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The projected population of the U.K. for 2005 ( by age ) Age (yrs) The median age is estimated as the age corresponding to a cumulative frequency of 30 million. The median age is 39 years ( Half the population of the U.K. will be over 39 in 2005. ) e.g. The projected population of the U.K. for 2005, by age: Source: USA IDB 9060280 – 89 8058470 – 79 7054660 – 69 6048850 – 59 5040940 – 49 4031930 – 39 3022720 – 29 2015810 – 19 10770 – 9 ( yrs )(m) ( yrs ) u.c.b.Cu.ffAGE Median of Discrete Data

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The projected population of the U.K. for 2005 ( by age ) Age (yrs) The quartiles are found similarly: lower quartile: 20 years upper quartile: 56 years e.g. The projected population of the U.K. for 2005, by age: The projected population of the U.K. for 2005 ( by age ) Source: USA IDB 9060280 – 89 8058470 – 79 7054660 – 69 6048850 – 59 5040940 – 49 4031930 – 39 3022720 – 29 2015810 – 19 10770 – 9 ( yrs )(m) ( yrs ) u.c.b.Cu.ffAGE The interquartile range is 36 years LQ = (n+1)th item of data UQ = (n+1)th item of data

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Median of Continuous Data This is the value which occurs half way so the probability that a value is less than the median is P(x m) = so Area under y = f(x) between ‘a’ and ‘m’ is where ‘a‘ is the lower boundary of the P.D.F and ‘m‘ is the median Quartiles : these are the values which divide the P.D.F up into and ranges. P(x < LQ) = 0.25P(x < UQ) = 0.75 So find the area as in finding the median.

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Ex1 p.d.ff(x) = x 0 < x < 4 Median divides the area in If x = m then f(x) = m Area = m m = 0.5As P(x < m) = 0.5 Area = base ht m 2 = 0.5m 2 = 8m = 8

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For a curve integration is again used Ex2 Find the median Area = As area = up to the median Area = A m = A 0 = 0 Area = A m - A 0 = - 0 = 0.5 as area = 0.5 up to the median Solving = 0.5m = = 1.587 m

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