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Quantiles Edexcel S1 Mathematics 2003

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Introduction- what is a quantile? Quantiles are used to divide data into intervals containing an equal number of values. For example: Deciles D1, …, D9 divide data into 10 parts Quartiles Q1, Q2, Q3 divide data into 4 parts Percentiles P1, …, P100 divide into 100 parts......... ….................................... … …...... D1D2D3D4D5D6D7D8D9

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Ungrouped data Treat data as individual values Use textbook method of rounding to next value or next.5 th value Grouped data Use linear interpolation to estimate quantile. Treat data as continuous within each group / class Assumes values are evenly distributed within each class. Introduction – Grouped / Ungrouped data

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Example – Ungrouped data Question: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16 Find the median, and first and ninth deciles of the number of appointments The median is the middle value: n/2 = 18/2 = 9 9.5 th value = = 10.5 appointments Whole number- so round up to.5 th 6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16 Answer: median Find average of 9 th and 10 th value

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Example – Ungrouped data Question: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16 Find the median, and first and ninth deciles of the number of appointments The median is the middle value: n/2 = 18/2 = 9 9.5 th value = = 10.5 appointments The first decile, D1, is the 1/10 th value: n/10 = 18/10 = 1.8 7 appointments 2 nd value = Not whole - so round up to whole Find the 2 nd value 6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16 Answer: D1median

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Example – Ungrouped data Question: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16 Find the median, and first and ninth deciles of the number of appointments The median is the middle value: n/2 = 18/2 = 9 9.5 th value = = 10.5 appointments The first decile, D1, is the 1/10 th value: n/10 = 18/10 = 1.8 7 appointments The ninth decile, D9, is the 9/10 th value: 9n/4 = 9x18/10 = 16.2 17 th value = 15 appointments 2 nd value = Not whole - so round up to whole Find the 17 th value 6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16 Answer: D1medianD9

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Example – Ungrouped data Question: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16 Find the median, and first and ninth deciles of the number of appointments The median is the middle value: n/2 = 18/2 = 9 9.5 th value = = 10.5 appointments The first decile, D1, is the 1/10 th value: n/10 = 18/10 = 1.8 7 appointments The ninth decile, D9, is the 9/10 th value: 9n/4 = 9x18/10 = 16.2 17 th value = 15 appointments 2 nd value = 6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16 Answer: D1medianD9

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Example – G rouped data Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Waiting times1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency81520152012100 Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value No rounding as interpolation is being used lies in class 15 - 19 9 + 15 + 20 = 44 so 50 th value is not in first 3 classes

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value Lower class boundary (lcb) class frequenc y median = 14.5 + (19.5 – 14.5) Median position Cumulative frequency to lcb uc b lcb 15 14.519.5 median 44 15 – 19 class lies in class 15 - 19

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value Lower class boundary (lcb) class frequenc y median = 14.5 + (19.5 – 14.5) Median position Cumulative frequency to lcb uc b lcb 9 14.519.5 6 median 44 15 – 19 class Linear interpolation: Assume 15 values are evenly distributed in class lies in class 15 - 19

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value Lower class boundary (lcb) class frequenc y median = 14.5 + Frequency in class up to median 96 median 44 15 – 19 class 14.519.5 Linear interpolation: Assume 15 values are evenly distributed in class lies in class 15 - 19 (19.5 – 14.5) uc b lcb

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value Lower class boundary (lcb) class frequenc y median = 14.5 +. (5) Frequency in class up to median class width 9 5 6 median 44 15 – 19 class 14.519.5 Linear interpolation: Assume 15 values are evenly distributed in class lies in class 15 - 19

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Example – G rouped data Waiting times1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value median = 14.5 + 9 2 6 median 44 15 – 19 class 14.519.5 Linear interpolation: Assume 15 values are evenly distributed in class 2 3 lies in class 15 - 19

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The median is the middle value: n/2 = 100/2 = 50 th value median = 14.5 + 96 median = 16.5 44 15 – 19 class 14.519.5 Linear interpolation: Assume 15 values are evenly distributed in class 2= 16.5 minutes lies in class 15 - 19

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q1 value is the 1/4 th value: n/4 = 100/4 = 25 th value Q1 lies in class 10 - 14

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q1 value is the 1/4 th value: n/4 = 100/4 = 25 th value Lower class boundary (lcb) class frequenc y Q1 = 9.5 + (14.5 – 9.5) Q1 position Cumulative frequency to lcb uc b lcb 19 9.514.5 1 Q1 24 10 – 14 class Q1 lies in class 10 - 14

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q1 value is the 1/4 th value: n/4 = 100/4 = 25 th value Q1 = 9.5 +. (5) 19 9.514.5 1 Q1 24 10 – 14 class Q1 lies in class 10 - 14 = 9.75 Lower class boundary (lcb) class frequenc y Frequency in class up to Q1 class width

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q1 value is the 1/4 th value: n/4 = 100/4 = 25 th value Q1 = 9.5 +. (5) 19 9.514.5 1 Q1 24 10 – 14 class Q1 lies in class 10 - 14 = 9.75

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q3 value is the 3/4 th value: n/4 = 100/4 = 75 th value Q3 lies in class 20 - 29 Q1 = 9.75

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q3 value is the 3/4 th value: n/4 = 100/4 = 75 th value Lower class boundary (lcb) class frequenc y Q3 = 19.5 + (19.5 – 29.5) Q3 position Cumulative frequency to lcb uc b lcb 4 19.529.5 16 Q3 59 20 – 29 class Q3 lies in class 20 - 29 Q1 = 9.75

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times The Q3 value is the 3/4 th value: n/4 = 100/4 = 75 th value Lower class boundary (lcb) Q3 = 19.5 +. (10) 4 19.529.5 16 Q3 59 20 – 29 class Q3 lies in class 20 - 29 Q1 = 9.75 class frequenc y Frequency in class up to Q3 class width = 27.5

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Example – G rouped data Waiting times 1 - 45 - 910 - 1415 - 1920 - 2930 - 3940 - 4950 + frequency91520152011100 Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded: Answer: Estimate the median and interquartile range of waiting times Q1 = 9.75Q3 = 27.5 IQR = Q3 – Q1 = 27.5 – 9.75 = 17.75 minutes

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G rouped data - summary Quantile = lcb +.(ucb – lcb) = lcb +. class width Use linear interpolation to estimate quantile. Treat data as continuous within each group / class Assumes values are evenly distributed within each class. rest of class freq lcbucb freq in class up to quantile Quantile cum. freq. to lcb class

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