The bell curve is literally, the symmetrical curve created on a graph when using a frequency distribution method for a set of data, splitting the mean symmetrically. There is a big difference between standard deviation and the bell curve! Standard deviation shows the difference in variation from the average; the bell curve, also normal distribution or Gaussian distribution, shows the standard deviation and is created by the normal or equal distribution of the mean among either half. The bell curve is an important distribution pattern occurring in many different forms every day, all around us. Some example include: height, blood pressure, lengths of manufactured objects, etc... Most data creates a bell-shaped curve when graphed on a histogram (hints the name :).Bell curves have a single central peak at the mean of the data. Fifty percent of the distribution lies to the left of the mean and fifty percent lies to the right of the mean. The distribution of the bell curve is controlled by the standard deviation (from that set of data). The smaller the standard deviation, the more concentrated the data in the bell curve will be. The bell curve/ normal distribution equation:

At the Grauer Algebra 2 Corporation, the ages of all new employees hired during the last 5 years are normally distributed. Within this curve, 95.4% of the ages, centered about the mean, are between 24.6 and 37.4 years. Find the mean age and the standard deviation of the data.| |

Solution: As was seen in Example 1, 95.4% implies a span of 2 standard deviations from the mean. The mean age is symmetrically located between -2 standard deviations (24.6) and +2 standard deviations (37.4). The mean age is _24.6 + 37.4 = 31 years of age. 2...

...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...

...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 (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]...

...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...

...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?...

...the information provided by the StandardDeviation.
2. The ability to use the StandardDeviation to calculate the percentage of occurrence of a variable either above or below a particular value.
3. The ability to describe a normal distribution as evidenced by a bell shaped curve as well as the ability to prepare a distribution chart from a set of data (module 3 Case).
Part 1
(1) To get the best deal on a CD player, Tom called eight appliance stores and asked the cost of a specific model. The prices he was quoted are listed below:
$ 298 $ 125 $ 511 $ 157 $ 231 $ 230 $ 304 $ 372 Find the Standarddeviation
$ 298 + $ 125+ $ 511+ $ 157+ $ 231+ $ 230+ $ 304+ $ 372= 2228/8 = 278.5(subtract from #s)
19,-153, 232, -121, -47, -48, 25, 93 (square numbers)
380, 2356, 54056, 14762, 2256, 2352, 650, 8742 = 106(added)
(Divide by 7) 15251 (take square root) StandardDeviation = approximately 123.
(2) When investigating times required for drive-through service, the following results (in seconds) were obtained. Find the range, variance, and standarddeviation for each of the two samples, and then compare the two sets of results.
Wendy's 120 123 153 128 124 118 154 110
MacDonald's 115 126 147 156 118 110 145 137
(2) Set 1:
Range :...

... 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...

...Name: Ashley Lee
Class: HLT-362 Applied Statistics for Healthcare Professionals
Date: 04/01/2015
EXERCISE 18 • Mean, StandardDeviation, and 95% and 99% of the Normal Curve
1. Assuming that the distribution is normal for weight relative to the ideal and 99% of the male participants scored between (–53.68, 64.64), where did 95% of the values for weight relative to the ideal lie? Round your answer to two decimal places.
In order to find where 95% of the values for the weight of relative to the ideal lies you would use the formula that is presented in the text on page 132 of Exercise 18. This formula is:. The = MEAN (5.48) and the (SD) =StandardDeviation (22.93). These numbers were derived from table 1 on pg.133 under the column labeled Male. The problem is worked out as such:
Formula: 5.48±1.96(22.93)
5.48-1.96(22.93) = 5.48-44.94
5.48-44.94= -39.46
5.48+1.96(22.93) = 5.48+44.94
5.48+44.94= 50.42
ANSWER= (-39.46,50.42)
2. Which of the following values from Table 1 tells us about variability of the scores in a distribution?
a. 60.22
b. 11.94
c. 22.57 ←Answer
d. 53.66
The answer for question number 2 is (C). The SD indicates the variability. In the answer set the only choice that was SD was choice (C) the other options were Mean scores listed in table 1.
3. Assuming that the distribution for General Health Perceptions is normal, 95% of the...

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