 averages. List those 10 averages.
- Does this histogram look like the graph in Part (c)?
- In one or two complete sentences, explain why the graphs either look the same or look different?
- Based on the theory of the central limit theorem, X ¯ X ¯ ~ _____(_____, ____).

The average wait time is

- two and a half hours

Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is

Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited.

The 90 th percentile sample average wait time (in minutes) for a sample of 100 riders is:

Would you be surprised, based on numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?

- There is not enough information.

What's the approximate probability that the average price for 16 gas stations is more than $4.69?

- almost zero

Find the probability that the average price for 30 gas stations is less than $4.55.

Suppose in a local kindergarten through 12 th grade (K–12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate the following using the normal approximation to the binomial distribution:

- Find the probability that less than 100 favor a charter school for grades K through 5.
- Find the probability that 170 or more favor a charter school for grades K through 5.
- Find the probability that no more than 140 favor a charter school for grades K through 5.
- Find the probability that there are fewer than 130 that favor a charter school for grades K through 5.
- Find the probability that exactly 150 favor a charter school for grades K through 5.

If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology.

Four friends, Janice, Barbara, Kathy, and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places.

- Find the probability that Janice is the driver at most 20 days.
- Find the probability that Roberta is the driver more than 16 days.
- Find the probability that Barbara drives exactly 24 of those 96 days.

X ~ N (60, 9). Suppose that you form random samples of 25 from this distribution. Let X ¯ X ¯ be the random variable of averages. Let ΣX be the random variable of sums. For parts (c) through (f), sketch the graph, shade the region, label and scale the horizontal axis for X ¯ X ¯ , and find the probability.

- Sketch the distributions of X and X ¯ X ¯ on the same graph.
- P ( x ¯ x ¯ < 60) = _____
- Find the 30 th percentile for the mean.
- P (56 < x ¯ x ¯ < 62) = _____
- P (18 < x ¯ x ¯ < 58) = _____
- Σx ~ _____(_____, _____)
- Find the minimum value for the upper quartile for the sum.
- P (1400 < Σx < 1550) = _____

Suppose that the length of research papers is uniformly distributed from 10 to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers.

- μ x = _____
- σ x = _____
- In words, X ¯ X ¯ = ______________.
- Without doing any calculations, do you think that it’s likely the professor will need to read a total of more than 1,050 pages? Why?
- Calculate the probability that the professor will need to read a total of more than 1,050 pages.
- Why is it so unlikely that the average length of the papers will be less than 12 pages?

Salaries for managers in a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from that district.

- Find the 90 th percentile for the average manager's salary.

The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.

- In words, X ¯ X ¯ = ___________________.
- Is it likely that an individual stayed more than five days in the hospital? Why or why not?
- Is it likely that the average stay for the 80 women was more than five days? Why or why not?
- An individual stayed more than five days.
- The average stay of 80 women was more than five days.
- If we were to sum up the women’s stays, is it likely that collectively, they spent more than a year in the hospital? Why or why not?

NeverReady batteries has engineered a newer, longer-lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean life span is 16.7 hours or less? Is the company’s claim reasonable?

Men have an average weight of 172 pounds with a standard deviation of 29 pounds.

- Find the probability that 20 randomly selected men will have a sum weight greater than 3,600 pounds.
- If 20 men have a sum weight greater than 3,500 pounds, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain.

Large bags of a brand of multicolored candies have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class:

The bag contained 465 candies and the listed weights in the table came from randomly selected candies. Count the weights.

- Find the mean sample weight and the standard deviation of the sample weights of candies in the table.
- Find the sum of the sample weights in the table and the standard deviation of the sum of the weights.
- If 465 candies are randomly selected, find the probability that their weights sum to at least 396.9 g.
- Is the candy company's labeling accurate?

The Screw Right Company claims their 3 4 3 4 inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded:

The screws were randomly selected from the local home repair store.

- Find the mean diameter and standard deviation for the sample.
- Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible?

Your company has a contract to perform preventive maintenance on thousands of air conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points?

Certain coins have an average weight of 5.201 g with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?

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## Central Limit Theorem

Central limit theorem assignment help | central limit theorem homework help, the central limit theorem.

- The central limit theorem states that if some certain conditions are satisfied, then the distribution of the arithmetic mean of a number of independent random variables approaches a normal distribution as the number of variables approaches infinity.
- In simple words, If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal.

This is Parent’s Age with µ = 45.64. Let’s imagine that this is the true population value. (Data is normal). As the population is normal, the drawn sample will have mean equal to the population mean, and a stand. dev. that decreases as the N of our sample increases

Taking samples from population:

For sample sizes with n > 30, the sample distribution (x-bar distribution) will be normally distributed with mean µ and standard deviation σ/√n.

Note: If we assume that the population is normally distributed, the sample size is not important.

Because of the central limit theorem, we can use the normal table for any random sample over 30 respondents to find the probability of sample averages!

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