The Central Limit Theorem for Sums (Optional), The Central Limit Theorem | Statistics

Learning objectives

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:

μ_X = the mean of Χ
σ_Χ = the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N[(n)(μ_Χ), ( $\sqrt{n}$ )(σ_Χ)].

The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.

The random variable ΣX has the following z-score associated with it:

Σx is one sum.
z = Σx–(n)( μ X ) ( n )( σ X )
1. (n)(μ_X) = mean of ΣX
2. $(\sqrt{n}) (σ_{X})$ = standard deviation of $Σ X$

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
Find the sum that is 1.5 standard deviations above the mean of the sums.

Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.

ΣX = the sum or total of 80 values. Because μ_X = 90, σ_X = 15, and n = 80, $Σ X$ ~ N[(80)(90),
( $\sqrt{80}$ )(15)]

mean of the sums = (n)(μ_X) = (80)(90) = 7200
standard deviation of the sums = $(\sqrt{n})(σ_{X}) = (\sqrt{80})$ (15)
sum of 80 values = Σx = 7500

a. Find P(Σx > 7500)

P(Σx > 7500) = 0.0127

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

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 7200 on the horizontal axis. The point 7500 is also labeled. A vertical line extends from point 7500 to the curve. The area to the right of 7500 below the curve is shaded.

b. Find Σx where z = 1.5.

Σx = (n)(μ_X) + (z) $(\sqrt{n})$ (σ_Χ) = (80)(90) + (1.5)( $\sqrt{80}$ )(15) = 7401.2

In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample size is 50.

What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
Find the probability that the sum of the ages is between 1,500 and 1,800 years.
Find the 80^th percentile for the sum of the 50 ages.

μ_Σx = nμ_x = 50(34) = 1,700 and σ_Σx = $\sqrt{n}$ σ_x = $(\sqrt{50})$ (15) = 106.01
The distribution is normal for sums by the central limit theorem.
P(1500 < Σx < 1800) = normalcdf (1500, 1800, (50)(34), $(\sqrt{50})$ (15)) = 0.7974
Let k = the 80^th percentile.
k = invNorm(0.80,(50)(34), $(\sqrt{50})$ (15)) = 1789.3

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 size of 70.

What are the mean and standard deviation for the sums?
Find the 95^th percentile for the sum of the sample. Interpret this value in a complete sentence.
Find the probability that the sum of the sample is at least 10 hours.

μ_Σx = nμ_x = 70(8.2) = 574 minutes and σ_Σx = $(\sqrt{n}) (σ_{x})$ = $(\sqrt{70})$ (1) = 8.37 minutes
Let k = the 95^th percentile.
k = invNorm (0.95,(70)(8.2), $(\sqrt{70})$ (1)) = 587.76 minutes
Ninety-five percent of the app engagement times are at most 587.76 minutes.
10 hours = 600 minutes
P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), $(\sqrt{70})$ (1)) = 0.0009

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 size of 70.

What is the probability that the sum of the sample is between seven hours and 10 hours? What does this mean in context of the problem?
Find the 84^th and 16^th percentiles for the sum of the sample. Interpret these values in context.

7 hours = 420 minutes
10 hours = 600 minutes
normalcdf $P (420 \leq Σ x \leq 600) = n o r m a l c d f (420, 600, (70) (8.2), \sqrt{70} (1)) = 0.9991$
This means that for this sample sums there is a 99.9% chance that the sums of usage minutes will be between 420 minutes and 600 minutes.
$i n v N o r m (0.84, (70) (8.2), \sqrt{70} (1)) = 582.32$
$i n v N o r m (0.16, (70) (8.2), \sqrt{70} (1)) = 565.68$
Since 84% of the app engagement times are at most 582.32 minutes and 16% of the app engagement times are at most 565.68 minutes, we may state that 68% of the app engagement times are between 565.68 minutes and 582.32 minutes.

References

Farago, P. (2012, Oct. 29). The truth about cats and dogs: Smartphone vs tablet usage differences. Flurry Analytics Blog. Retrieved from http://flurrymobile.tumblr.com/post/113379683050/the-truth-about-cats-and-dogs-smartphone-vs

Chapter Review

The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution, even if the original population is not normally distributed. Additionally, if the original population has a mean of μ_X and a standard deviation of σ_x, the mean of the sums is nμ_x and the standard deviation is $(\sqrt{n})$ (σ_x), where n is the sample size.

Formula Review

Central limit theorem for sums: ∑X ~ N[(n)(μ_x),( $\sqrt{n}$ )(σ_x)]

Mean for sums (∑X): (n)(μ_x)

Central limit theorem for sums z-score and standard deviation for sums: $z for the sample mean = \frac{Σ x - (n) (μ_{X})}{(\sqrt{n}) (σ_{X})}$

Standard deviation for sums (∑X): $(\sqrt{n})$ (σ_x)

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Find the probability that the sum of the 95 values is greater than 7,650.

0.3345

Find the probability that the sum of the 95 values is less than 7,400.

Find the sum that is two standard deviations above the mean of the sums.

7833.92

Find the sum that is 1.5 standard deviations below the mean of the sums.

Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Find the probability that the sum of the 40 values is greater than 7,500.

0.0089

Find the probability that the sum of the 40 values is less than 7,000.

Find the sum that is one standard deviation above the mean of the sums.

7326.49

Find the sum that is 1.5 standard deviations below the mean of the sums.

Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

77.45%

Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Find the probability that the sum of the 100 values is greater than 3,910.

Find the probability that the sum of the 100 values is less than 3,900.

0.4207

Find the probability that the sum of the 100 values falls between the numbers you found in eip-311 (16) and eip-886 (17).

Find the sum with a z-score of –2.5.

3,888.5

Find the sum with a z-score of 0.5.

Find the probability that the sums will fall between the z-scores –2 and 1.

0.8186

Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

What is the mean of ΣX?

What is the standard deviation of ΣX?

What is P(Σx = 290)?

What is P(Σx > 290)?

0.9772

True or False: Only the sums of normal distributions are also normal distributions.

In order for the sums of a distribution to approach a normal distribution, what must be true?

The sample size, n, gets larger.

What three things must you know about a distribution to find the probability of sums?

An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42?

An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200?

Use the following information to answer the next three exercises: A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

What is the z-score for Σx = 840?

26.00

What is the z-score for Σx = 1,186?

What is P(Σx < 1186)?

0.1587

Use the following information to answer the next three exercises: An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest.

What is the mean of ΣX?

What is the standard deviation of ΣX?

1000

What is P(Σx > 9000)?

Homework

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

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

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?

the total length of time for nine criminal trials
N(189, 21)
0.0432
162.09; 90 percent of the total nine trials of this type will last 162 days or more.

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 = _______________.
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 = _____________.
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)?

X = the salary of one elementary school teacher in the district
X ~ N(44000, 6500)
ΣX ~ sum of the salaries of 10 elementary school teachers in the sample
ΣX ~ N(44,000, 20,554.80)
0.9742
$52,330.09
466,342.04
Sampling 70 teachers instead of 10 would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.

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