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 standard deviation for a table: The table shows the number of children living in a sample of households:

We find the mean for such frequency data using the revised formula: [pic], where n = [pic].

So, we find that [pic]

The revised formula for the variance is: [pic].

So we find that: [pic]=

Therefore, s.d. = [pic]

Example 2: Grouped frequency data (examination style question) A survey of 500 coaches arriving at a bus station found that 400 were delayed. The delay times, in minutes, for these 400 delayed coaches are summarised in the table.

...I'll be honest. Standarddeviation 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 "standarddeviation" in a story. But that doesn't mean you should ignore this concept.
The standarddeviation 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......

...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...

...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...

...13. Variance and StandardDeviation (expected). Using the data from problem 13, calculate the variance and standarddeviation of the three investments, stock, corporate bond, and government bond. If the estimates for both the probabilities of the economy and the returns in each state of the economy are correct, which investment would you choose considering both risk and return? Why?
ANSWER
Variance of Stock = 0.10 x (0.25 – 0.033)2 + 0.15 x (0.12 – 0.033)2
+ 0.50 x (0.04 – 0.033)2 + 0.25 x (-0.12 – 0.033)2
= 0.10 x 0.0471 + 0.15 x 0.0076 + 0.50 x 0.0000 + 0.25 x 0.0234
= 0.0047 + 0.0011 + 0.0000 + 0.0059 = 0.0117 or 1.17%
StandardDeviation of Stock = (0.0117)1/2 = 0.1083 or 10.83%
Variance of Corp. Bond = 0.10 x (0.09 – 0.052)2 + 0.15 x (0.07 – 0.052)2
+ 0.50 x (0.05 – 0.052)2 + 0.25 x (0.03 – 0.052)2
= 0.10 x 0.0014 + 0.15 x 0.0003 + 0.50 x 0.0000 + 0.25 x 0.0005
= 0.0001 + 0.0000 + 0.0000 + 0.0001 = 0.000316 or 0.00316%
StandardDeviation of Corp. Bond = (0.000316)1/2 = 0.017776 or 1.78%
Variance of Gov. Bond = 0.10 x (0.08 – 0.042)2 + 0.15 x (0.06 – 0.042)2
+ 0.50 x (0.04 – 0.042)2 + 0.25 x (0.02 – 0.042)2
= 0.10 x 0.0014 + 0.15 x 0.0003 + 0.50 x 0.0000 + 0.25 x 0.0005
= 0.0001 + 0.0000 + 0.0000 + 0.0001 = 0.000316 or 0.0316%
Standard...