THE CENTRAL LIMIT THEOREM:LESSON 24

Please get out your homework from Lesson 23 so we can

review the answers.

OBJECTIVE:

Understand the Central Limit Theorem and how it is applied to statistical process control.

NOTES ON CENTRAL LIMIT THEOREM:

1) “Limit” refers to the fact that the distribution of the sample mean becomes normal in the limit, as the sample size increases.

2) “Central” is indicative of the central importance of this theorem in statistics.

SAMPLING DISTRIBUTION OF THE SAMPLE MEAN:

• The sample mean as the population, = 𝜇• A standard deviation equal to

the standard deviation of the population divided by the square root of the sample size,

=

• A shape that becomes more approximately normal with larger and larger sample sizes.

Let’s take a look at an example!

As we work through this investigation, look for answers to the following question:

What is the Central Limit Theorem and how does it allow you to use control charts even when individual measurements

come from a skewed distribution?

Suppose the carnival wheel below is designed to produce random digits from the set .1) Suppose the

wheel is operatingCorrectly and produces digits at random.

a) Describe the shape of the probability distribution of all possible outcomes.

b) What is the mean of the distribution?

c) What is the standard deviation of the distribution?

2) Look at the eight tests on the worksheet given to you in Lesson 23 and assume the observations come from a distribution that is approximately normal. Can Test 1 ever signal that this process may be out of control? Explain. What about Test 5?

Your company has been successful in adjusting the machine that fills cereal boxes so that the mean weight in the boxes is 16.1 ounces. After filling several hundred boxes, it is obvious that the distribution of the weights is somewhat skewed.

To ensure that you are warned if the process goes out of control, you decide to use an x-bar chart with samples of size 3. You set the upper control limit at 16.39 and the lower limit at 15.81. Several days later, seven samples of the weights of three individual boxes were taken, as shown in the table.

SAMPLE MEASUREMENTSFirst 16.05, 16.27, 15.68

Second 16.10, 15.80, 15.75Third 16.98, 16.12, 15.73

Fourth 15.93, 15.92, 15.92Fifth 16.47, 16.04, 16.93Sixth 15.58, 15.90, 16.25

Seventh 15.84, 15.60, 15.77

SAMPLE NUMBER AVERAGESFirst 16.05 16.27 15.68 16.0000

Second 16.10 15.80 15.75 15.8833Third 16.98 16.12 15.73 16.2767

Fourth 15.93 15.92 15.91 15.9200Fifth 16.47 16.04 16.93 16.4800Sixth 15.58 15.90 16.25 15.9100

Seventh 15.84 15.60 15.77 15.7367GRAND MEAN 16.1357 15.9500 16.0029 16.0295STANDARD DEVIATION 0.4604 0.2190 0.4521 0.2576UPPER CONTROL LIMIT 17.5169 16.6070 17.3590 16.8025LOWER CONTROL LIMIT 14.7545 15.2930 14.6467 15.2566

MEASUREMENTS

a) Compute the mean of each sample of size 3. Plot all seven means on an x-bar chart that shows the upper and lower control limits.

b) Has the process gone out of control?

c) Why were the samples of size 3 used rather than size 1?

1 2 3 4 5 6 715.7

15.9

16.1

16.3

16.5

16.0000

15.8833

16.2767

15.9200

16.4800

15.9100

15.7367

AVERAGES

SAMPLE NUMBER

SA

MP

LE M

EA

N

UCL = 16.38

LCL = 15.81

= 16.10𝜒

Homework1. “Summarize the

Mathematics” worksheet.2. “Los Angeles Rainfall 1878 -

2006 ” worksheet.