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ASSIGNMENT/PROJECT VII MATH 566
1. (Assessing normality). Many times in statistics it is necessary to see if a
distribution of data values is approximately normal. One technique is to draw a
histogram for the data and see if it is approximately bell-shaped. The following
table lists the number of branches of the 50 top libraries:
67, 84, 80, 77, 97, 59, 62, 37, 33, 42, 36, 54, 18, 12, 19, 33, 49, 24,
25, 22, 24, 29, 9, 21, 21, 24, 31, 17, 15, 21, 13, 19, 19, 22, 22, 30,
41, 18, 20, 22, 26, 33, 14, 14, 16, 22, 26, 10, 16, 25
(i) Construct a frequency distribution for the data using class width of 10.
The answer for this one is what you answered for ii. The range is supposed to be
10 in which case 0-9 is a width of 10. 0-10 would be a width of 11.
Bin
Range Frequency
9
1
19
14
29
17
39
7
49
3
59
2
69
2
79
1
89
2
99
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(ii) Draw a histogram from the distribution given in (i).
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0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
1
14
17
7
3
2
2
1
2
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(iii) Describe the shape of the distribution.
The shape is approximately normal; it is skewed up and to the right.
The shape is skewed to the left, not right.
(iv) Based on your answer to (iii), do you think that the distribution is approximately normal?
Yes, it is approximately normal.
No, I would not call this shape approximately normal.
In addition to the histogram, distributions that are normal have about 68% of
the values fall within 1 standard deviation of the mean, about 95% of the values
fall within 2 standard deviation of the mean, and almost 100% of the values fall
within 3 standard deviation of the mean (see Figure 6-5, page 268).
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(v) Find the mean and standard deviation for the data.
Mean
31.4
Stddev
20.61
(vi) What percent of data values fall within 1 standard deviation of the mean?
Obs in
Sigma
Range
Mean Stdev
Mean +
Stdev
1 Sigma
10.79
52.01
40
80%
2 Sigma
(9.83)
72.63
46
92%
3 Sigma
(30.44)
93.24
49
98%

Percentage
of Sample
82% is one standard deviation of the mean – Incorrect – Would
probably by 82% if you counted the 2 observations that are less than
10.
(vii) What percent of data values fall within 2 standard deviation of the mean?

92% is within two standard deviation of the mean
(viii) What percent of data values fall within 3 standard deviation of the mean?

98% is within three standard deviation of the mean
(ix) How do your answers to (vi), (vii), and (viii) compare to 68%, 95%, and
100%, respectively?

Comparing 68% to 82% there is a significant gap, 95% to 92 the gap has
minimized and same with comparing 100% to 98%
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2. (Smart people). Assume that you are thinking about starting a Mensa chapter
in your home town, which has a population of 10,000 people. You need to know
how many people would qualify for Mensa, which requires an IQ score of at least
130. You realize that IQ scores is normally distributed with a mean of 100 and a
standard deviation of 15. Complete the following:
(i) Find the approximate number of people in your home town that are eligible
for Mensa.
Get the z score:

z = (130-100)/15 = 2

Use the z table to get probability (z > 2) = 0.02275

Multiply by the 10000 people:
227.5 about 228 people
(ii) Is it reasonable to continue your quest for a Mensa chapter in your home
town?
Yes, it is reasonable, since there are enough people in the home town to start
a chapter
(iii) How would you proceed to find out about how many of the eligible people
would actually join the new chapter? Be specific about your methods of gathering
data.

You can put up flyers at the library, bus stops, local colleges, schools, and
other populated areas.

You can encourage people to get IQ tested.

You can ask your friends to spread the word.

You can send mass e-mails about the project to family and friends
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(iv) What would be the minimum IQ score needed if you wanted to start an
Ultra-Mensa club that include only the top 1% of IQ scores?
The z score for the top 1% is (from a table):
2.3263
Use the formula:
x = sigma x z + mu
x = 2.3263 x 15 + 100
x = 134.8945
3. p. 314 Using the standard normal distribution find each
Probability.
a. P(0 < z < 2.07) =
0.4808
b. P(-1.83 < z < 0) =
0..4664
c. P(-1.59 < z < 2.01) =
0.9219
d. P(1.33 < z < 1.88) =
0.0617
e. P(-2.56 < z < 0.37) =
.06391
f. P(z > 1.66) =
0.0485
g. P(z < 2.03) =
0.9788
h. P(z > 1.19) =
0.1170
i. P(z < 1.93) =
0.9732
j. P(z > 1.77) =
0.0383
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-x has a normal distribution with a mean of 30 and a standard deviation of 4
-z has a normal distribution with a mean of 0 and a standard deviation of 1
4. p. 316
The time it takes for a certain pain reliever to begin to reduce symptoms is 30
minutes, with a standard deviation of 4 minutes. Assuming the variable is
normally distributed, find the probability that it will take the medication
a. Between 34 and 35 minutes to begin to work.
a. P(34<x<35) = P(34-30)/4<Z<35-30)/4) = P(1<Z<1.25) = P(Z<1.25)P(Z<1) = 0.894 – 0.841 = 0.053
b. More than 35 minutes to begin to work.
b. P(x>) =P(Z>1.25)=1-P(Z<1.25) = 1-0.894 = 0.106
c. Less than 25 minutes to begin to work.
c. P(x<25) = P(z<1.25) = 0.106
d. Between 35 and 40 minutes to begin to work.
d. P(35<x<40) = P(1.25<Z<2.5) = P(Z<2.5)-P(Z<1.25)=0.994-0.894 = 0.1
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5. P.317
The average number of years a person takes to complete a graduate degree
program is 3. The standard deviation is 4 months. Assume the variable is
normally distributed. If an individual enrolls in the program, find the probability
that it will take:
4 months = 0.333333 years
a. More than 4 years to complete the program.
Z= (4-3)/0.333333 = 3
P(z > 3) from a table = 0.00135
b. Less than 3 years to complete the program.
Z = 0, since 3 years in the mean
P(z < 0) = 0.5 , because half the data is below the mean
c. Between 3.8 and 4.5 years to complete the program.
Z (3.8) = (3.8-3)/0.3333 = 2.4
Z (4.5) = (4.5-3)/0.3333 = 4.5
P(24 < z < 4.5) from a table = 0.008194
d. Between 2.5 and 3.1 years to complete the program.
Z (2.5) = (2.5-3)/0.33333 = -1.5
Z (3.1) = (3.1-3)/.33333 = 0.3
P(-1.5 < z < 0.3) from a table = 0.5511
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6. Data Project 1 p. 319.
Select a variable (interval or ratio) and collect 30 data values.
a. Construct a frequency distribution for the variable.
(a) Collect a data of size n=30 or more and construct
a frequency distribution table.
b. Use the procedure described in the Critical Thinking
Challenge on page 317 to graph the distribution on
normal probability paper.
(b) Find the mean and standard deviation of the
data.
c. Can you conclude that the data are approximately normally distributed?
Explain your answer.
(c) Find the approximate probability for each
class interval given in part (a).
(d) refer to the critical thinking challenges
problem for drawing the graph on normal probability paper.
use the cdf (cumulative distribution function)
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1. Make a table, as shown.
Boundaries
Frequency
Frequency Cumulative Frequency
89.5–104.5
104.5–119.5
119.5–134.5
134.5–149.5
149.5–164.5
164.5–179.5
Low
Cumulative Percent
24
62
72
26
12
_4__
200
Freq
89.5-104.5
104.5-119.5
Center
Point
97
112
Cumm Freq Cumm Freq %
24
62
24
86
12%
43%
2,328
6,944
Probability
of Rages
9%
28%
119.5-134.5
127
72
158
79%
9,144
37%
134.5-149.5
142
26
184
92%
3,692
15%
149.5-164.5
164.5-179.5
157
172
12
4
196
200
98%
100%
Total
Wtd Avg
1,884
688
24,680
4,113
8%
3%
Mean
Std Dev of Freq
108.36
27.47
Range
Cumm
Normal
Distribution
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Wtd Avg
Gap to reach
Mean
Percentage of
2nd range
Offset into 2nd
Range
Wtd Mean
1,785
25.7%
3.86
108.36
34%
55%
75%
89%
96%
99%
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