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Published byMegan Crabtree Modified over 4 years ago

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HISTOGRAMS Representing Data Module S1

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Why use a Histogram When there is a lot of data When data is Continuous a mass, height, volume, time etc Presented in a Grouped Frequency Distribution usually in groups or classes that are UNEQUAL

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Continuous data NO GAPS between Bars

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Bars are different in width Determined by Grouped Frequency Distribution

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AREA is proportional to FREQUENCY NOT height, because of UNEQUAL classes! So we use FREQUENCY DENSITY = Frequency Class width

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Grouped Frequency Distribution Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq1491835 Speed, kph0< v 4040< v 5050< v 6060< v 9090< v 110 Frequency8015259030 Classes No gaps GAPS!Need to adjust to Continuous Ready to graph

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Adjusting Classes Class Widths Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq1491835 9½4½19½29½39½59½ 105 20

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Frequency Density Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq 1491835 Class width 510 20 Frequency Density 2.80.91.80.30.25

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Drawing Sensible Scales Bases correctly aligned Plot the Class Boundaries Heights correct Frequency Density

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4.519.59.529.539.549.559.5 3.0 2.0 1.0 Freq Dens Time (Mins)

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Estimating a Frequency Imagine we want to Estimate the number of people with a time between 12 and 25 mins Because rounded to nearest minute Consider the interval 11.5 to 25.5

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4.519.59.529.539.549.559.5 3.0 2.0 1.0 Freq Dens Time (Mins) 11.5 25.5 Frequency = 0.9 x 8 = 7.2 Frequency = 1.8 x 6 = 10.8 Total Frequency = 18

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…and the other one? Simpler to plot No adjustments required – class widths friendly No ½ values Estimation from the EXACT values given No adjustment required Estimate 15 to 56 would use 15 and 56! Appear LESS OFTEN in the exam Speed, kph0< v 4040< v 5050< v 6060< v 9090< v 110 Frequency8015259030

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