Statistics I · Unit 2 · 6 hrs
Descriptive Statistics
Exam-focused notes for Descriptive Statistics (Statistics I, STA169): what the TU syllabus asks and how it has actually been tested, with 23 solved past questions from this unit.
What this unit covers
- Measures of central tendency
- Measures of dispersion
- Measures of skewness
- Measures of kurtosis
- Moments
- Steam and leaf display
- five number summary
- box plot
- Problems and illustrative examples related to computer Science and IT
Measures of central tendency
Measures of Central Tendency and Dispersion in Descriptive Statistics
Grouped frequency distribution, $n = 50$: Mass (lbs) 1-2 2-3 3-4 4-5 5-6 6-7 --------------------- Frequency $f$ 8 10 15 9 6 2 $\sum f = 8+10+15+9+6+2 = 50$ ✓ Required: mean, standard deviation, variance, coefficient of variation, plus conceptual explanatio...
Full solved answer →What are the requisites for good average?
From the following distribution of marks of 200 students of a college:
$$\begin{array}{c|ccccccc} \text{Marks} & 30\text{-}40 & 40\text{-}50 & 50\text{-}60 & 60\text{-}70 & 70\text{-}80 & 80\text{-}90 \ \hline \text{No. of students} & 14 & 50 & 60 & 45 & 20 & 11 \ \end{array}$$
Compute:
i. The minimum marks obtained by top 10% students.
ii. Modal marks.
[5]
1. Rigidly defined - It should have a clear, unambiguous definition. 2. Based on all observations - It should use every value in the data. 3. Easy to understand and compute - Simple to calculate and interpret. 4. Capable of further algebraic treatment - Sui...
Full solved answer →The following table shows the marks obtained by 130 students in computer science.i. Find the appropriate measure of central tendency. ii. Compute the minimum marks obtained by the top 20% of students. [5]
- Total number of students: $N = 130$ - Subject: Computer Science marks - Required: (i) appropriate measure of central tendency, (ii) minimum marks of top 20% of students Missing data: The actual frequency distribution table (class intervals and frequencies...
Full solved answer →If 50 image of your website, 10 have black and white image, and their average scanned image occupies with 2.5 megabytes of memory. The total image occupies by the entire work 281 megabytes. Find the average occupies megabytes of those color images. [5]
Parameter Value ------------------ Total images 50 Black and white (B&W) images 10 Average size of each B&W image 2.5 MB Total memory occupied by all images 281 MB --- $$\text{Color images} = 50 - 10 = 40$$ $$= 10 \times 2.5 = 25 \text{ MB}$$ $$= 281 - 25 =...
Full solved answer →Calculate Q1, D7 and P58 from the following data and interpret the results.
| Weight | 0-10 | 10-15 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-60 |
|---|---|---|---|---|---|---|---|---|---|
| No. of person | 4 | 8 | 30 | 15 | 13 | 6 | 4 | 4 | 1 |
[5]
Weight 0-10 10-15 20-25 25-30 30-35 35-40 40-45 45-50 50-60 ---------------------------------------------------------------------- No. of persons 4 8 30 15 13 6 4 4 1 Note: The class 15-20 is absent from the data. To keep the classes continuous, I insert it...
Full solved answer →The following table gives the installation time (in minutes) for hardware on 50 different computers. If the average installation time is 30.2 minutes, find missing frequencies.
$$\begin{array}{|c|ccccc|c|}\hline \text{Installation Time} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & \text{Total} \ \text{Number of computers} & 4 & - & 10 & - & 10 & 50 \ \hline \end{array}$$
[5]
Installation Time 0-10 10-20 20-30 30-40 40-50 Total :---::---::---::---::---::---::---: No. of computers 4 $f1$ 10 $f2$ 10 50 - Total frequency $N = 50$ - Mean $\bar{x} = 30.2$ minutes Missing frequencies: $f1$ (class 10-20) and $f2$ (class 30-40). --- $$4...
Full solved answer →Power Failure Duration Analysis
The length of power failure in minutes are recorded in the following table. Find $Q_3$, $D_2$ and $P_{40}$ and interpret the results.
| Power failure time | 22 | 23 | 24 | 25 | 26 | 27 | 28 | Total |
|---|---|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | 7 | 10 | 4 | 3 | 2 | 33 |
[5]
Power Failure Time (x) 22 23 24 25 26 27 28 ------------------------ Frequency (f) 2 5 7 10 4 3 2 Total $N = 33$. x f cf --------- 22 2 2 23 5 7 24 7 14 25 10 24 26 4 28 27 3 31 28 2 33 For a discrete series we use position $\frac{k(N+1)}{n}$ (with $N+1 = 3...
Full solved answer →Measurement of computer chip's thickness (in nanometers) is recorded below. Find the mode of thickness of computer chips and interpret the result.
| Thickness of chips in n.m. | 34-39 | 39-44 | 44-49 | 49-54 | 54-59 | Total |
|---|---|---|---|---|---|---|
| No. of computers | 3 | 11 | 16 | 25 | 5 | 60 |
[5]
Thickness (n.m.) No. of computers (f) -------------------------------------- 34 - 39 3 39 - 44 11 44 - 49 16 49 - 54 25 54 - 59 5 Total 60 Class width $h = 5$, continuous class intervals. --- The modal class is the class with the highest frequency. Highest ...
Full solved answer →Calculate Q3, D6 and P80 from the following data and interpret the results.
| Respiratory rate | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
|---|---|---|---|---|---|---|---|---|---|
| No. of Person | 8 | 12 | 36 | 25 | 28 | 18 | 9 | 12 | 6 |
[5]
Discrete (ungrouped) frequency distribution: Respiratory rate (x) 10 15 20 25 30 35 40 45 50 ------------------------------ No. of persons (f) 8 12 36 25 28 18 9 12 6 x f cf --------- 10 8 8 15 12 20 20 36 56 25 25 81 30 28 109 35 18 127 40 9 136 45 12 148 ...
Full solved answer →Define statistics and discuss its importance in the field of computational sciences. The following are the numbers of minutes that a person had to wait for the bus to work on 20 working days: 15, 10, 2, 17, 5, 8, 3, 10, 2, 9, 5, 9, 13, 1, 10, 12, 5, 10, 8, 4. Compute mean, median, mode, standard, variance and coefficient of variation.[10]
Statistics is the branch of science that deals with the collection, organization, presentation, analysis, and interpretation of numerical data in order to draw valid conclusions and make rational decisions under conditions of uncertainty. 1. Fast computatio...
Full solved answer →Partition Values
What are partition values?
Partition values are values that divide a distribution into equal parts. Common partition values include quartiles (dividing into 4 parts), deciles (dividing into 10 parts), and percentiles (dividing into 100 parts).
From the following distribution of scores of 200 students of a college, compute:
| Scores | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
|---|---|---|---|---|---|---|
| Number of students | 14 | 50 | 60 | 45 | 20 | 11 |
i. The minimum scores obtained by top 10% students.
The top 10% corresponds to the 90th percentile. We need to find the score below which 90% of students fall.
Position of 90th percentile = $\frac{90}{100} \times 200 = 180$
Cumulative frequencies:
- 30-40: 14
- 40-50: 64
- 50-60: 124
- 60-70: 169
- 70-80: 189
- 80-90: 200
The 180th student falls in the 70-80 class.
Using the formula: $P_{90} = L + \frac{\frac{90N}{100} - CF}{f} \times h$
$$P_{90} = 70 + \frac{180 - 169}{20} \times 10 = 70 + \frac{11}{20} \times 10 = 70 + 5.5 = 75.5$$
The minimum score obtained by top 10% students is 75.5
ii. The range of middle 60% students.
The middle 60% corresponds to the 20th to 80th percentiles.
For 20th percentile: Position = $\frac{20}{100} \times 200 = 40$
The 40th student falls in the 40-50 class.
$$P_{20} = 40 + \frac{40 - 14}{50} \times 10 = 40 + \frac{26}{50} \times 10 = 40 + 5.2 = 45.2$$
For 80th percentile: Position = $\frac{80}{100} \times 200 = 160$
The 160th student falls in the 60-70 class.
$$P_{80} = 60 + \frac{160 - 124}{45} \times 10 = 60 + \frac{36}{45} \times 10 = 60 + 8 = 68$$
The range of middle 60% students is 45.2 to 68
[5]
Partition values are values that divide a distribution (arranged in order) into a number of equal parts. Common ones: - Quartiles ($Q$): divide data into 4 equal parts - Deciles ($D$): divide data into 10 equal parts - Percentiles ($P$): divide data into 10...
Full solved answer →five number summary
What aspect of summary measures of data can be explained by the measures of skewness? Kelvin Hota is the national sales manager for National Text Books. He has a sales staff of 10 who visit college professors all over the United States. Each Sunday morning he requires his sales staffs to send him a report. Listed below are the number of visits last week. Compute the five number summary. 25, 6, 10, 13, 15, 2, 18, 5, 20, 30 [5]
Summary measures describe several aspects of data: central tendency (mean, median, mode), dispersion (range, variance), and shape. Measures of skewness explain the shape aspect, specifically the lack of symmetry of a distribution. Skewness tells us: - Wheth...
Full solved answer →Write notes on any two: i. Nominal and ordinal scale ii. Kurtosis iii. Five number summary [5]
--- Nominal scale is the simplest and lowest level of measurement scale. It is a system of assigning numbers or symbols to objects or events in order to distinguish one from another and label them. Key characteristics: - The symbols or numbers assigned have...
Full solved answer →Moments
The first four moments of a distribution about x=2 are 1,2.5,5.5 and 16. Calculate the first four moments about the mean. Test the skewness and kurtosis. Interpret the results. [5]
Moments about $x = 2$ (arbitrary origin $a = 2$): $$\mu1' = 1, \quad \mu2' = 2.5, \quad \mu3' = 5.5, \quad \mu4' = 16$$ --- First: $$\mu1 = 0$$ Second: $$\mu2 = \mu2' - (\mu1')^2 = 2.5 - 1 = 1.5$$ Third: $$\mu3 = \mu3' - 3\mu2'\mu1' + 2(\mu1')^3 = 5.5 - 3(2...
Full solved answer →Compute first four moments about arbitrary point 4 from following distribution and describe the characteristics of data.
$$\begin{array}{c|ccccc} X & 2 & 3 & 4 & 5 & 6 \ \hline f & 1 & 3 & 7 & 2 & 1 \ \end{array}$$
[5]
X 2 3 4 5 6 ------------------ f 1 3 7 2 1 $N = \sum f = 14$, Arbitrary point $A = 4$. Moment formula about $A$: $\mur' = \dfrac{\sum f d^r}{N}$, where $d = X - A = X - 4$. X f d fd fd² fd³ fd⁴ ---------------------------- 2 1 -2 -2 4 -8 16 3 3 -1 -3 3 -3 3...
Full solved answer →Measures of dispersion
Measures of Dispersion
Measures of dispersion quantify the spread or variability of data values about a central value. Central tendency tells us where data centers; dispersion tells us how scattered the data is. Small dispersion means values cluster tightly; large dispersion mean...
Full solved answer →Question
Two batsmen A and B made the following runs in a series of cricket matches. Who is a more consistent player, and why?
$$\begin{array}{c|ccccc} A & 10 & 0 & 56 & 80 & 24 \ B & 36 & 37 & 45 & 28 & 29 \ \end{array}$$
[5]
- Batsman A runs: $10, 0, 56, 80, 24$ (n = 5) - Batsman B runs: $36, 37, 45, 28, 29$ (n = 5) Consistency is measured by the Coefficient of Variation (CV); the lower CV indicates the more consistent player. $$CV = \frac{\sigma}{\bar{x}} \times 100$$ --- $$\b...
Full solved answer →What are different methods of measuring dispersion?
Sample of polythene bags from two manufacturers, A, B, are tested by a prospective buyer for bursting pressure and the results are as follows. Which set of bags has more uniform pressure? If prices are the same, which manufacturer's bags would be preferred by buyer? Use appropriate statistical tool.
| Bursting Pressure | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
|---|---|---|---|---|---|---|
| A | 2 | 9 | 29 | 54 | 11 | 5 |
| B | 9 | 11 | 18 | 32 | 27 | 13 |
[10]
Bursting Pressure Mid-value $x$ $fA$ $fB$ :---::---::---::---: 5-10 7.5 2 9 10-15 12.5 9 11 15-20 17.5 29 18 20-25 22.5 54 32 25-30 27.5 11 27 30-35 32.5 5 13 $\sum fA = 110$, $\sum fB = 110$ --- Absolute Measures - Range: $R = L - S$ - Quartile Deviation: ...
Full solved answer →What are the roles of measure of dispersion in descriptive statistics?
Following table gives the frequency distribution of thickness of computer chips (in nanometer) manufactured by two companies.
$$\begin{array}{|c|cccccc|}\hline \text{Thickness of computer chips} & 5 & 10 & 15 & 20 & 25 & 30 \ \hline \text{Number of chips} & & & & & & \ \text{Company A} & 10 & 15 & 24 & 20 & 18 & 13 \ \text{Company B} & 12 & 18 & 20 & 22 & 24 & 4 \ \hline \end{array}$$
[10]
A measure of dispersion quantifies the spread or scatter of observations around a central value. Its main roles are: 1. Judging reliability of an average: Small dispersion means the average represents the data well; large dispersion weakens the average's re...
Full solved answer →Distinction Between Absolute and Relative Measures of Dispersion
Absolute Measure of Dispersion: An absolute measure expresses the dispersion in the same units as the original data. Examples include range, variance, standard deviation, and mean deviation. These measures are useful when comparing datasets with similar means and units.
Relative Measure of Dispersion: A relative measure is a dimensionless quantity expressed as a ratio or percentage of a central tendency measure (usually the mean). Examples include coefficient of variation, coefficient of range, and coefficient of quartile deviation. These measures are useful for comparing the consistency of datasets with different means or different units.
Computer Consistency Analysis
| Time (in seconds) | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 |
|---|---|---|---|---|---|---|
| A | 5 | 16 | 13 | 7 | 5 | 4 |
| B | 2 | 7 | 12 | 19 | 9 | 1 |
For Company A:
Mean: $\bar{x}_A = \frac{1 \times 5 + 3 \times 16 + 5 \times 13 + 7 \times 7 + 9 \times 5 + 11 \times 4}{50} = \frac{5 + 48 + 65 + 49 + 45 + 44}{50} = \frac{256}{50} = 5.12$ seconds
Variance: $\sigma_A^2 = \frac{\sum f(x - \bar{x})^2}{n} = \frac{5(1-5.12)^2 + 16(3-5.12)^2 + 13(5-5.12)^2 + 7(7-5.12)^2 + 5(9-5.12)^2 + 4(11-5.12)^2}{50} = 6.8896$
Standard Deviation: $\sigma_A = \sqrt{6.8896} = 2.624$ seconds
Coefficient of Variation: $CV_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{2.624}{5.12} \times 100 = 51.25%$
For Company B:
Mean: $\bar{x}_B = \frac{1 \times 2 + 3 \times 7 + 5 \times 12 + 7 \times 19 + 9 \times 9 + 11 \times 1}{50} = \frac{2 + 21 + 60 + 133 + 81 + 11}{50} = \frac{308}{50} = 6.16$ seconds
Variance: $\sigma_B^2 = \frac{2(1-6.16)^2 + 7(3-6.16)^2 + 12(5-6.16)^2 + 19(7-6.16)^2 + 9(9-6.16)^2 + 1(11-6.16)^2}{50} = 5.8704$
Standard Deviation: $\sigma_B = \sqrt{5.8704} = 2.423$ seconds
Coefficient of Variation: $CV_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{2.423}{6.16} \times 100 = 39.37%$
Conclusion: Since $CV_B (39.37%) < CV_A (51.25%)$, Company B's computers are more consistent as they have a lower coefficient of variation, indicating less relative variability in execution time.
Basis Absolute Measure Relative Measure ----------------------------------------- Definition Expresses dispersion in the same units as the original data Expresses dispersion as a pure number (ratio/percentage), free of units Unit Has units (seconds, kg, cm,...
Full solved answer →Measurement of Dispersion and Consistency Analysis
Measurement of dispersion refers to statistical measures describing the spread or variability of data values around a central value. It indicates how much individual observations deviate from the average. Common measures: Range, Mean Deviation, Standard Dev...
Full solved answer →Measures of kurtosis
Compute percentile coefficient of kurtosis from the following data and interpret the result.
| Hourly wages (Rs) | 23-27 | 28-32 | 33-37 | 38-42 | 43-47 | 48-52 |
|---|---|---|---|---|---|---|
| Number of workers | 22 | 16 | 9 | 4 | 3 | 1 |
[5]
Class (Wages) Boundaries f cf ------------ 23-27 22.5-27.5 22 22 28-32 27.5-32.5 16 38 33-37 32.5-37.5 9 47 38-42 37.5-42.5 4 51 43-47 42.5-47.5 3 54 48-52 47.5-52.5 1 55 Total N = 55 Class width $h = 5$. $$k = \frac{QD}{P{90} - P{10}}, \qquad QD = \frac{Q3...
Full solved answer →Measures of skewness
Define skewness and kurtosis. The first four moments about mean are 0, 14.75, 39.75 and 152.31. Compute skewness and kurtosis and interpret the results. [5]
Skewness measures the lack of symmetry in a frequency distribution. It indicates the direction and degree of departure from symmetry. A perfectly symmetrical distribution has skewness = 0. Kurtosis measures the degree of peakedness or flatness of a distribu...
Full solved answer →Make Unit 2 stick
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