# Introduction

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This module provides a brief introduction to the Central Limit Theorem.

## Student learning outcomes

By the end of this chapter, the student should be able to:

• Recognize the Central Limit Theorem problems.
• Classify continuous word problems by their distributions.
• Apply and interpret the Central Limit Theorem for Means.
• Apply and interpret the Central Limit Theorem for Sums.

## Introduction

Why are we so concerned with means? Two reasons are that they give us a middle ground for comparison and they are easy tocalculate. In this chapter, you will study means and the Central Limit Theorem.

The Central Limit Theorem (CLT for short) is one of the most powerful and useful ideas in all of statistics. Both alternatives are concerned with drawing finitesamples of size $n$ from a population with a known mean, $\mu$ , and a known standard deviation, $\sigma$ . The first alternative says that if we collect samples of size $n$ and $n$ is "large enough," calculate each sample's mean, and create a histogram of those means,then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are "largeenough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.

In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact isthat the sample means and the sums tend to follow the normal distribution. And, the rest you will learn in this chapter.

The size of the sample, $n$ , that is required in order to be to be 'large enough' depends on the original population from which the samples are drawn. If the original population is far from normal then more observations are neededfor the sample means or the sample sums to be normal. Sampling is done with replacement.

## Optional collaborative classroom activity

Do the following example in class: Suppose 8 of you roll 1 fair die 10 times, 7 of you roll 2 fair dice 10 times, 9 of you roll 5 fair dice 10 times, and 11 of you roll 10 fair dice10 times.

Each time a person rolls more than one die, he/she calculates the sample mean of the faces showing. For example, one person might roll 5 fair dice and get a 2, 2, 3, 4, 6 on oneroll.

The mean is $\phantom{\rule{10pt}{0ex}}\frac{2+2+3+4+6}{5}=3.4$ . $\phantom{\rule{10pt}{0ex}}$ The 3.4 is one mean when 5 fair dice are rolled. This same person would roll the 5 dice 9 more times and calculate 9 more means for a total of 10 means.

Your instructor will pass out the dice to several people as described above. Roll your dice 10 times. For each roll, record the faces and find the mean. Round to the nearest0.5.

Your instructor (and possibly you) will produce one graph (it might be a histogram) for 1 die, one graph for 2 dice, one graph for 5 dice, and one graph for 10 dice.Since the "mean" when you roll one die, is just the face on the die, what distribution do these means appear to be representing?

Draw the graph for the means using 2 dice. Do the sample means show any kind of pattern?

Draw the graph for the means using 5 dice. Do you see any pattern emerging?

Finally, draw the graph for the means using 10 dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice?

As the number of dice rolled increases from 1 to 2 to 5 to 10, the following is happening:

1. The mean of the sample means remains approximately the same.
2. The spread of the sample means (the standard deviation of the sample means) gets smaller.
3. The graph appears steeper and thinner.

You have just demonstrated the Central Limit Theorem (CLT).

The Central Limit Theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution).

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
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yes
Asali
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Samantha
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Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
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Damian
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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