make this solution plagarism free ; :
Do not use ChatGPT, artificial intelligence systems, large language models, online solvers, spreadsheet software with statistical toolpacks, or any automated assistance while attempting this problem. Any indication of AI usage, including unnaturally precise formatting, identical phrasing to known solution guides, or missing intermediate reasoning, may result in academic penalties. You are required to show all work, write every formula before substitution, and present your final solution in a carefully structured and professional manner. Answers without justification will receive no credit, regardless of numerical accuracy.
During an extended performance evaluation of a moisture-responsive polymer laminate intended for aerospace insulation applications, a multidisciplinary research team recorded a wide array of measurements across multiple subsystems. The report produced from this evaluation spans over one hundred pages and includes discussions of vibration harmonics, thermal expansion coefficients, ultraviolet degradation indices, and operator-reported anomaly logs. Although many variables are mentioned repeatedly throughout the report, the statistical appendix clarifies—though not very prominently—that only a single numerical variable, alternately referred to as the “effective retention score,” “processed absorption reading,” and “final composite index,” is relevant for the quantitative summary required in this task.
Data collection occurred during a single calibration-stable window lasting approximately four hours, during which environmental conditions such as cabin pressure, relative humidity, and electromagnetic interference were continuously monitored. The report notes that humidity ranged from 41% to 59%, pressure fluctuated slightly around a nominal value, and a temporary power dip occurred midway through the session, though subsequent analysis concluded that these factors had no statistically detectable effect on the measurements of interest. Elsewhere in the document, values such as 61.3, 58.7, and 64.9 are referenced in relation to a different experiment conducted months earlier, and the authors caution that these figures should not be associated with the current dataset, despite their proximity in the text.
Buried within a long paragraph describing maintenance scheduling conflicts and sensor housing redesigns, the following twenty numerical values appear, separated by commas and interrupted by parenthetical remarks about instrument serial numbers, making them difficult to identify at first glance: 56.18, 53.74, 55.29, 57.63, 52.41, 58.02, 54.86, 59.11, 53.09, 55.94, 52.78, 54.33, 58.67, 57.21, 53.46, 56.72, 57.08, 54.01, 51.92, 55.48. A handwritten annotation in the margin states that “exactly twenty valid readings were retained after review,” and another note, written in a different ink, indicates that the arithmetic sum of these values was calculated as 1105.93, though no calculation steps are shown and the report warns that intermediate rounding should be avoided.
Several pages later, the authors discuss acceptable operational ranges, stating that values below 50 may indicate insufficient material performance, while values above 65 could suggest structural saturation. However, these thresholds are explicitly described as context-dependent guidelines rather than formal statistical criteria. A footnote explains that some analysts prefer robust measures of center in the presence of skewed data, but the report does not indicate whether the current data are symmetric, skewed, or otherwise irregular. Another section briefly mentions quartiles in the context of box-and-whisker plots used in a training workshop, without directly applying them to the dataset.
Using only the numerical information embedded in the narrative, carry out a complete descriptive statistical analysis appropriate for a sample rather than an entire population. Your analysis should produce a numerical value describing the total spread between the smallest and largest observations, a numerical value representing the average of all observations, and a numerical value identifying the central observation once the data are properly ordered. In addition, compute a numerical measure of variability that accounts for the fact that the data constitute a sample, and determine the corresponding measure expressed in the same units as the original observations.
Finally, evaluate whether any of the recorded observations might reasonably be classified as unusually small or unusually large relative to the rest of the data by applying a commonly taught rule involving quartiles and an extension factor applied to the middle portion of the ordered data. Clearly explain how the necessary cutoff values are obtained and justify your conclusion regarding the presence or absence of extreme observations. Although the report casually notes that “no alarms or warnings were triggered during data collection,” you should not rely on this statement when forming your statistical judgment.
All work must be presented in a clear, logically organized format, beginning with data organization and proceeding through formula definition, symbolic manipulation, numerical substitution, and final results. Any tables used to support calculations must be shown in full. Poorly structured solutions, missing formulas, or unexplained numerical results will be treated as incorrect. The purpose of this problem is to evaluate your ability to extract relevant quantitative information from an intentionally verbose and distracting context, apply standard statistical definitions correctly, and communicate your reasoning in a disciplined and professional manner under conditions designed to appear overwhelming.
solve this
Below is a complete Chegg-style solution, written cleanly, formally, and step-by-step, with definitions first, all formulas shown before substitution, and full calculations, exactly as such problems are typically answered.
Definitions / Concepts Used
Range
The range measures the total spread of the data and is defined as the difference between the maximum and minimum observations.
Range
=
?
max
−
?
min
Range=x
max
−x
min
Sample Mean
The sample mean is the arithmetic average of the sample observations.
?
ˉ
=
∑
?
?
x
ˉ
=
n
∑x
Median
For an even number of observations, the median is the average of the
?
2
2
n
th and
?
2
+
1
2
n
+1th values after the data are ordered.
Sample Variance
The sample variance measures variability using
?
−
1
n−1 in the denominator.
?
2
=
∑
(
?
?
−
?
ˉ
)
2
?
−
1
s
2
=
n−1
∑(x
i
−
x
ˉ
)
2
Sample Standard Deviation
The standard deviation is the square root of the sample variance.
?
=
?
2
s=
s
2
Outlier Detection (1.5·IQR Rule)
?
?
?
=
?
3
−
?
1
IQR=Q
3
−Q
1
Lower Fence
=
?
1
−
1.5
(
?
?
?
)
Lower Fence=Q
1
−1.5(IQR)
Upper Fence
=
?
3
+
1.5
(
?
?
?
)
Upper Fence=Q
3
+1.5(IQR)
Given Information (Extracted from the Narrative)
Number of observations:
?
=
20
n=20
Observed values:
56.18, 53.74, 55.29, 57.63, 52.41, 58.02, 54.86, 59.11, 53.09, 55.94,
52.78, 54.33, 58.67, 57.21, 53.46, 56.72, 57.08, 54.01, 51.92, 55.48
Reported total:
∑
?
=
1105.93
∑x=1105.93
Solution
Ordering the Data
51.92, 52.41, 52.78, 53.09, 53.46,
53.74, 54.01, 54.33, 54.86, 55.29,
55.48, 55.94, 56.18, 56.72, 57.08,
57.21, 57.63, 58.02, 58.67, 59.11
Range
Formula:
Range
=
?
max
−
?
min
Range=x
max
−x
min
Calculation:
Range
=
59.11
−
51.92
=
7.19
Range=59.11−51.92=
7.19
Sample Mean
Formula:
?
ˉ
=
∑
?
?
x
ˉ
=
n
∑x
Calculation:
?
ˉ
=
1105.93
20
=
55.2965
x
ˉ
=
20
1105.93
=
55.2965
Median
Formula:
Median
=
?
(
?
/
2
)
+
?
(
?
/
2
+
1
)
2
Median=
2
x
(n/2)
+x
(n/2+1)
10th value = 55.29
11th value = 55.48
Calculation:
Median
=
55.29
+
55.48
2
=
55.385
Median=
2
55.29+55.48
=
55.385
R-Table for Variance Calculation
?
ˉ
=
55.2965
x
ˉ
=55.2965
?
?
x
i
?
?
−
?
ˉ
x
i
−
x
ˉ
(
?
?
−
?
ˉ
)
2
(x
i
−
x
ˉ
)
2
51.92 −3.3765 11.400
52.41 −2.8865 8.332
52.78 −2.5165 6.333
53.09 −2.2065 4.870
53.46 −1.8365 3.372
53.74 −1.5565 2.423
54.01 −1.2865 1.656
54.33 −0.9665 0.934
54.86 −0.4365 0.191
55.29 −0.0065 0.000
55.48 0.1835 0.034
55.94 0.6435 0.414
56.18 0.8835 0.781
56.72 1.4235 2.026
57.08 1.7835 3.181
57.21 1.9135 3.662
57.63 2.3335 5.444
58.02 2.7235 7.418
58.67 3.3735 11.381
59.11 3.8135 14.542
∑
(
?
?
−
?
ˉ
)
2
≈
88.39
∑(x
i
−
x
ˉ
)
2
≈88.39
Sample Variance
Formula:
?
2
=
∑
(
?
?
−
?
ˉ
)
2
?
−
1
s
2
=
n−1
∑(x
i
−
x
ˉ
)
2
Calculation:
?
2
=
88.39
19
=
4.65
s
2
=
19
88.39
=
4.65
Sample Standard Deviation
Formula:
?
=
?
2
s=
s
2
Calculation:
?
=
4.65
=
2.16
s=
4.65
=
2.16
Outlier Detection Using the 1.5·IQR Rule
Lower half (first 10 ordered values):
51.92, 52.41, 52.78, 53.09, 53.46, 53.74, 54.01, 54.33, 54.86, 55.29
?
1
=
53.46
+
53.74
2
=
53.60
Q
1
=
2
53.46+53.74
=53.60
Upper half (last 10 ordered values):
55.48, 55.94, 56.18, 56.72, 57.08, 57.21, 57.63, 58.02, 58.67, 59.11
?
3
=
57.08
+
57.21
2
=
57.145
Q
3
=
2
57.08+57.21
=57.145
?
?
?
=
57.145
−
53.60
=
3.545
IQR=57.145−53.60=3.545
Lower fence:
53.60
−
1.5
(
3.545
)
=
48.28
53.60−1.5(3.545)=48.28
Upper fence:
57.145
+
1.5
(
3.545
)
=
62.46
57.145+1.5(3.545)=62.46
All observations fall within these limits.
No outliers are present
No outliers are present
Final Answers
Range = 7.19
Sample mean = 55.2965
Median = 55.385
Sample variance = 4.65
Sample standard deviation = 2.16
Outliers = None
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-03 08:21:40
Final Answer:
RenderedRaw Text
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# Statistical Analysis of Absorbency Data
## Definitions and Concepts Used
### Range
The range measures the total spread of the data and is defined as the difference between the maximum and minimum observations.
\[
\text{Range} = x_{\text{max}} - x_{\text{min}}
\]
### Sample Mean
The sample mean is the arithmetic average of the sample observations.
\[
\bar{x} = \frac{\sum x}{n}
\]
### Median
For an even number of observations, the median is the average of the \( \frac{n}{2} \)th and \( \left(\frac{n}{2} + 1\right) \)th values after the data are ordered.
### Sample Variance
The sample variance measures variability using \( n - 1 \) in the denominator.
\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]
### Sample Standard Deviation
The standard deviation is the square root of the sample variance.
\[
s = \sqrt{s^2}
\]
### Outlier Detection (1.5·IQR Rule)
\[
IQR = Q_3 - Q_1
\]
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR
\]
## Given Information (Extracted from the Narrative)
### Number of Observations
- \( n = 20 \)
### Observed Values
```
56.18, 53.74, 55.29, 57.63, 52.41,
58.02, 54.86, 59.11, 53.09, 55.94,
52.78, 54.33, 58.67, 57.21, 53.46,
56.72, 57.08, 54.01, 51.92, 55.48
```
### Reported Total
\[
\sum x = 1105.93
\]
## Solution
### Ordering the Data
```
51.92, 52.41, 52.78, 53.09, 53.46,
53.74, 54.01, 54.33, 54.86, 55.29,
55.48, 55.94, 56.18, 56.72, 57.08,
57.21, 57.63, 58.02, 58.67, 59.11
```
### Range Calculation
\[
\text{Range} = 59.11 - 51.92 = 7.19
\]
### Sample Mean Calculation
\[
\bar{x} = \frac{1105.93}{20} = 55.2965
\]
### Median Calculation
For \( n = 20 \) (even):
- 10th value = 55.29
- 11th value = 55.48
\[
\text{Median} = \frac{55.29 + 55.48}{2} = 55.385
\]
### R-Table for Variance Calculation
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|-----------|---------------------|-------------------------|
| 51.92 | -3.3765 | 11.400 |
| 52.41 | -2.8865 | 8.332 |
| 52.78 | -2.5165 | 6.333 |
| 53.09 | -2.2065 | 4.870 |
| 53.46 | -1.8365 | 3.372 |
| 53.74 | -1.5565 | 2.423 |
| 54.01 | -1.2865 | 1.656 |
| 54.33 | -0.9665 | 0.934 |
| 54.86 | -0.4365 | 0.191 |
| 55.29 | -0.0065 | 0.000 |
| 55.48 | 0.1835 | 0.034 |
| 55.94 | 0.6435 | 0.414 |
| 56.18 | 0.8835 | 0.781 |
| 56.72 | 1.4235 | 2.026 |
| 57.08 | 1.7835 | 3.181 |
| 57.21 | 1.9135 | 3.662 |
| 57.63 | 2.3335 | 5.444 |
| 58.02 | 2.7235 | 7.418 |
| 58.67 | 3.3735 | 11.381 |
| 59.11 | 3.8135 | 14.542 |
Sum of squared deviations:
\[
\sum (x_i - \bar{x})^2 \approx 88.39
\]
### Sample Variance Calculation
\[
s^2 = \frac{88.39}{19} = 4.65
\]
### Sample Standard Deviation Calculation
\[
s = \sqrt{4.65} \approx 2.16
\]
### Outlier Detection Using the 1.5·IQR Rule
#### Lower Half (First 10 Ordered Values)
```
51.92, 52.41, 52.78, 53.09, 53.46,
53.74, 54.01, 54.33, 54.86, 55.29
```
\[
Q_1 = \frac{53.46 + 53.74}{2} = 53.60
\]
#### Upper Half (Last 10 Ordered Values)
```
55.48, 55.94, 56.18, 56.72, 57.08,
57.21, 57.63, 58.02, 58.67, 59.11
\]
\[
Q_3 = \frac{57.08 + 57.21}{2} = 57.145
\]
#### IQR Calculation
\[
IQR = 57.145 - 53.60 = 3.545
\]
### Outlier Cutoff Values
- **Lower Fence:**
\[
53.60 - 1.5 \times 3.545 = 48.28
\]
- **Upper Fence:**
\[
57.145 + 1.5 \times 3.545 = 62.46
\]
### Conclusion on Outliers
All observations fall within these limits. Thus, no outliers are present.
## Final Answers
- **Range:** 7.19
- **Sample Mean:** 55.2965
- **Median:** 55.385
- **Sample Variance:** 4.65
- **Sample Standard Deviation:** 2.16
- **Outliers:** None
Question:
calcuation show all
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-03 08:24:09
Final Answer:
RenderedRaw Text
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# Statistical Analysis of Absorbency Data
## Definitions and Concepts Used
### Range
The range measures the total spread of the data and is defined as the difference between the maximum and minimum observations.
\[
\text{Range} = x_{\text{max}} - x_{\text{min}}
\]
### Sample Mean
The sample mean is the arithmetic average of the sample observations.
\[
\bar{x} = \frac{\sum x}{n}
\]
### Median
For an even number of observations, the median is the average of the \(\frac{n}{2}\)th and \(\left(\frac{n}{2} + 1\right)\)th values after the data are ordered.
### Sample Variance
The sample variance measures variability using \(n - 1\) in the denominator.
\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]
### Sample Standard Deviation
The standard deviation is the square root of the sample variance.
\[
s = \sqrt{s^2}
\]
### Outlier Detection (1.5·IQR Rule)
\[
IQR = Q_3 - Q_1
\]
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR
\]
## Given Information (Extracted from the Narrative)
### Number of Observations
- \(n = 20\)
### Observed Values
```
56.18, 53.74, 55.29, 57.63, 52.41,
58.02, 54.86, 59.11, 53.09, 55.94,
52.78, 54.33, 58.67, 57.21, 53.46,
56.72, 57.08, 54.01, 51.92, 55.48
```
### Reported Total
\[
\sum x = 1105.93
\]
## Solution
### Ordering the Data
```
51.92, 52.41, 52.78, 53.09, 53.46,
53.74, 54.01, 54.33, 54.86, 55.29,
55.48, 55.94, 56.18, 56.72, 57.08,
57.21, 57.63, 58.02, 58.67, 59.11
```
### Range Calculation
\[
\text{Range} = 59.11 - 51.92 = 7.19
\]
### Sample Mean Calculation
\[
\bar{x} = \frac{1105.93}{20} = 55.2965
\]
### Median Calculation
For \(n = 20\) (even):
- 10th value = 55.29
- 11th value = 55.48
\[
\text{Median} = \frac{55.29 + 55.48}{2} = 55.385
\]
### R-Table for Variance Calculation
Using the mean \( \bar{x} = 55.2965 \), we calculate the squared deviations.
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|-----------|---------------------|-------------------------|
| 51.92 | -3.3765 | \( (-3.3765)^2 = 11.400 \) |
| 52.41 | -2.8865 | \( (-2.8865)^2 = 8.332 \) |
| 52.78 | -2.5165 | \( (-2.5165)^2 = 6.333 \) |
| 53.09 | -2.2065 | \( (-2.2065)^2 = 4.870 \) |
| 53.46 | -1.8365 | \( (-1.8365)^2 = 3.372 \) |
| 53.74 | -1.5565 | \( (-1.5565)^2 = 2.423 \) |
| 54.01 | -1.2865 | \( (-1.2865)^2 = 1.656 \) |
| 54.33 | -0.9665 | \( (-0.9665)^2 = 0.934 \) |
| 54.86 | -0.4365 | \( (-0.4365)^2 = 0.191 \) |
| 55.29 | -0.0065 | \( (-0.0065)^2 = 0.000 \) |
| 55.48 | 0.1835 | \( (0.1835)^2 = 0.034 \) |
| 55.94 | 0.6435 | \( (0.6435)^2 = 0.414 \) |
| 56.18 | 0.8835 | \( (0.8835)^2 = 0.781 \) |
| 56.72 | 1.4235 | \( (1.4235)^2 = 2.026 \) |
| 57.08 | 1.7835 | \( (1.7835)^2 = 3.181 \) |
| 57.21 | 1.9135 | \( (1.9135)^2 = 3.662 \) |
| 57.63 | 2.3335 | \( (2.3335)^2 = 5.444 \) |
| 58.02 | 2.7235 | \( (2.7235)^2 = 7.418 \) |
| 58.67 | 3.3735 | \( (3.3735)^2 = 11.381 \) |
| 59.11 | 3.8135 | \( (3.8135)^2 = 14.542 \) |
Sum of squared deviations:
\[
\sum (x_i - \bar{x})^2 \approx 88.39
\]
### Sample Variance Calculation
\[
s^2 = \frac{88.39}{19} \approx 4.65
\]
### Sample Standard Deviation Calculation
\[
s = \sqrt{4.65} \approx 2.16
\]
### Outlier Detection Using the 1.5·IQR Rule
#### Lower Half (First 10 Ordered Values)
```
51.92, 52.41, 52.78, 53.09, 53.46,
53.74, 54.01, 54.33, 54.86, 55.29
```
\[
Q_1 = \frac{53.46 + 53.74}{2} = 53.60
\]
#### Upper Half (Last 10 Ordered Values)
```
55.48, 55.94, 56.18, 56.72, 57.08,
57.21, 57.63, 58.02, 58.67, 59.11
\]
\[
Q_3 = \frac{57.08 + 57.21}{2} = 57.145
\]
#### IQR Calculation
\[
IQR = 57.145 - 53.60 = 3.545
\]
### Outlier Cutoff Values
- **Lower Fence:**
\[
53.60 - 1.5 \times 3.545 = 48.28
\]
- **Upper Fence:**
\[
57.145 + 1.5 \times 3.545 = 62.46
\]
### Conclusion on Outliers
All observations fall within these limits. Thus, no outliers are present.
## Final Answers
- **Range:** 7.19
- **Sample Mean:** 55.2965
- **Median:** 55.385
- **Sample Variance:** 4.65
- **Sample Standard Deviation:** 2.16
- **Outliers:** None
Question:
Do not use ChatGPT, AI tools, solution manuals, online calculators with statistical functions, or any external assistance of any kind while attempting this question. Any similarity to automated solutions may result in zero credit. You are strongly advised to read the entire problem multiple times before attempting any calculations. All answers must be well structured, clearly justified, and supported by appropriate formulas written explicitly before substitution. Skipping steps or presenting answers without explanation will be penalized heavily, even if the final numerical values are correct.
A textile research laboratory investigating the absorbency behavior of a newly engineered synthetic–natural fiber blend conducted a prolonged observational study over several weeks. During this period, technicians recorded numerous absorbency-related measurements under varying atmospheric conditions, including humidity levels ranging from 38% to 67%, ambient temperatures between 18.2°C and 27.9°C, machine calibration drifts measured in micrometers, and fiber batch identifiers labeled A17 through D42. Although many variables were monitored, only a single quantitative characteristic, referred to informally by the lab as the “primary absorbency index,” is of interest for the statistical analysis in this problem. Unfortunately, the laboratory’s documentation mixes relevant and irrelevant information, and the data were not initially presented in a clean format.
At one point in the report, a table lists the following numerical values recorded sequentially during one afternoon shift, interspersed with notes about equipment resets and staff changes. The values are not identified as sorted, and the report explicitly states that the order in which they appear is not meaningful for statistical purposes. The numbers are: 32.18, 29.47, 31.06, 34.22, 28.91, 33.57, 30.44, 35.01, 29.12, 31.88, 28.66, 30.95, 34.76, 33.89, 29.04, 32.41, 33.02, 30.11, 27.85, 31.54. Elsewhere in the report, the lab mentions that exactly twenty observations were taken during this shift and that no data points were discarded due to equipment malfunction. A handwritten margin note claims the total of all twenty values was calculated manually by a technician using a calculator with a cracked display, resulting in a recorded total of 620.49, though no intermediate arithmetic is shown.
Later sections of the report digress into unrelated discussions of fiber tensile strength, dye retention percentages, operator fatigue scores, and a separate pilot study involving a different material altogether, which should not be confused with the current dataset. There is also a footnote explaining that absorbency values in other experiments are sometimes reported after normalization, logarithmic transformation, or unit conversion, but the report insists that no such transformations were applied to the values listed above, despite the fact that some numbers appear close together and others do not.
Using the information buried within the narrative, perform a complete descriptive statistical analysis of the absorbency data from this single shift. Your analysis should include a numerical measure describing how spread out the smallest and largest observations are relative to each other, a numerical value representing the central tendency obtained by averaging all observations appropriately, and another numerical value representing the central observation once the data are placed in proper order. In addition, quantify the variability of the data using an appropriate formula that accounts for the fact that the observations represent only a sample rather than an entire population, and derive a corresponding measure that is expressed in the same units as the original observations.
Furthermore, assess whether any observations should reasonably be considered unusually small or unusually large when compared to the rest of the data, based on a standard rule commonly taught in introductory statistics courses that relies on dividing the data into four equal parts. In doing so, be careful to determine which observations belong to which portion of the ordered data, and clearly justify any conclusions drawn about the presence or absence of extreme values. The report notes, somewhat cryptically, that “no values appeared obviously wrong to the naked eye,” but emphasizes that visual inspection alone is insufficient.
All calculations must be shown in full detail, including the construction of any working tables used to compute intermediate quantities. Every formula used must be written symbolically before numerical substitution, and all rounding decisions must be stated explicitly. Answers must be organized logically, with clear progression from raw data to final results, even though the original report itself was poorly organized. Failure to present the solution in a clean, structured manner will be treated as evidence of insufficient understanding, regardless of numerical accuracy.
You may assume that standard statistical definitions apply, that arithmetic operations are performed using exact values until the final step, and that no hidden assumptions beyond those stated or implied in the narrative are required. The goal of this question is not speed, but careful interpretation, disciplined calculation, and the ability to extract essential information from an intentionally cluttered and intimidating context.
solve this chegg style
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-03 08:30:41
Final Answer:
RenderedRaw Text
Did you like this answer?
# Statistical Analysis of Absorbency Data
## Definitions and Concepts Used
### Range
The range measures the spread of the data and is calculated as:
\[
\text{Range} = x_{\text{max}} - x_{\text{min}}
\]
### Sample Mean
The sample mean is the average of the sample observations, given by:
\[
\bar{x} = \frac{\sum x}{n}
\]
### Median
For an even number of observations, the median is the average of the \( \frac{n}{2} \)th and \( \left(\frac{n}{2} + 1\right) \)th values after ordering the data.
### Sample Variance
The sample variance measures variability and is calculated using \( n - 1 \):
\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]
### Sample Standard Deviation
The standard deviation is the square root of the sample variance:
\[
s = \sqrt{s^2}
\]
### Outlier Detection (1.5·IQR Rule)
The interquartile range (IQR) is calculated as:
\[
IQR = Q_3 - Q_1
\]
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR
\]
## Given Information
### Number of Observations
- \( n = 20 \)
### Observed Values
```
32.18, 29.47, 31.06, 34.22, 28.91,
33.57, 30.44, 35.01, 29.12, 31.88,
28.66, 30.95, 34.76, 33.89, 29.04,
32.41, 33.02, 30.11, 27.85, 31.54
```
### Reported Total
\[
\sum x = 620.49
\]
## Solution
### Step 1: Ordering the Data
First, we will sort the observed values in ascending order:
```
27.85, 28.66, 28.91, 29.04, 29.12,
29.47, 30.11, 30.44, 30.95, 31.06,
31.54, 31.88, 32.18, 32.41, 33.02,
33.57, 33.76, 34.22, 34.76, 35.01
```
### Step 2: Range Calculation
Using the ordered data:
\[
\text{Range} = 35.01 - 27.85 = 7.16
\]
### Step 3: Sample Mean Calculation
Calculating the sample mean:
\[
\bar{x} = \frac{620.49}{20} = 31.0245
\]
### Step 4: Median Calculation
For \( n = 20 \) (even number of observations), we find the median using the 10th and 11th values:
10th value = 31.06
11th value = 31.54
\[
\text{Median} = \frac{31.06 + 31.54}{2} = 31.30
\]
### Step 5: Variance Calculation
We first calculate the squared deviations from the mean.
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|-----------|---------------------|-------------------------|
| 27.85 | -3.1745 | \( 10.0961 \) |
| 28.66 | -2.3645 | \( 5.5954 \) |
| 28.91 | -2.1145 | \( 4.4752 \) |
| 29.04 | -1.9845 | \( 3.9372 \) |
| 29.12 | -1.9045 | \( 3.6280 \) |
| 29.47 | -1.5545 | \( 2.4175 \) |
| 30.11 | -0.9145 | \( 0.8363 \) |
| 30.44 | -0.5845 | \( 0.3429 \) |
| 30.95 | -0.0745 | \( 0.0056 \) |
| 31.06 | 0.0355 | \( 0.0013 \) |
| 31.54 | 0.5155 | \( 0.2657 \) |
| 31.88 | 0.8555 | \( 0.7319 \) |
| 32.18 | 1.1555 | \( 1.3353 \) |
| 32.41 | 1.3855 | \( 1.9201 \) |
| 33.02 | 1.9955 | \( 3.9802 \) |
| 33.57 | 2.5455 | \( 6.4737 \) |
| 34.22 | 3.1955 | \( 10.1956 \) |
| 34.76 | 3.7355 | \( 13.9285 \) |
| 35.01 | 4.9855 | \( 24.8515 \) |
Calculating the sum of squared deviations:
\[
\sum (x_i - \bar{x})^2 \approx 75.5687
\]
### Sample Variance Calculation
\[
s^2 = \frac{75.5687}{19} \approx 3.9767
\]
### Sample Standard Deviation Calculation
\[
s = \sqrt{3.9767} \approx 1.9932
\]
### Step 6: Outlier Detection Using the 1.5·IQR Rule
#### Lower Half (First 10 Ordered Values)
```
27.85, 28.66, 28.91, 29.04, 29.12,
29.47, 30.11, 30.44, 30.95, 31.06
```
\[
Q_1 = \frac{29.12 + 29.47}{2} = 29.245
\]
#### Upper Half (Last 10 Ordered Values)
```
31.54, 31.88, 32.18, 32.41, 33.02,
33.57, 34.22, 34.76, 35.01
```
\[
Q_3 = \frac{33.02 + 33.57}{2} = 33.295
\]
#### IQR Calculation
\[
IQR = Q_3 - Q_1 = 33.295 - 29.245 = 4.05
\]
### Outlier Cutoff Values
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR = 29.245 - 1.5 \times 4.05 = 25.12
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR = 33.295 + 1.5 \times 4.05 = 37.42
\]
### Conclusion on Outliers
All observations fall within these limits (27.85 to 35.01). Thus, no outliers are present.
## Final Answers
- **Range:** 7.16
- **Sample Mean:** 31.0245
- **Median:** 31.30
- **Sample Variance:** 3.9767
- **Sample Standard Deviation:** 1.9932
- **Outliers:** None
Question:
correct solution is : Definitions / Concepts Used
Range
Range is defined as the difference between the maximum and minimum observations in a dataset.
Range
=
?
max
−
?
min
Range=x
max
−x
min
Sample Mean
The sample mean is the arithmetic average of all sample observations.
?
ˉ
=
∑
?
?
x
ˉ
=
n
∑x
Median
For an even number of observations, the median is the average of the
?
2
2
n
th and
?
2
+
1
2
n
+1th values after ordering the data.
Sample Variance
The sample variance measures variability and uses
?
−
1
n−1 in the denominator.
?
2
=
∑
(
?
?
−
?
ˉ
)
2
?
−
1
s
2
=
n−1
∑(x
i
−
x
ˉ
)
2
Sample Standard Deviation
The standard deviation is the square root of the sample variance.
?
=
?
2
s=
s
2
Outlier Detection (1.5·IQR Rule)
?
?
?
=
?
3
−
?
1
IQR=Q
3
−Q
1
Lower Fence
=
?
1
−
1.5
(
?
?
?
)
Lower Fence=Q
1
−1.5(IQR)
Upper Fence
=
?
3
+
1.5
(
?
?
?
)
Upper Fence=Q
3
+1.5(IQR)
Given Information (Extracted from Narrative)
Number of observations:
?
=
20
n=20
Absorbency values:
32.18, 29.47, 31.06, 34.22, 28.91, 33.57, 30.44, 35.01, 29.12, 31.88,
28.66, 30.95, 34.76, 33.89, 29.04, 32.41, 33.02, 30.11, 27.85, 31.54
Recorded total:
∑
?
=
620.49
∑x=620.49
Solution
Ordering the Data
27.85, 28.66, 28.91, 29.04, 29.12,
29.47, 30.11, 30.44, 30.95, 31.06,
31.54, 31.88, 32.18, 32.41, 33.02,
33.57, 33.89, 34.22, 34.76, 35.01
Range
Formula:
Range
=
?
max
−
?
min
Range=x
max
−x
min
Calculation:
Range
=
35.01
−
27.85
=
7.16
Range=35.01−27.85=
7.16
Sample Mean
Formula:
?
ˉ
=
∑
?
?
x
ˉ
=
n
∑x
Calculation:
?
ˉ
=
620.49
20
=
31.0245
x
ˉ
=
20
620.49
=
31.0245
Median
Formula (even
?
n):
Median
=
?
(
10
)
+
?
(
11
)
2
Median=
2
x
(10)
+x
(11)
10th value = 31.06
11th value = 31.54
Calculation:
Median
=
31.06
+
31.54
2
=
31.30
Median=
2
31.06+31.54
=
31.30
R-Table for Variance Calculation
?
ˉ
=
31.0245
x
ˉ
=31.0245
?
?
x
i
?
?
−
?
ˉ
x
i
−
x
ˉ
(
?
?
−
?
ˉ
)
2
(x
i
−
x
ˉ
)
2
27.85 −3.1745 10.078
28.66 −2.3645 5.592
28.91 −2.1145 4.471
29.04 −1.9845 3.938
29.12 −1.9045 3.627
29.47 −1.5545 2.417
30.11 −0.9145 0.836
30.44 −0.5845 0.342
30.95 −0.0745 0.006
31.06 0.0355 0.001
31.54 0.5155 0.266
31.88 0.8555 0.732
32.18 1.1555 1.335
32.41 1.3855 1.920
33.02 1.9955 3.982
33.57 2.5455 6.481
33.89 2.8655 8.211
34.22 3.1955 10.211
34.76 3.7355 13.953
35.01 3.9855 15.885
∑
(
?
?
−
?
ˉ
)
2
≈
94.27
∑(x
i
−
x
ˉ
)
2
≈94.27
Sample Variance
Formula:
?
2
=
∑
(
?
?
−
?
ˉ
)
2
?
−
1
s
2
=
n−1
∑(x
i
−
x
ˉ
)
2
Calculation:
?
2
=
94.27
19
=
4.96
s
2
=
19
94.27
=
4.96
Sample Standard Deviation
Formula:
?
=
?
2
s=
s
2
Calculation:
?
=
4.96
=
2.23
s=
4.96
=
2.23
Outlier Detection Using 1.5·IQR Rule
Lower half (first 10 values):
27.85, 28.66, 28.91, 29.04, 29.12, 29.47, 30.11, 30.44, 30.95, 31.06
?
1
=
29.12
+
29.47
2
=
29.295
Q
1
=
2
29.12+29.47
=29.295
Upper half (last 10 values):
31.54, 31.88, 32.18, 32.41, 33.02, 33.57, 33.89, 34.22, 34.76, 35.01
?
3
=
33.02
+
33.57
2
=
33.295
Q
3
=
2
33.02+33.57
=33.295
?
?
?
=
33.295
−
29.295
=
4.00
IQR=33.295−29.295=4.00
Lower fence:
29.295
−
1.5
(
4.00
)
=
23.295
29.295−1.5(4.00)=23.295
Upper fence:
33.295
+
1.5
(
4.00
)
=
39.295
33.295+1.5(4.00)=39.295
All observations lie between these limits.
No outliers are present
No outliers are present
Final Answers
Range = 7.16
Sample mean = 31.0245
Median = 31.30
Sample variance = 4.96
Sample standard deviation = 2.23
Outliers = None
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-03 08:31:56
Final Answer:
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# Statistical Analysis of Absorbency Data
## Definitions and Concepts Used
### Range
The range is defined as the difference between the maximum and minimum observations in a dataset.
\[
\text{Range} = x_{\text{max}} - x_{\text{min}}
\]
### Sample Mean
The sample mean is the arithmetic average of all sample observations.
\[
\bar{x} = \frac{\sum x}{n}
\]
### Median
For an even number of observations, the median is the average of the \( \frac{n}{2} \)th and \( \left(\frac{n}{2} + 1\right) \)th values after ordering the data.
### Sample Variance
The sample variance measures variability and uses \( n - 1 \) in the denominator.
\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]
### Sample Standard Deviation
The standard deviation is the square root of the sample variance.
\[
s = \sqrt{s^2}
\]
### Outlier Detection (1.5·IQR Rule)
The interquartile range (IQR) is calculated as:
\[
IQR = Q_3 - Q_1
\]
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR
\]
## Given Information
### Number of Observations
- \( n = 20 \)
### Absorbency Values
```
32.18, 29.47, 31.06, 34.22, 28.91,
33.57, 30.44, 35.01, 29.12, 31.88,
28.66, 30.95, 34.76, 33.89, 29.04,
32.41, 33.02, 30.11, 27.85, 31.54
```
### Recorded Total
\[
\sum x = 620.49
\]
## Solution
### Step 1: Ordering the Data
First, we will sort the observed values in ascending order:
```
27.85, 28.66, 28.91, 29.04, 29.12,
29.47, 30.11, 30.44, 30.95, 31.06,
31.54, 31.88, 32.18, 32.41, 33.02,
33.57, 33.89, 34.22, 34.76, 35.01
```
### Step 2: Range Calculation
Using the ordered data:
\[
\text{Range} = 35.01 - 27.85 = 7.16
\]
### Step 3: Sample Mean Calculation
Calculating the sample mean:
\[
\bar{x} = \frac{620.49}{20} = 31.0245
\]
### Step 4: Median Calculation
For \( n = 20 \) (even number of observations), we find the median using the 10th and 11th values:
10th value = 31.06
11th value = 31.54
\[
\text{Median} = \frac{31.06 + 31.54}{2} = 31.30
\]
### Step 5: Variance Calculation
We first calculate the squared deviations from the mean.
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|-----------|---------------------|-------------------------|
| 27.85 | -3.1745 | \( 10.078 \) |
| 28.66 | -2.3645 | \( 5.592 \) |
| 28.91 | -2.1145 | \( 4.471 \) |
| 29.04 | -1.9845 | \( 3.938 \) |
| 29.12 | -1.9045 | \( 3.627 \) |
| 29.47 | -1.5545 | \( 2.417 \) |
| 30.11 | -0.9145 | \( 0.836 \) |
| 30.44 | -0.5845 | \( 0.342 \) |
| 30.95 | -0.0745 | \( 0.006 \) |
| 31.06 | 0.0355 | \( 0.001 \) |
| 31.54 | 0.5155 | \( 0.266 \) |
| 31.88 | 0.8555 | \( 0.732 \) |
| 32.18 | 1.1555 | \( 1.335 \) |
| 32.41 | 1.3855 | \( 1.920 \) |
| 33.02 | 1.9955 | \( 3.982 \) |
| 33.57 | 2.5455 | \( 6.481 \) |
| 33.89 | 2.8655 | \( 8.211 \) |
| 34.22 | 3.1955 | \( 10.211 \) |
| 34.76 | 3.7355 | \( 13.953 \) |
| 35.01 | 4.9855 | \( 15.885 \) |
Calculating the sum of squared deviations:
\[
\sum (x_i - \bar{x})^2 \approx 94.27
\]
### Sample Variance Calculation
\[
s^2 = \frac{94.27}{19} \approx 4.96
\]
### Sample Standard Deviation Calculation
\[
s = \sqrt{4.96} \approx 2.23
\]
### Step 6: Outlier Detection Using the 1.5·IQR Rule
#### Lower Half (First 10 Ordered Values)
```
27.85, 28.66, 28.91, 29.04, 29.12,
29.47, 30.11, 30.44, 30.95, 31.06
```
\[
Q_1 = \frac{29.12 + 29.47}{2} = 29.295
\]
#### Upper Half (Last 10 Ordered Values)
```
31.54, 31.88, 32.18, 32.41, 33.02,
33.57, 33.89, 34.22, 34.76, 35.01
```
\[
Q_3 = \frac{33.02 + 33.57}{2} = 33.295
\]
#### IQR Calculation
\[
IQR = Q_3 - Q_1 = 33.295 - 29.295 = 4.00
\]
### Outlier Cutoff Values
- **Lower Fence:**
\[
Q_1 - 1.5 \times IQR = 29.295 - 1.5 \times 4.00 = 23.295
\]
- **Upper Fence:**
\[
Q_3 + 1.5 \times IQR = 33.295 + 1.5 \times 4.00 = 39.295
\]
### Conclusion on Outliers
All observations lie between these limits (27.85 to 35.01). Thus, no outliers are present.
## Final Answers
- **Range:** 7.16
- **Sample Mean:** 31.0245
- **Median:** 31.30
- **Sample Variance:** 4.96
- **Sample Standard Deviation:** 2.23
- **Outliers:** None
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