1. Created by: Tonya Jagoe

2. Measures of Central Tendency & Spread 252018 385629 502821 18 Input the data for these test scores into your calculator to find the appropriate statistics. If the statistic does not apply – put NA. You must determine if this is population or sample data. Mean = µ or Sum of x’s = ∑ Sample standard deviation = Sx Population standard deviation = σx Minimum Value = minX = Quartile 1 = Q1= Median = Med = Quartile 3 = Q3 = Maximum Value = maxX = IQR = Q3 – Q1 Range = Max - Min Sample Variance = (Sx) 2 Pop. Variance = (σx) 2 Then, create a box plot and the standard deviation graphical representation. Ages of all volunteers at a hospital. 29.2 321 NA 12.7 18 25 38 56 (12.70517562) 2 = 161.4 NA 56 - 18 = 38 38 – 18 = 20 Outliers (?) = 1.5(20) = 30 38 + 30 = 68 18 – 30 = -12 No data above 68 or below -12, therefore, NO OUTLIERS

3. 60 40 20100 30 50 29.2 +12.7 41.9 -12.7 16.5 Mean (µ) ± 1 std. dev. (σ) 29.2 +2(12.7) 54.6 -2(12.7) 3.8 Mean (µ) ± 2 std. dev. (σ) 60 40 20100 30 50 Min & Q1 Med Q3 Max Box Plot Standard Deviation Distribution

4. Measures of Central Tendency & Spread 7899910 5756 9 Input the data for these test scores into your calculator to find the appropriate statistics. If the statistic does not apply – put NA. You have to determine if this is population or sample data. Mean = µ or Sum of x’s = ∑ Sample standard deviation = Sx Population standard deviation = σx Minimum Value = minX = Quartile 1 = Q1= Median = Med = Quartile 3 = Q3 = Maximum Value = maxX = IQR = Q3 – Q1 Range = Max - Min Sample Variance = (Sx) 2 Pop. Variance = (σx) 2 Then, create a box plot and the standard deviation graphical representation. Quiz scores from randomly selected group. 7.8 94 1.8 NA 5 6.5 8.5 9 10 (1.800673275) 2 = 3.2 NA 10 - 5 = 5 9 – 6.5 = 2.5 Outliers (?) = 1.5(2.5) = 3.75 9 + 3.75 = 12.75 6.5 – 3.75 = 2.75 No data above 12.75 or below 2.75, therefore, NO OUTLIERS

5. 12 8 420 6 10 7.8 +1.8 9.6 -1.8 6.0 Mean (µ) ± 1 std. dev. (σ) 7.8 +2(1.8) 11.4 -2(1.8) 4.2 Mean (µ) ± 2 std. dev. (σ) 12 8 420 6 10 MinQ1 Med Q3Max Box Plot Standard Deviation Distribution

6. Measures of Central Tendency & Spread 606162665968 606672 6368 707160846665 Input the data for these test scores into your calculator to find the appropriate statistics. If the statistic does not apply – put NA. You have to determine if this is population or sample data. Then, create a box plot and the standard deviation graphical representation. Height in inches of students in a class. Mean = µ or Sum of x’s = ∑ Sample standard deviation = Sx Population standard deviation = σx Minimum Value = minX = Quartile 1 = Q1= Median = Med = Quartile 3 = Q3 = Maximum Value = maxX = IQR = Q3 – Q1 Range = Max - Min Sample Variance = (Sx) 2 Pop. Variance = (σx) 2 66.3 1193 NA 6.0 59 61 66 70 84 (6.01669488) 2 = 36.2 NA 84 - 59 = 25 70 – 61 = 9 Outliers (?) = 1.5(9) = 13.5 70 + 13.5 = 83.5 61 – 13.5 = 47.5 84 is an outlier.84 > 83.5

7. 66.3 +6.0 72.3 -6.0 60.3 Mean (µ) ± 1 std. dev. (σ) 66.3 +2(6.0) 78.3 -2(6.0) 54.3 Mean (µ) ± 2 std. dev. (σ) MinQ1MedQ3 Max Box Plot Standard Deviation Distribution 85 75656055 70 80 50 Last X Before outlier 85 75656055 70 80 50