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 crimes committed, for example. And the y-axis (the vertical one) is the number of datapoints for each value on the x-axis... in other words, the number of people who eat x calories, the number of households that earn x dollars, or the number of cities with x crimes committed.

Now, not all sets of data will have graphs that look this perfect. Some will have relatively flat curves, others will be pretty steep. Sometimes the mean will lean a little bit to one side or the other. But all normally distributed data will have something like this...

...StandardDeviation
objective
• Describe standarddeviation and
it’s importance in biostatistics.
Measure of Dispersion
• Indicates how widely the scores
are dispersed around the central
point (or mean.)
-StandarddeviationStandardDeviation.
• The most commonly used method
of dispersion in oral hygiene.
• The larger the standarddeviation,
the wider the distribution curve.
StandardDeviation
• 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
StandardDeviation
• The spread of scores around the
mean:
• For example, if the mean is 60 and
the standarddeviation 10, the
lowest score might be around 30,
and the highest score might be
around 90.
StandardDeviation &
Variance
Usefulness
• When comparing the amount of dispersion in
two data sets.
• Greater variance = greater dispersion
• Standarddeviation--”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...

...StandardDeviation (continued)
L.O.: To find the mean and standarddeviation from a frequency table.
The formula for the standarddeviation 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 standarddeviation 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]...

...the new point on the standarddeviation?
The new point has made the standarddeviation 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 standarddeviation.
What I did differently was to have two outliners on both ends of the outline so I can create the largerstandarddeviation 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 standarddeviation is 0. Explain why the standarddeviation for this one is zero. Don’t show just the calculation. Explain in words why the standarddeviation 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,...

... 119,988.00 | 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 standarddeviation 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...

...all adult men 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 standarddeviation 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?...

...equal to 3.9 milligrams of tar per cigarette and a standarddeviation equal to 1.0 milligram. Suppose a sample of 100 low-tar cigarettes is randomly selected from a day’s production and the tar content is measured in each. Assuming that the tobacco company’s claim is true, what is the probability that the mean tar content of the sample is greater than 4.15 milligrams?
[0.00621]
2. The safety limit of a crane is known to be 32 tons. The mean weight and the standarddeviation of a large number of iron rods are 0.3 ton and 0.2 ton respectively. One hundred rods are lifted at a time. Compute the probability of an accident.
[0.1587]
3. A soft –drink vending machine is set so that the amount of drink dispensed is a random variable with a mean of 200 milliliters and a standarddeviation of 15 milliliters. What is the probability that the mean amount dispensed in a random sample of size 36 is at least 204 milliliters?
[0.0548]
4. An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with mean (μ) = 117 cm and standarddeviation (σ) = 6.1 cm. Find the probability that if four subcomponents are randomly selected, their mean length exceeds 120 cm.
[0.16354]
5. The number of pizzas consumed per month by university students is...

...Mean and StandardDeviation
The mean, indicated by μ (a lower case Greek mu), is the statistician's jargon for the average value of a signal. It is found just as you would expect: add all of the samples together, and divide by N. It looks like this in mathematical form:
In words, sum the values in the signal, xi, by letting the index, i, run from 0 to N-1. Then finish the calculation by dividing the sum by N. This is identical to the equation: μ =(x0 + x1 + x2 + ... + xN-1)/N. If you are not already familiar with Σ (upper case Greek sigma) being used to indicate summation, study these equations carefully, and compare them with the computer program in Table 2-1. Summations of this type are abundant in DSP, and you need to understand this notation fully. In electronics, the mean is commonly called the DC (direct current) value. Likewise, AC (alternating current) refers to how the signal fluctuates around the mean value. If the signal is a simple repetitive waveform, such as a sine or square wave, its excursions can be described by its peak-to-peak amplitude. Unfortunately, most acquired signals do not show a well defined peak-to-peak value, but have a random nature, such as the signals in Fig. 2-1. A more generalized method must be used in these cases, called the standarddeviation, denoted by σ (a lower case Greek sigma).
As a starting point, the expression,|xi-μ|, describes how far the ith sample deviates...