Edexcel Mathematics Higher Tier, November 2009 (1380/4H) (Paper 4, calculator)


Question 18 a) There are 100 students so to find the median we would usually add one and divide by two. This would give ((100 + 1)/2 = 50.5). However as we have an even number it will be good enough to just divide by two. Half of 100 is 50 so we need to see where the 50th piece of data (50th student) lies. By the time we have got to 130 cm we have seen 8 students. By the time we have got to 140 cm we have seen 24 (8+16) students. By the time we have got to 150 cm we have seen 49 (8 + 16 + 25) students. The next category (150 – 160) covers the 50th student to the 79th student so this is where our median lies.


This is also demonstrated using the following cumulative frequency table: Height (h cm)


Frequency


120h 130 8 130h 140 16 140h 150 25 150h 160 30 160h 170 21


Cumulative Frequency 8


24 49 79


100 Here we can see that the 50th student and therefore the median is the in category 140  h  150


b) Height (h cm)


Frequency


120h 130 8 130h 140 16 140h 150 25 150h 160 30 160h 170 21 Total


Midpoint 125


135 145 155 165


100


Frequency x midpoint 1000


2160 3625 4650 3465


14900


Now to find the mean we divide the total of (frequency x midpoint) by the total frequency 14900 ÷ 100 = 149


www.chattertontuition.co.uk


0775 950 1629


Page 8


Page 1  |  Page 2  |  Page 3  |  Page 4  |  Page 5  |  Page 6  |  Page 7  |  Page 8  |  Page 9  |  Page 10  |  Page 11  |  Page 12  |  Page 13  |  Page 14  |  Page 15