Make one example and solution for each of the following chapters.
Each example must be your original work.However, students may use the examples from PowerPoint lecture notes as a reference.
Ch6.5 Central limit theorem
Ch7.1 Estimation for µ: When the standard deviation is known
Ch7.2 Estimation for µ: When the standard deviation is unknown
Ch7.3 Estimation for p in the Binomial Distribution
Ch8.2 Testing for the mean
6.2
Standard Units and Areas Under the Standard Normal Distribution
Tree-ring dates were used extensively in archaeological studies at Burnt Mesa Pueblo. Supp
ose at one site on the mesa, tree-ring dates (for many samples) gave a mean date of 𝜇1 = y
ear 1274 with standard deviation 𝜎1 = 32 years. At a second, removed site, the tree-ring d
ates gave a mean of 𝜇2 = year 1123 with standard deviation 𝜎2 = 41 years. Assume that both
sites had dates that were approximately normally distributed. In the first area, an object
was found and dated as x1 = year 1249. In the second area, another object was found and da
ted as x2 = year 1238.
6.3 Areas under the curve
Assume that x has a normal distribution with the specified mean and standard devia
tion. Find the indicated probability.
P(4 ≤ x ≤ 9);
𝜇 = 5; 𝜎 = 4
6.5
Central limit theorem
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unpl
anned) shopping interval. Based on a newspaper article, the mean of the x distribution is
about $28 and the estimated standard deviation is about $10.
7.1
Estimation for µ: When the standard deviation is known
What price do farmers get for their watermelon crops? In the third week of July, a random
sample of 42 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon.
Assume that 𝜎 is known to be $1.90 per 100 pounds.
7.2
Estimation for µ: When the standard deviation is unknown
Do you want to own your own candy store? With some interest in running your own business a
nd a decent credit rating, you can probably get a bank loan on startup costs for franchise
s such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Fact
ory. Startup costs (in thousands of dollars) for a random sample of candy stores are given
below. Assume that the population of x values has an approximately normal distribution.
7.3
Estimation for p in the Binomial Distribution
Let’s return to our flu shot experiment described at the beginning of this
section. Suppose that 800 students were selected at random from a student b
ody of 20,000 and given shots to prevent a certain type of flu. All 800 stu
dents were exposed to the flu, and 600 of them did not get the flu. Let p r
epresent the probability that the shot will be successful for any single st
udent selected at random from the entire population of 20,000. Let q be the
probability that the shot is not successful.
8.2 Testing for the mean
The sample mean is 47.0 Previous studies of sunspot activity during this p
eriod indicate that = 35. It is thought that for thousands of years, the m
ean number of sunspots per four-week period was about = 41. Sunspot activi
ty above this level may (or may not) be linked to gradual climate change. D
o the data indicate that the mean sunspot activity during the Spanish colon
ial period was higher than 41? Use = 0.05.
(a) Establish the null and alternate hypotheses. Since we want to know whet
her the average sunspot activity during the Spanish colonial period was hig
her than the long-term average of
= 41,
Normal Curves and
Sampling
Distributions
6
Copyright © Cengage Learning. All rights reserved.
Section
6.2
Standard Units and
Areas Under the
Standard Normal
Distribution
Copyright © Cengage Learning. All rights reserved.
z Scores and Raw Scores
3
z Scores and Raw Scores
Normal distributions vary from one another in two ways:
The mean may be located anywhere on the x axis, and
the bell shape may be more or less spread according to the
size of the standard deviation .
The differences among the normal distributions cause
difficulties when we try to compute the area under the
curve in a specified interval of x values and, hence, the
probability that a measurement will fall into that interval.
4
z Scores and Raw Scores
It would be a futile task to try to set up a table of areas
under the normal curve for each different and
combination. We need a way to standardize the
distributions so that we can use one table of areas for all
normal distributions.
We achieve this standardization by considering how many
standard deviations a measurement lies from the mean. In
this way, we can compare a value in one normal
distribution with a value in another, different normal
distribution. The next situation shows how this is done.
5
z Scores and Raw Scores
Suppose Tina and Jack are in two different sections of the
same course. Each section is quite large, and the scores
on the midterm exams of each section follow a normal
distribution. In Tina’s section, the average (mean) was 64
and her score was 74. In Jack’s section, the mean was 72
and his score was 82.
Both Tina and Jack were pleased that their scores were
each 10 points above the average of each respective
section. However, the fact that each was 10 points above
average does not really tell us how each did with respect to
the other students in the section.
6
z Scores and Raw Scores
In Figure 6-14, we see the normal distribution of grades for
each section.
Distributions of Midterm Scores
Figure 6-14
7
z Scores and Raw Scores
Tina’s 74 was higher than most of the other scores in her
section, while Jack’s 82 is only an upper-middle score in
his section. Tina’s score is far better with respect to her
class than Jack’s score is with respect to his class.
The preceding situation demonstrates that it is not sufficient
to know the difference between a measurement (x value)
and the mean of a distribution. We need also to consider
the spread of the curve, or the standard deviation.
What we really want to know is the number of standard
deviations between a measurement and the mean. This
“distance” takes both and into account.
8
z Scores and Raw Scores
We can use a simple formula to compute the number z of
standard deviations between a measurement x and the
mean of a normal distribution with standard deviation :
9
z Scores and Raw Scores
The mean is a special value of a distribution. Let’s see
what happens when we convert x = to a z value:
The mean of the original distribution is always zero, in
standard units. This makes sense because the mean is
zero standard variations from itself.
An x value in the original distribution that is above the
mean has a corresponding z value that is positive.
10
z Scores and Raw Scores
Again, this makes sense because a measurement above
the mean would be a positive number of standard
deviations from the mean. Likewise, an x value below the
mean has a negative z value. (See Table 6-2.)
x Values and Corresponding z Values
Table 6-2
11
Example
In Tina’s section, the average (mean) was 64 and her
score was 74 and the standard deviation was 8.
In Jack’s section, the mean was 72 and his score was 82
and the standard deviation was 6.
74 − 64
= 1.25
8
82 − 72
Z ( jack ) =
= 1.67
6
Z (tina ) =
Since Jack’s Z score is higher than Tina’s Z score, Jack
did better with respect to the other students in the section.
12
z Scores and Raw Scores
13
Standard Normal Distribution
14
Standard Normal Distribution
If the original distribution of x values is normal, then the
corresponding z values have a normal distribution as well.
The z distribution has a mean of 0 and a standard deviation
of 1. The normal curve with these properties has a special
name.
Any normal distribution of x values can be converted to the
standard normal distribution by converting all x values to
their corresponding z values. The resulting standard
distribution will always have mean = 0 and standard
deviation = 1.
15
Areas Under the Standard Normal Curve
16
Areas Under the Standard Normal Curve
The standard normal distribution can be a tremendously
helpful tool.
Addition of 3 and 4
The Standard Normal Distribution ( = 0, = 1)
FIGURE 6-15
17
Using a Standard Normal Distribution Table
18
19
20
Using a Standard Normal Distribution Table
Using a table to find areas and probabilities associated with
the standard normal distribution is a fairly straightforward
activity.
However, it is important to first observe the range of z
values for which areas are given. This range is usually
depicted in a picture that accompanies the table.
In this text, we will use the left-tail style table. This style
table gives cumulative areas to the left of a specified z.
21
Using a Standard Normal Distribution Table
Determining other areas under the curve utilizes the fact
that the area under the entire curve is 1.
Taking advantage of the symmetry of the normal
distribution is also useful.
The procedures you learn for using the left-tail style normal
distribution table apply directly to cumulative normal
distribution areas found on calculators and in computer
software packages such as Excel and Minitab.
22
Example 5(a) – Standard normal distribution table
Use Table 5 of Appendix II to find the described areas
under the standard normal curve.
(a) Find the area under the standard normal curve to the
left of z = –1.00.
Solution:
First, shade the area to be found
on the standard normal distribution
curve, as shown in Figure 6-16.
Area to the Left of z = –1.00
Figure 6-16
23
Example 5(a) – Solution
cont’d
Notice that the z value we are using is negative. This
means that we will look at the portion of Table 5 of
Appendix II for which the z values are negative. In the
upper-left corner of the table we see the letter z.
The column under z gives us the units value and tenths
value for z. The other column headings indicate the
hundredths value of z. Table entries give areas under the
standard normal curve to the left of the listed z values.
24
Example 5(a) – Solution
cont’d
To find the area to the left of z = –1.00, we use the row
headed by –1.0 and then move to the column headed by
the hundredths position, .00. This entry is shaded in Table
6-3. We see that the area is 0.1587.
Excerpt from Table 5 of Appendix II Showing Negative z Values
Table 6-3
25
Example 5(b) – Standard normal distribution table
cont’d
(b) Find the area to the left of z = 1.18, as illustrated in
Figure 6-17.
Area to the Left of z = 1.18
Figure 6-17
Solution:
In this case, we are looking for an area to the left of a
positive z value, so we look in the portion of Table 5 that
shows positive z values.
26
Example 5(b) – Solution
cont’d
Again, we first sketch the area to be found on a standard
normal curve, as shown in Figure 6-17.
Excerpt from Table 5 of Appendix II Showing Positive z Values
Table 6-4
Look in the row headed by 1.1 and move to the column
headed by .08. The desired area is shaded (see Table 6-4).
We see that the area to the left of 1.18 is 0.8810.
27
Using a Standard Normal Distribution Table
Procedure:
28
Using a Standard Normal Distribution Table
29
Example 6(a) – Using table to find areas
Use Table 5 of Appendix II to find the specified areas.
Find the area between z = 1.00 and z = 2.70.
Solution:
First, sketch a diagram showing the area (see Figure 6-19).
Area from z = 1.00 to z = 2.70
Figure 6-19
30
Example 6(a) – Solution
cont’d
Because we are finding the area between two z values, we
subtract corresponding table entries.
(Area between 1.00 and 2.70) = (Area left of 2.70)
– (Area left of 1.00)
= 0.9965 – 08413
= 0.1552
31
Example 6(b) – Using table to find areas cont’d
Find the area to the right of z = 0.94.
Solution:
First, sketch the area to be found (see Figure 6-20).
Area to the Right of z = 0.94.
Figure 6-20
32
Example 6(b) – Solution
cont’d
(Area to right of 0.94) = (Area under entire curve)
– (Area to left of 0.94)
= 1.000 – 0.8264
= 0.1736
Alternatively,
(Area to right of 0.94) = (Area to left of – 0.94)
= 0.1736
33
Normal Curves and
Sampling
Distributions
6
Copyright © Cengage Learning. All rights reserved.
Section
6.5
The Central Limit
Theorem
Copyright © Cengage Learning. All rights reserved.
The
Distribution, Given x is normal
3
Theorem 6.1
4
The x Distribution, Given x is normal.
In Section 6.4, we began a study of the distribution of x
values, where x was the (sample) mean length of five trout
caught by children at the Pinedale children’s fishing pond.
Let’s consider this example again in the light of a very
important theorem of mathematical statistics.
5
Example 12(a) – Probability regarding x and x
Suppose a team of biologists has been studying the
Pinedale children’s fishing pond.
Let x represent the length of a single trout taken at random
from the pond.
This group of biologists has determined that x has a normal
distribution with mean = 10.2 inches and standard
deviation = 1.4 inches.
What is the probability that a single trout taken at
random from the pond is between 8 and 12 inches long?
6
Example 12(a) – Solution
We use the methods of Section 6.3, with = 10.2 and
= 1.4 ,to get
7
Example 12(a) – Solution
cont’d
Therefore,
Therefore, the probability is about 0.8433 that a single trout
taken at random is between 8 and 12 inches long.
8
Example 12(b) – Probability regarding x and x
What is the probability that the mean x length of five trout
taken at random is between 8 and 12 inches?
Solution:
If we let
represent the mean of the distribution, then
Theorem 6.1, part (b), tells us that
If
represents the standard deviation of the
distribution, then Theorem 6.1, part (c), tells us that
9
Example 12 – Solution
To create a standard z variable from
and divide by :
cont’d
, we subtract
To standardize the interval 8 < < 12, we use 8 and then
12 in place of in the preceding formula for z.
8<
< 12
–3.49 < z < 2.86
10
Example 12 – Solution
Theorem 6.1, part (a), tells us that
distribution. Therefore,
P(8 <
cont’d
has a normal
< 12) = P(–3.49 < z < 2.86)
= 0.9979 – 0.0002
= 0.9977
The probability is about 0.9977 that the mean length based
on a sample size of 5 is between 8 and 12 inches.
11
Example 12(c) – Probability regarding x and x
Looking at the results of parts (a) and (b), we see that the
probabilities (0.8433 and 0.9977) are quite different. Why is
this the case?
Solution:
According to Theorem 6.1, both x and have a normal
distribution, and both have the same mean of 10.2 inches.
The difference is in the standard deviations for x and .
The standard deviation of the x distribution is = 1.4.
12
Example 12 – Solution
The standard deviation of the
cont’d
distribution is
The standard deviation of the distribution is less than half
the standard deviation of the x distribution.
13
Example 12 – Solution
Figure 6-33 shows the distributions of x and
(a) The x distribution with = 10.2 and
= 1.4
cont’d
.
(b) The x distribution with
= 10.2
and
= 0.63 for samples of size n = 5
14
The x Distribution, Given x is normal.
Looking at Figure 6-33(a) and (b), we see that both curves
use the same scale on the horizontal axis. The means are
the same, and the shaded area is above the interval from 8
to 12 on each graph.
It becomes clear that the smaller standard deviation of the
distribution has the effect of gathering together much more
of the total probability into the region over its mean.
Therefore, the region from 8 to 12 has a much higher
probability for the distribution.
15
The x Distribution, Given x Follows
any Distribution
16
The x Distribution, Given x Follows any Distribution
17
Example 13
18
Example 14
19
The x Distribution, Given x Follows any Distribution
Procedure:
20
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