...
Standard Deviation Abstract
QRB/501
Standard Deviation Abstract
Standard Deviations Are Not Perverse
Purpose:
The purpose of this article is to illustrate how using statistical data, such as standard deviation, can help a cattleman choose the best lot of calf’s at auction. The statistical data used in these decision making processes can also help the cattleman with future analysis of the lots purchased and existing stock.
Research Question:
How can understanding the standard deviation of weights in a lot of calf’s be used to determine which lot should be purchased?
Hypothesis:
The hypothesis proposed in this article is that in using the group average and standard deviation a cattleman can choose a lot of calf’s that is larger in number and closer in average size, thereby, increase the likelihood of a good purchase.
Findings:
The finding in this article prove that choosing a smaller standard deviation will ensure that the lot of calf’s will have a more consistent weight amount the group. The example provided in the text outlines two groups of calves and the standard deviation of both groups, one being 150 lbs and the other at 40 lbs. The second group with the smaller standard deviation helps the cattleman to understand that in the group 66% of the calf’s will fall into the weight range of plus or minus 40 pounds of the...

...Deviation
Definition:
Behavior commonly seen in children that is the result of some obstacle to normal development such behavior may be commonly understand as negative (a timid child, a destructive child) or positive (a quite child), both positive and negative deviation will disappear once the child begins to concentrate on a piece of work freely chosen by him.
The physical deforms are easier to identify. This can be by birth due to an accident etc… and most such physical deforms can be either cured. However, deforms that take place in development of psychological aspects of a child are not only threat to building the character and the personality of the child also you find certain physical deforms in curable in medicine.
Dr. Montessori, according to her she fugues deviated children are.
The naughty children act and react very strong as a result of severe treatment they have received. So their behavior is cruel from others.
The weak children are always mistaken for good children and parent are happy because of their timidly and but they are lazy and afraid of everything.
The bright children are very imaginative and live in their own fantasy world.
Deviations shown by the strong and weak children are:
In the absorbent mind Montessori discussed deviation shown by the strong, meaning those who resist and overcome the obstacles they meet and deviation...

...I'll be honest. Standard deviation is a more difficult concept than the others we've covered. And unless you are writing for a specialized, professional audience, you'll probably never use the words "standard deviation" in a story. But that doesn't mean you should ignore this concept.
The standard deviation is kind of the "mean of the mean," and often can help you find the story behind the data. To understand this concept, it can help to learn about what statisticians call normal distribution of data.
A normal distribution of data means that most of the examples in a set of data are close to the "average," while relatively few examples tend to one extreme or the other.
Let's say you are writing a story about nutrition. You need to look at people's typical daily calorie consumption. Like most data, the numbers for people's typical consumption probably will turn out to be normally distributed. That is, for most people, their consumption will be close to the mean, while fewer people eat a lot more or a lot less than the mean.
When you think about it, that's just common sense. Not that many people are getting by on a single serving of kelp and rice. Or on eight meals of steak and milkshakes. Most people lie somewhere in between.
If you looked at normally distributed data on a graph, it would look something like this:
The x-axis (the horizontal one) is the value in question... calories consumed, dollars earned or...

...in Laramie, Wyoming.
b) all residential addresses in Laramie, Wyoming.
c) the members of the marketing firm that actually conducted the survey.
d) the 100 addresses to which the survey was mailed.
7. The chance that all 100 homes in a particular neighborhood in Laramie end up being the sample of residential addresses selected is
a) the same as for any other set of 100 residential addresses.
b) exactly 0. Simple random samples will spread out the addresses selected.
c) reasonably large due to the “cluster” effect.
d) 100 divided by the size of the population of Laramie.
Costs for standard veterinary services at a local animal hospital follow a Normal distribution with a mean of $80 and a standard deviation of $20. Answer the next three questions.
8. Give the sample space for the costs of standard veterinary services.
a) {X ≥ 0}
b) { 0 ≤ X ≤ 80}
c) {0 ≤ X ≤ 160}
d) None of these.
9. What is the probability that one bill for veterinary services costs less than $95?
a) 0.75
a) 0.7734
b) 0.2266
c) 0.15
10. What is the probability that one bill for veterinary services costs between $75 and $105?
a) 1
a) 0.25
b) 0.4013
c) 0.4931
11. In an instant lottery, your chances of winning are 0.2. If you play the lottery five times and outcomes are independent, what is the probability that you win at least once?
a. 0.2
a) 0.08192
b) 0.32768
c) 0.67232...

...Standard Deviation
objective
• Describe standard deviation and
it’s importance in biostatistics.
Measure of Dispersion
• Indicates how widely the scores
are dispersed around the central
point (or mean.)
-Standard deviation
Standard Deviation.
• The most commonly used method
of dispersion in oral hygiene.
• The larger the standard deviation,
the wider the distribution curve.
Standard Deviation
• SD, , (sigma)
• Indicates how subjects differ from
the average of the group/ the more
they spread out, the larger the
deviation
• Based upon ALL scores, not just
high/low or middle half
• Analyzes descriptively the spread of
scores around the mean
– 14+ 2.51 = Mean of 14 and SD of
2.51
Standard Deviation
• The spread of scores around the
mean:
• For example, if the mean is 60 and
the standard deviation 10, the
lowest score might be around 30,
and the highest score might be
around 90.
Standard Deviation &
Variance
Usefulness
• When comparing the amount of dispersion in
two data sets.
• Greater variance = greater dispersion
• Standard deviation--”average” difference
between the mean of a sample and each data
value in the sample
14+ 2.51 = Mean of 14 and SD of 2.51
Distribution Shape
• Normal
• Skewed
• Multimodial
NORMAL CURVE
• Mean is the focal point from which all assumptions made
• Area under curve = 100%
•...

...Standard Deviation (continued)
L.O.: To find the mean and standard deviation from a frequency table.
The formula for the standard deviation of a set of data is [pic]
Recap question
A sample of 60 matchboxes gave the following results for the variable x (the number of matches in a box):
[pic].
Calculate the mean and standard deviation for x.
Introductory example for finding the mean and standarddeviation for a table:
The table shows the number of children living in a sample of households:
|Number of children, x |Frequency, f |xf |x2f |
|0 |14 |0 × 14 = 0 |02 × 14 = 0 |
|1 |12 |1 × 12 = 12 | |
|2 |8 | | |
|3 |6 | |32 × 6 = 54 |
|TOTAL |[pic] |[pic] |[pic]...

...988.00 | 29,997.00 | - | 149,985.00 |
σ² = 149,985.00 - 299.97²
= 60,003
σ = [pic]
= 244.9551
P([pic]≥320) = 1 - P ( Z < [pic] - µ[pic] )
σ[pic]
= 1 - P ( Z < 320-299.97 )
244.9551/[pic]
= 1 - P ( Z < 20.03 )
14.1425
= 1 - P ( Z < 1.42)
= 1 - 0.9222
= 0.0778 is the probability that at least 320 will attend.
1 - NORMDIST(320,299.97,244.9551/SQRT(300),1) = 0.0783
(c) A circuit contains three resistors wired in series. Each is rated at 6 ohms. Suppose, however, that the true resistance of each one is a normally distributed random variable with a mean of 6 ohms and a standard deviation of 0.3 ohm. What is the probability that the combined resistance will exceed 19 ohms? How "precise" would the manufacturing process have to be to make the probability less than 0.005 that the combined resistance of the circuit would exceed 19 ohms?
n = 3
µ = 6 µ[pic] = 6
σ = 0.3 σ[pic] = 0.3 /[pic]
[pic]= 19 / 3
P ([pic]≥ 6.33) = 1- P ( Z < [pic] - µ[pic] )
σ[pic]
= 1 - P (Z < 6.33-6 )
0.3 /[pic]
= 1 - P ( Z < 0.33 )
0.1732
= 1 – P ( Z < 1.91
= 1- 0.9719
= 0.0281 is the probability that the combined resistance will exceed 19 ohms.
1-NORMDIST(6.33,6,0.3/SQRT(3),1) = 0.0284
0.005 = -2.58
-2.58...

...new point on the standard deviation?
The new point has made the standard deviation to go up to over 2.07
b) Follow the instructions to create the next two graphs then answer the following question: What did you do differently to create the data set with the larger standard deviation.
What I did differently was to have two outliners on both ends of the outline so I can create the larger standard deviation and also to keep the mean at five.
2. Go back to the applet and put points matching each of the following data set into the first graph of the applet and clear the other two graphs. Set the lower limit to 0 and the upper limit to 100.
50, 50, 50, 50, 50
Notice that the standard deviation is 0. Explain why the standard deviation for this one is zero. Don’t show just the calculation. Explain in words why the standard deviation is zero when all of the points are the same.
There’s not a deviation from this sample because all the data points are equal to each other.
3. Go back to the applet one last time and set all 3 of the lower limits to 0 and upper limits to 100. Then put each of the following three data sets into one of the graphs.
Data set 1: 0, 25, 50, 75, 100
Data set 2: 30, 40, 50, 60, 70
Data set 3: 40, 45, 50, 55, 60
Note that all three data sets have a median of 50. Notice...