1. Researcher conducted by a tobacco company indicates that the relative frequency distribution of tar content of its newly developed low-tar cigarette has a mean equal to 3.9 milligrams of tar per cigarette and a standard deviation 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 standard deviation 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 standard deviation 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 standard deviation (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 normally distributed with a mean of 10 and a standard deviation of 3. What is the probability that in a random sample of 25 students, more than 275 pizzas are consumed? [0.04746]

6.The number of customers who enter a supermarket each hour is normally distributed with a mean of 600 and a standard deviation of 200. The supermarket is open 16 hours per day. What is the probability that the total number of customers...

...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...

...charity work.
| Frequently | Occasionally | Not at all | Total |
Male | 221 | 456 | 795 | 1472 |
Female | 207 | 430 | 741 | 1378 |
Total | 428 | 886 | 1536 | 2850 |
A person is selected at random from the sample.
a) What is the probability the person is female or occasionally involved in charity work?
b) Are the events “being female and occasionally involved in charity work” and “being frequently involved in charity work” mutually exclusive?
yes
6. A company gave psychological tests to perspective employees. The random variable x represents the possible test scores.
a) Use the histogram to find the probability that a person selected at random from the survey’s sample had a test score of more than two.
b) Find the probability that the person had a test score of at most 2.
7. The following table is a frequency distribution for the number of dogs per household in a small town.
Dogs | 0 | 1 | 2 |
Households | 931 | 297 | 180 |
a) Construct the probability distribution. (round to the thousandths place)
x | 0 | 1 | 2 |
P(x) | 931/1408=0.661 | 297/1408=0.211 | 180/1408=0.128 |
b) Find the mean of the probability distribution. (round to the nearest tenth)
c) Using the mean from part b find the standarddeviation of the probability distribution.
8. A computer password consists of two letters followed by a...

...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...

...A survey is conducted in Chicago (population 2,800,000) using random-digit-dialing equipment which place calls at random to residential phones, both mobile and landline. The purpose of the survey is to determine the percentage of adults who would favor a half-cent increase in the sales tax to help fund public transportation. Four hundred adults are interviewed and 36% of them favor the proposal. Answer the next two questions.
1. The sample size for this sample survey appears to be
a) 400
b) 2,800,000
c) 144
d) 1,008,000
2. The 36% is a
a) Parameter
b) Margin of error
c) Chance of 144 people agreeing to the statement
d) Statistic
3. Event A occurs with probability 0.05. Event B occurs with probability 0.75. If A and B are disjoint, which statement is true?
a) P(A and B) = 0
b) P(A or B) = 0.80
c) P(A and B) = 0.0375
d) Both (a) and (b) are true.
4. Event A occurs with probability 0.05. Event B occurs with probability 0.75. If A and B are independent, which statement is true.
e) P(A and B) = 0
a) P(A or B) = 0.80
b) P(A and B) = 0.0375
c) Both (a) and (b) are true.
A marketing research firm wishes to determine if the adult men in Laramie, Wyoming would be interested in a new upscale men's clothing store. From a list of all residential addresses in Laramie, the firm selects a simple randomsample...

...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 : maximum - minimum =...

...we take a randomsample of size 100 from a discrete distribution in this manner: A green die and a red die are thrown simultaneously 100 times and let Xi denote the sum of the spots on the two dice on the ith throw, i = 1, 2,...100. Find the probability that the sample mean number of spots on the two dice is less than 7.5.
n = 100
µ = 7 µ[pic] = 7
σ = 2.41 σ[pic] = 2.41 /[pic]
|X |2 |3 |4 |5 |
|P(X=x) |0.3333 |0.3333 |0.3333 | 1.00 |
|P.X |199.9800 |99.9900 |0.0000 | 299.97 |
|P².X | 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...

...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:...

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