Section 1: Measures of Central Tendency (Mean, Median, Mode) / 集中趨勢的量度（平均數、中位數、眾數）

Example 1.1: Ungrouped Data / 不分組數據

• Question / 題目:
• Five students scored 45, 60, 60, 75, and 90 in a quiz; calculate the mean, median, and mode of their scores.
• 五名學生在一項測驗中分別取得 45、60、60、75 及 90 分；計算他們分數的平均數、中位數和眾數。
• Solution / 題解:
• The mean is calculated by summing all scores and dividing by 5, which equals $(45+60+60+75+90)\mathrm{/}5=66$(45+60+60+75+90)/5=66 marks.
• 平均數是把所有分數相加後除以 5，計算得出 $(45+60+60+75+90)\mathrm{/}5=66$(45+60+60+75+90)/5=66 分。
• The median is the 3rd score when ordered, which is 60 marks.
• 按次序排列後，中位數是第 3 個數據，即 60 分。
• The mode is 60 marks because it occurs most frequently.
• 眾數是 60 分，因為它出現的次數最多。

Example 1.2: Grouped Data / 分組數據

• Question / 題目:
• The table shows the daily study hours of 20 students: 1–3 hours (class mark 2) has 5 students, 4–6 hours (class mark 5) has 11 students, and 7–9 hours (class mark 8) has 4 students; determine the mean study hours and the modal class.
• 下表顯示 20 名學生的每日學習時數：1–3 小時（組中點 2）有 5 人，4–6 小時（組中點 5）有 11 人，7–9 小時（組中點 8）有 4 人；求平均學習時數及眾數組。
• Solution / 題解:
• The mean is calculated as $[(2\times 5)+(5\times 11)+(8\times 4)]\mathrm{/}20=97\mathrm{/}20=4.85$[(2×5)+(5×11)+(8×4)]/20=97/20=4.85 hours.
• 平均數的計算方式為 $[(2\times 5)+(5\times 11)+(8\times 4)]\mathrm{/}20=97\mathrm{/}20=4.85$[(2×5)+(5×11)+(8×4)]/20=97/20=4.85 小時。
• The modal class is 4–6 hours because it has the highest frequency of 11.
• 眾數組為 4–6 小時，因為其頻數最高，達 11 人。

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Section 2: Effect of Extreme Values (Outliers) / 極端值（離群值）的影響

Example 2.1: Impact on Mean vs. Median / 對平均數與中位數的影響

• Question / 題目:
• A group of workers have monthly salaries of $15,000, $16,000, $17,000, $18,000, and $120,000; explain how the extreme value affects the mean and why the median is a better measure.
• 一組工人的月薪分別為 $15,000、$16,000、$17,000、$18,000 及 $120,000；解釋極端值如何影響平均數，以及為何中位數是更佳的量度。
• Solution / 題解:
• The mean salary is pulled up significantly to $37,200 due to the outlier of $120,000.
• 由於存在 $120,000 的離群值，平均薪酬被大幅拉高至 $37,200。
• The median is $17,000, which is not distorted by the outlier and better represents a typical worker’s salary.
• 中位數為 $17,000，不會受離群值扭曲，更能代表一般工人的薪酬水平。

Example 2.2: Trimming Outliers / 移除離群值

• Question / 題目:
• The scores of 6 participants in a speed test are 12, 13, 14, 15, 16, and 80 seconds; find the new mean if the outlier is removed.
• 6 名參加者在速度測試中的成績分別為 12、13、14、15、16 及 80 秒；若剔除離群值，求新的平均數。
• Solution / 題解:
• The original mean is $(12+13+14+15+16+80)\mathrm{/}6=25$(12+13+14+15+16+80)/6=25 seconds, heavily distorted by the outlier 80.
• 原本的平均數為 $(12+13+14+15+16+80)\mathrm{/}6=25$(12+13+14+15+16+80)/6=25 秒，受到離群值 80 的嚴重扭曲。
• Removing the outlier 80 yields a new mean of $(12+13+14+15+16)\mathrm{/}5=14$(12+13+14+15+16)/5=14 seconds.
• 剔除離群值 80 後，新的平均數為 $(12+13+14+15+16)\mathrm{/}5=14$(12+13+14+15+16)/5=14 秒。

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Section 3: Effect of Adding a Common Constant on Central Tendency / 數據全體加上常數對集中趨勢的影響

Example 3.1: Uniform Mark Adjustment / 全體加分調整

• Question / 題目:
• A class test had a mean of 52, a median of 50, and a mode of 48; if the teacher adds 5 bonus marks to every student, find the new mean, median, and mode.
• 某次測驗的平均數為 52 分，中位數為 50 分，眾數為 48 分；若老師為每位學生額外增加 5 分，求新的平均數、中位數和眾數。
• Solution / 題解:
• When a constant $k=5$k=5 is added to all data, each central tendency measure increases by 5.
• 當所有數據都加上常數 $k=5$k=5 時，每項集中趨勢量度都會增加 5。
• Therefore, the new mean is $52+5=57$52+5=57, the new median is $50+5=55$50+5=55, and the new mode is $48+5=53$48+5=53.
• 因此，新的平均數為 $52+5=57$52+5=57 分，新的中位數為 $50+5=55$50+5=55 分，新的眾數為 $48+5=53$48+5=53 分。

Example 3.2: Reversing a Shift / 反求原始量度

• Question / 題目:
• After adding 10 minutes of reading time to all students’ recorded daily times, the new median is 45 minutes; what was the original median?
• 每位學生記錄的每日閱讀時間均增加 10 分鐘後，新的中位數變為 45 分鐘；原本的中位數是多少？
• Solution / 題解:
• Since adding a constant $k$k increases the median by $k$k, the original median is found by subtracting $k$k.
• 由於加上常數 $k$k 會使中位數增加 $k$k，原本的中位數可透過減去 $k$k 求得。
• The original median was $45-10=35$45−10=35 minutes.
• 原本的中位數為 $45-10=35$45−10=35 分鐘。

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Section 4: Effect of Inserting a Datum “ZERO” on Central Tendency / 加入數值「零」對集中趨勢的影響

Example 4.1: Effect on the Mean / 對平均數的影響

• Question / 題目:
• Four students completed 8, 10, 12, and 14 homework exercises; calculate the change in the mean if a fifth student who completed 0 exercises is included.
• 四名學生分別完成了 8、10、12 及 14 份功課練習；若加入第五名完成 0 份功課的學生，計算其平均數的變化。
• Solution / 題解:
• The original mean of the four students is $(8+10+12+14)\mathrm{/}4=11$(8+10+12+14)/4=11 exercises.
• 原本四名學生的平均數為 $(8+10+12+14)\mathrm{/}4=11$(8+10+12+14)/4=11 份練習。
• With the zero added, the new mean becomes $(0+8+10+12+14)\mathrm{/}5=8.8$(0+8+10+12+14)/5=8.8 exercises, showing a clear decrease.
• 加入零之後，新的平均數變為 $(0+8+10+12+14)\mathrm{/}5=8.8$(0+8+10+12+14)/5=8.8 份練習，顯示平均數明顯下降。

Example 4.2: Effect on the Median / 對中位數的影響

• Question / 題目:
• Given the ordered data set $\{4,6,8,10\}${4,6,8,10}, find the median before and after inserting a value of 0.
• 給定已排序的數據組 $\{4,6,8,10\}${4,6,8,10}，求加入數值 0 之前及之後的中位數。
• Solution / 題解:
• The original median of the 4 values is $(6+8)\mathrm{/}2=7$(6+8)/2=7.
• 原本 4 個數值的中位數為 $(6+8)\mathrm{/}2=7$(6+8)/2=7。
• After inserting 0, the new ordered set is $\{0,4,6,8,10\}${0,4,6,8,10} with 5 values, so the new median becomes the 3rd datum, which is 6.
• 加入 0 後，新的有序數據組為包含 5 個數值的 $\{0,4,6,8,10\}${0,4,6,8,10}，因此新的中位數變為第 3 個數據，即 6。

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Section 5: Sources of Deception in Misuse of Means / 誤用平均數的做假根源

Example 5.1: Selective Sampling Bias / 蓄意抽樣偏差

• Question / 題目:
• A tutoring centre claims that its students achieve an average score of 95 marks, but the survey was conducted solely on the top 10 scholarship winners; identify the source of deception.
• 一家補習社聲稱其學生的平均分為 95 分，但該調查僅以獲得獎學金的 10 名頂尖學生為樣本；指出其中的做假根源。
• Solution / 題解:
• This is an example of deliberately sampling only a specific part of the population that benefited from the centre.
• 這是屬於蓄意只抽取受惠於該中心特徵的特定母體為樣本的例子。
• The reported mean is biased and cannot represent the performance of all general students.
• 所公佈的平均數存有偏差，無法代表全體一般學生的真實表現。

Example 5.2: Hiding Distribution Characteristics / 隱藏分佈特徵與離群值

• Question / 題目:
• A company boasts that its average staff bonus is $50,000, while most employees only received $5,000 and the top executive received $500,000; explain the flaw in this report.
• 一家公司吹噓其員工的平均花紅為 $50,000，但絕大多數員工只拿到 $5,000，而最高管理層一人獨得 $500,000；解釋該報告的漏洞。
• Solution / 題解:
• The report only quotes the arithmetic mean without revealing extreme values or multi-modal patterns.
• 該報告僅引用算術平均數，而沒有揭示極端值或多峰分佈的真實情況。
• A few massive outliers heavily inflate the mean, giving a false impression of what typical employees receive.
• 少數龐大的離群值大幅拉高了平均數，給人對一般員工實質所得的虛假印象。

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Section 6: Measures of Dispersion (Range, IQR, Q.D.) / 離差的量度（分佈域、四分位數間距、四分位差）

Example 6.1: Ungrouped Range, IQR, and Q.D. / 不分組數據的分佈域、四分位數間距及四分位差

• Question / 題目:
• For the ordered data set $\{10,14,18,22,26,30,34\}${10,14,18,22,26,30,34}, find the range, inter-quartile range (IQR), and quartile deviation (Q.D.).
• 對於已排序的數據組 $\{10,14,18,22,26,30,34\}${10,14,18,22,26,30,34}，求分佈域、四分位數間距（IQR）及四分位差（Q.D.）。
• Solution / 題解:
• The range is the maximum value minus the minimum value: $34-10=24$34−10=24.
• 分佈域為最大值減去最小值：$34-10=24$34−10=24。
• For $n=7$n=7, the lower quartile $Q_1$Q1​ is the 2nd datum (14) and the upper quartile $Q_3$Q3​ is the 6th datum (30).
• 當 $n=7$n=7 時，第一四分位數 $Q_1$Q1​ 為第 2 個數據（14），第三四分位數 $Q_3$Q3​ 為第 6 個數據（30）。
• The IQR is $Q_3-Q_1=30-14=16$Q3​−Q1​=30−14=16, and the Q.D. is $(Q_3-Q_1)\mathrm{/}2=16\mathrm{/}2=8$(Q3​−Q1​)/2=16/2=8.
• 四分位數間距為 $Q_3-Q_1=30-14=16$Q3​−Q1​=30−14=16，而四分位差為 $(Q_3-Q_1)\mathrm{/}2=16\mathrm{/}2=8$(Q3​−Q1​)/2=16/2=8。

Example 6.2: Grouped Data Boundaries / 分組數據的分佈域計算

• Question / 題目:
• A grouped frequency table has classes 10–19, 20–29, and 30–39; determine the range of this grouped data.
• 某分組頻數表的組別為 10–19、20–29 及 30–39；求該組分組數據的分佈域。
• Solution / 題解:
• The lowest class boundary is 9.5 and the highest class boundary is 39.5.
• 最低組界為 9.5，而最高組界為 39.5。
• The range is calculated as highest boundary minus lowest boundary: $39.5-9.5=30$39.5−9.5=30.
• 分佈域的計算方式為最高組界減去最低組界：$39.5-9.5=30$39.5−9.5=30。

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Section 7: Box-and-Whisker Diagram & Skewness / 箱形圖與偏態

Example 7.1: Reading a Five-Number Summary / 讀取五數綜合摘要

• Question / 題目:
• A box plot shows a minimum of 15, $Q_1=25$Q1​=25, median $Q_2=40$Q2​=40, $Q_3=55$Q3​=55, and a maximum of 70; find the range and IQR.
• 一個箱形圖顯示最小值為 15、$Q_1=25$Q1​=25、中位數 $Q_2=40$Q2​=40、$Q_3=55$Q3​=55 以及最大值為 70；求其分佈域和四分位數間距。
• Solution / 題解:
• The range is the length between the two whiskers: $\text{Max}-\text{Min}=70-15=55$Max−Min=70−15=55.
• 分佈域為兩條鬚狀線之間的距離：$\text{最大值}-\text{最小值}=70-15=55$最大值−最小值=70−15=55。
• The inter-quartile range is the width of the central box: $Q_3-Q_1=55-25=30$Q3​−Q1​=55−25=30.
• 四分位數間距為中間方箱的寬度：$Q_3-Q_1=55-25=30$Q3​−Q1​=55−25=30。

Example 7.2: Identifying Skewness / 判斷分佈的偏態

• Question / 題目:
• In a box plot, the left whisker spans from 10 to 30 (length 20) while the right whisker spans from 50 to 90 (length 40); determine whether the distribution is positively or negatively skewed.
• 在一個箱形圖中，左鬚由 10 延伸至 30（長度為 20），而右鬚由 50 延伸至 90（長度為 40）；判斷該分佈屬於正偏態還是負偏態。
• Solution / 題解:
• The right whisker is longer than the left whisker, indicating more large extreme values on the high end.
• 由於右邊的鬚狀線比左邊長，表示高數值端存在較多較大的極端值。
• Therefore, this data distribution is positively skewed.
• 因此，此數據分佈呈現正偏態。

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Section 8: Standard Deviation and Variance / 標準差與方差

Example 8.1: Calculating Ungrouped Variance and Standard Deviation / 計算不分組數據的方差與標準差

• Question / 題目:
• Given the data set $\{2,4,6,8\}${2,4,6,8}, compute the variance ${\sigma }^2$σ2 and the standard deviation $\sigma$σ.
• 給定數據組 $\{2,4,6,8\}${2,4,6,8}，計算其方差 ${\sigma }^2$σ2 及標準差 $\sigma$σ。
• Solution / 題解:
• The mean of the data is $\bar{x}=(2+4+6+8)\mathrm{/}4=5$xˉ=(2+4+6+8)/4=5.
• 數據的平均數為 $\bar{x}=(2+4+6+8)\mathrm{/}4=5$xˉ=(2+4+6+8)/4=5。
• The variance is ${\sigma }^2=[(2-5{)}^2+(4-5{)}^2+(6-5{)}^2+(8-5{)}^2]\mathrm{/}4=(9+1+1+9)\mathrm{/}4=5$σ2=[(2−5)2+(4−5)2+(6−5)2+(8−5)2]/4=(9+1+1+9)/4=5.
• 方差為 ${\sigma }^2=[(2-5{)}^2+(4-5{)}^2+(6-5{)}^2+(8-5{)}^2]\mathrm{/}4=(9+1+1+9)\mathrm{/}4=5$σ2=[(2−5)2+(4−5)2+(6−5)2+(8−5)2]/4=(9+1+1+9)/4=5。
• The standard deviation is the square root of the variance: $\sigma =\sqrt{5}\approx 2.24$σ=5​≈2.24.
• 標準差為方差的平方根：$\sigma =\sqrt{5}\approx 2.24$σ=5​≈2.24。

Example 8.2: Interpreting Consistency / 比較數據的穩定性

• Question / 題目:
• Machine X produces parts with a mean length of 50 mm and standard deviation of 0.8 mm, while Machine Y produces parts with a mean length of 50 mm and standard deviation of 3.2 mm; compare their consistency.
• 機器 X 生產零件的平均長度為 50 毫米、標準差為 0.8 毫米；而機器 Y 生產零件的平均長度同為 50 毫米、標準差為 3.2 毫米；比較兩台機器的穩定性。
• Solution / 題解:
• Although both machines have identical mean dimensions, Machine X has a substantially smaller standard deviation.
• 雖然兩台機器的平均尺寸完全相同，但機器 X 的標準差明顯小得多。
• This implies that Machine X produces parts with far greater consistency and less variation than Machine Y.
• 這表示機器 X 所生產零件的穩定性遠高於機器 Y，波動幅度亦較小。

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Section 9: Effect of Adding a Common Constant on Dispersion / 數據全體加上常數對離差的影響

Example 9.1: Unchanged Standard Deviation / 標準差保持不變

• Question / 題目:
• A set of weights has a standard deviation of 4.5 kg; if each object is placed inside a 1 kg container, what is the new standard deviation?
• 一組物體的重量標準差為 4.5 公斤；若將每個物體放入一個重 1 公斤的容器中，新的標準差是多少？
• Solution / 題解:
• Adding a common constant $k=1$k=1 shifts every datum by the same amount without changing the distance between them.
• 全體加上常數 $k=1$k=1 會將所有數據平移相同的數值，但數值之間的相對間距維持不變。
• Therefore, the standard deviation remains unchanged at 4.5 kg.
• 因此，標準差保持不變，仍然是 4.5 公斤。

Example 9.2: Impact on Range and IQR / 對分佈域和四分位數間距的影響

• Question / 題目:
• A test originally has a range of 40 marks and an IQR of 18 marks; what will be the new range and IQR if every student is awarded 10 extra marks?
• 某測驗原來的分佈域為 40 分，四分位數間距為 18 分；若每位學生獲加 10 分，新的分佈域和四分位數間距各是多少？
• Solution / 題解:
• Measures of dispersion describe the spread of the data and are unaffected by adding a constant.
• 離差的量度是用以描述數據的分散程度，不受全體加上常數所影響。
• The new range remains 40 marks and the new IQR remains 18 marks.
• 新的分佈域維持在 40 分，新的四分位數間距亦維持在 18 分。

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Section 10: Effect of Multiplying a Common Constant on Dispersion / 數據全體乘以常數對離差的影響

Example 10.1: Scaling Standard Deviation and Variance / 放大標準差與方差

• Question / 題目:
• A data set has a standard deviation $\sigma =6$σ=6 and a variance ${\sigma }^2=36$σ2=36; if every value is multiplied by 3, find the new standard deviation and variance.
• 某數據組的標準差 $\sigma =6$σ=6，方差 ${\sigma }^2=36$σ2=36；若將每個數值均乘以 3，求新的標準差和方差。
• Solution / 題解:
• Multiplying each value by $k=3$k=3 scales the standard deviation by $k$k, so the new standard deviation is $6\times 3=18$6×3=18.
• 當每個數值乘以 $k=3$k=3 時，標準差會擴大 $k$k 倍，因此新的標準差為 $6\times 3=18$6×3=18。
• The variance scales by $k^2$k2, resulting in a new variance of $36\times 3^2=324$36×32=324 (or ${18}^2=324$182=324).
• 方差則會擴大 $k^2$k2 倍，得出新的方差為 $36\times 3^2=324$36×32=324（亦即 ${18}^2=324$182=324）。

Example 10.2: Unit Conversion / 單位換算對離差的影響

• Question / 題目:
• The lengths of steel rods measured in meters have a range of 1.5 m and an IQR of 0.4 m; determine the new range and IQR if all measurements are converted to centimeters ($\times 100$×100).
• 以米為單位量度的一組鋼筋長度，其分佈域為 1.5 米，四分位數間距為 0.4 米；若將所有測量值換算為厘米（$\times 100$×100），求新的分佈域和四分位數間距。
• Solution / 題解:
• Multiplying each datum by 100 scales linear measures of dispersion by a factor of 100.
• 將每個數據乘以 100，會使所有線性的離差量度同等放大 100 倍。
• The new range is $1.5\times 100=150\text{ cm}$1.5×100=150 cm, and the new IQR is $0.4\times 100=40\text{ cm}$0.4×100=40 cm.
• 新的分佈域為 $1.5\times 100=150\text{ 厘米}$1.5×100=150 厘米，新的四分位數間距為 $0.4\times 100=40\text{ 厘米}$0.4×100=40 厘米。