Created by: Tonya Jagoe

Measures of Central Tendency & Spread 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 NA ( ) 2 = NA = – 18 = 20 Outliers (?) = 1.5(20) = = – 30 = -12 No data above 68 or below -12, therefore, NO OUTLIERS

Mean (µ) ± 1 std. dev. (σ) (12.7) (12.7) 3.8 Mean (µ) ± 2 std. dev. (σ) Min & Q1 Med Q3 Max Box Plot Standard Deviation Distribution

Measures of Central Tendency & Spread 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 NA ( ) 2 = 3.2 NA = 5 9 – 6.5 = 2.5 Outliers (?) = 1.5(2.5) = = – 3.75 = 2.75 No data above or below 2.75, therefore, NO OUTLIERS

Mean (µ) ± 1 std. dev. (σ) (1.8) (1.8) 4.2 Mean (µ) ± 2 std. dev. (σ) MinQ1 Med Q3Max Box Plot Standard Deviation Distribution

Measures of Central Tendency & Spread 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) NA ( ) 2 = 36.2 NA = – 61 = 9 Outliers (?) = 1.5(9) = = – 13.5 = is an outlier.84 > 83.5

Mean (µ) ± 1 std. dev. (σ) (6.0) (6.0) 54.3 Mean (µ) ± 2 std. dev. (σ) MinQ1MedQ3 Max Box Plot Standard Deviation Distribution Last X Before outlier