Upcoming SlideShare
×

# FCA GCSE MMMR From Frequency Tables (Linear)

1,269 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,269
On SlideShare
0
From Embeds
0
Number of Embeds
214
Actions
Shares
0
8
0
Likes
0
Embeds 0
No embeds

No notes for slide

### FCA GCSE MMMR From Frequency Tables (Linear)

1. 1. Grade C Mean, Median, Mode and Range from Frequency Tables
2. 2. Dr Payne is trialling a new painkiller.350 patients take the tablets when they have a headacheThey record how long the tablet works for, to the nearest hour. Score Frequency 0 42 1 42 2 54 3 99 4 64 5 28 6 21 Total: 350
3. 3. The MODE is the score with the highest frequencyDon’t write down the value of the frequency.Here the mode is 3 Score Frequency 0 42 1 42 2 54 3 99 4 64 5 28 6 21 Total: 350
4. 4. The RANGE is the biggest score – the smallest scoreHere the range is 6 – 0 = 6 Score Frequency 0 42 1 42 2 54 3 99 4 64 5 28 6 21 Total: 350
5. 5. The MEDIAN is the middle value.For this we need an additional column for the cumulative frequency. This is the ongoingtotals of the frequencies and should total the same as the frequency total.To locate the position of the middle value work out (n+1) where n is the frequency total. 2 Score Frequency Cumulative Here, the position of the frequency middle is: 0 42 42 350+1 = 175.5 1 42 84 2 2 54 138 Look down the cumulative 3 99 237 frequency column for the 4 64 301 smallest value bigger than 175.5 and read across to 5 28 329 the score. 6 21 350 Total: 350 The median here is 3.
6. 6. To calculate the MEAN we need an additional column.For each row work out the value of the score x the frequencyThe mean is the Total of the score x frequency divided by the Total frequencyHere the mean is 969÷ 350 = 2.768571429Round your answer to a suitable number of decimal places (i.e 2.8) Score Frequency Score x frequency 0 42 0 1 42 42 2 54 108 3 99 297 4 64 256 5 28 140 6 21 126 Total: 350 969
7. 7. A survey is taken at a night club to ask people their age last birthday.The data was put into categories called class marks (grouped data) Ages Frequency 16-18 56 19-21 64 22-25 92 26-29 12 Total: 224
8. 8. The MEDIAN will be a group and is found the same way as before.Look for the group with the highest frequency.Here the median group is 22-25 Ages Frequency 16-18 56 19-21 64 22-25 92 26-29 12 Total: 224
9. 9. The RANGE will have two answers – an upper and lower bound.The upper bound range is the biggest value – the smallest valueThe lower bound range is the smallest value of the biggest group – the biggest value of thesmallest group. Ages Frequency Here the upper bound range is 29 – 16 = 13 16-18 56 The lower bound range is 26 – 18 = 8 19-21 64 22-25 92 26-29 12 Total: 224
10. 10. The MEDIAN is given as a group and is calculated in the same way as before.Work out the cumulative frequencyFind the position of the middle by (n+1) 2Read off the group that is in the middle position Ages Frequency Cumulative Here the position of the middle frequency is found by 224+1 = 112.5 16-18 56 56 2 19-21 64 120 Look down the cumulative 22-25 92 212 frequency column for the 26-29 12 224 smallest value bigger than 112.5 and read across to the group Total: 224 The median is the 19-21 group
11. 11. The MEAN cannot be calculated exactly as the data is grouped. So we can only work out anestimate for the mean.This is done in a similar way as before however, we first must work out the midpoint of eachgroup.To do this add together the smallest and largest value in each group and divide by 2Now multiply together the midpoint values with the corresponding frequency values.The mean estimate is the Total of the frequency x midpoint divided by the Total of the frequency.Here the mean is 4724 ÷ 224 =21.08928571Which to a suitable degree of accuracy is 21.1 Ages Frequency Midpoint Frequency x Midpoint 16-18 56 17 952 19-21 64 20 1280 22-25 92 23.5 2162 26-29 12 27.5 330 Total: 224 4724
12. 12. Exam Question
13. 13. Exam Question Money x Frequency 0 32 24 40 30 Total = 30 Total = 126Mean = 126 ÷ 30 = 4.2 4.20
14. 14. Exam Question
15. 15. Exam Question Midpoint Midpoint x frequency 65 780 75 1650 85 1955 95 2280 105 1995 Total = 100 Total = 8660 90 < t < 1008660 ÷ 100 = 86.6 86.6
16. 16. Exam Question Cumulative frequency Midpoint Midpoint x Frequency 20 5 100 37 15 255 49 25 300 81 35 1120 106 45 1125 Total = 106 Total = 2900 30 < t ≤ 40 (106+1) = 53.5 22900 ÷ 106 = 27.35849057 27.4