The Central Limit Theorem for Sample Means (Averages)

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 $\overline{X}$ = the mean of a sample of size 25. Because μ_x = 90, σ_x = 15, and n = 25,

Find P(85 < $\overline{x}$ < 92). Draw a graph.

P(85 < $\overline{x}$ < 92) = 0.6997

The probability that the sample mean is between 85 and 92 is 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 $\frac{\sigma x}{\sqrt{n}}$ = $\frac{15}{\sqrt{25}}$ = 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.

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 $\overline{X}$ = 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$\left(\text{2, }\frac{0.5}{\sqrt{50}}\right)$

Find P(1.8 < $\overline{x}$ < 2.3). Draw a graph.

normalcdf$\left(1.\text{8,2}\text{.3,2,}\frac{.5}{\sqrt{50}}\right)\text{= 0.9977}$

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

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.

1. What are the mean and standard deviation for the sample mean ages of tablet users?
2. What does the distribution look like?
3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
4. Find the 95^th percentile for the sample mean age (to one decimal place).

1. Because the sample mean tends to target the population mean, we have μ_χ = μ = 34. The sample standard deviation is given by ${\text{σ}}_{\text{χ}}$ = $\frac{\sigma }{\sqrt{n}}$ = $\frac{15}{\sqrt{100}}$ = $\frac{15}{10}$ = 1.5.
   
2. The central limit theorem states that for large sample sizes (n), the sampling distribution will be approximately normal.
   
3. 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.
   
4. Let k = the 95^th percentile.
   k = invNorm$\left(0.\text{95,34,}\frac{15}{\sqrt{100}}\right)$ = 36.5

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.

1. What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user?
2. What is the standard error of the mean?
3. Find the 90^th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.

1. ${\mu }_{\overline{x}}=\mu =8.2{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{1}{\sqrt{60}}=0.13$
2. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
3. Let k = the 90^th percentile.
   k = invNorm$\left(0.\text{90,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
4. P(8 < $\overline{x}$ < 8.5) = normalcdf$\left(\text{8,8}\text{.5,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 0.9293

What is the mean, standard deviation, and sample size?

mean = 4 hours, standard deviation = 1.2 hours, sample size = 16

Complete the distributions.

1. X ~ _____(_____, _____)
2. $\overline{X}$ ~ _____(_____, _____)

Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

1. View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.

2. P(________ < x < ________) = _______

a. Check student's solution.
b. 3.5, 4.25, 0.2441

Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

1. View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.

2. P(________________) = _______

What causes the probabilities in element-446 and element-322 to be different?

The fact that the two distributions are different accounts for the different probabilities.

Find the 95^th percentile for the mean time to complete one month's reviews. Sketch the graph.

1. View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.

2. The 95^th percentile =____________

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.

1. In words, Χ = ____________.
2. Χ ~ _____(_____, _____)
3. In words, $\overline{X}$ = ____________.
4. $\overline{X}$ ~ ______ (______, ______)
5. Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
6. 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.
7. Explain why there is a difference in part (e) and part (f).

1. Χ = amount of change students carry
2. Χ ~ Exp(1/0.88) or approximately Χ ~ Exp(1.1364)
3. $\overline{X}$ = average amount of change carried by a sample of 25 students.
4. $\overline{X}$ ~ N(0.88, 0.176)
5. 0.0819
6. 0.4276
7. The probability in part (e) represents the probability of an individual value. In part (f), the probability describes the mean of a sample of 25. Part (f) relies on the central limit theorem, so the distributions are different. Part (e) is exponential and part (f) is normal.

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.

1. If $\overline{X}$ = average distance in feet for 49 fly balls, then $\overline{X}$ ~ _______(_______, _______).
2. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for $\overline{X}$. Shade the region corresponding to the probability. Find the probability.
3. 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.

1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. 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.
5. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

1. length of time for an individual to complete IRS form 1040, in hours
2. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours
3. N$\left(\text{10}\text{.53, }\frac{1}{3}\right)$
4. Yes, I would be surprised, because the probability is almost 0.
5. No, I would not be totally surprised because the probability is 0.2312.

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 $\overline{X}$ be the average of the 49 races.

1. $\overline{X}$ ~ _____(_____, _____)
2. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
3. Find the 80^th percentile for the average of these 49 marathons.
4. 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.

1. In words, Χ = _________.
2. Χ ~ _____________
3. In words, $\overline{X}$ = _____________.
4. $\overline{X}$ ~ _____(_____, _____)
5. Find the first quartile for the average song length.
6. The IQR for the average song length is _______–_______.

1. the length of a song, in minutes, in the collection
2. U(2, 3.5)
3. the average length, in minutes, of the songs from a sample of five albums from the collection
4. N(2.75, 0.0220)
5. 2.74 minutes
6. 0.03 minutes

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.

1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. The IQR for $\overline{X}$ is from _______ acres to _______ acres.

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

1. When the sample size is large, the mean of $\overline{X}$ is approximately equal to the mean of Χ.
2. When the sample size is large, $\overline{X}$ is approximately normally distributed.
3. When the sample size is large, the standard deviation of $\overline{X}$ is approximately the same as the standard deviation of Χ.

1. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
2. True. According to the central limit theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
3. The standard deviation of the sampling distribution of the means will decrease, making it approximately the same as the standard deviation of X as the sample size increases.

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 $\overline{X}$ = average percentage of fat calories.

1. $\overline{X}$ ~ ______(______, ______)
2. 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.
3. 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.

1. In words, Χ = _____________.
2. In words, $\overline{X}$ = _____________.
3. $\overline{X}$ ~ _____(_____, _____)
4. How is it possible for the standard deviation to be greater than the average?
5. 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?

1. X = the yearly income of someone in a Third World country
2. the average salary from samples of 1,000 residents of a Third World country
3. $\overline{X}$ ∼ N$\left(\text{2,000, }\frac{\text{8,000}}{\sqrt{\text{1,000}}}\right)$
4. Very wide differences in data values can have averages smaller than standard deviations.
5. The distribution of the sample mean will have higher probabilities closer to the population mean.
   P(2,000 < $\overline{X}$ < 2,100) = 0.1537
   P(2,100 < $\overline{X}$ < 2,200) = 0.1317

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

1. The mean, median, and mode are equal.
2. The area under the curve is 1.
3. The curve never touches the x-axis.
4. 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:

1. $\overline{X}$ ~ N(4.59, 0.10)
2. $\overline{X}$ ~ N$\left(\text{4}\text{.59, }\frac{0.10}{\sqrt{16}}\right)$
3. $\overline{X}$ ~ N$\left(\text{4}\text{.59, }\frac{16}{0.10}\right)$
4. $\overline{X}$ ~ N$\left(\text{4}\text{.59, }\frac{\sqrt{16}}{0.10}\right)$

b