Frequency Tables

A crossed group contains five marks.

Count complete groups of five, then any extra marks.

Add the frequencies, not the category names.

Find the category with the largest frequency.

The three frequencies must add to 32.

Count how many entries equal 2; do not add them.

1. The value 2 appears four times, so its frequency is 4.

Compare the two relevant frequencies.

1. Art has 18 pupils and music has 11. Difference =18-11=7 pupils.

Check which class interval includes 10.

1. The class interval 10≤t<20 includes 10. The interval 0≤t<10 does not include 10, because t must be less than 10.

Include 10 but exclude 20.

1. The times in this class interval are 10, 16, 18, 10 and 14. The frequency is 5.

A possible value can have no observations.

1. There are no weeks with 3 absences, so the frequency is 0. Do not leave it as an unknown value.

Work systematically through the list and check the total.

Part (a)

1. Water appears 5 times.

Part (b)

1. Juice appears 4 times.

Part (c)

1. Milk appears 3 times. The total check is 5+4+3=12.

Use the chess frequency over the total number of pupils.

1. Total =9+7+8=24.
2. Probability of chess =9/24=3/8.

At least 20 means 20 or more.

1. Use the last two classes: 11+6=17 journeys.

Count each tally separately and compare it with the printed frequency.

1. Red has 6 marks and blue has 10, matching their frequencies. Green has 9 marks but its frequency is written as 8.

Test the boundary signs carefully.

1. The class interval 10≤m<15 includes 10 but not 15.
2. A mass of 5 kg is not in 0≤m<5, because m must be less than 5. It goes only in 5≤m<10.

Use the tennis count to recover the total number of pupils.

1. Tennis represents 40% of the total, so total =24÷0.4=60.
2. Football frequency =60-24-14=22.

For each class, use the rows for 2 and 3 books.

1. Class A: 6+3=9 pupils. Class B: 9+5=14 pupils.
2. There are 9+14=23 pupils who borrowed at least 2 books. The class totals are 20 and 22, so the required fraction is 23/42.

Move one observation from the wrong class to the correct class.

1. The count for 10≤t<20 rises from 9 to 10. The count for 20≤t<30 falls from 6 to 5.
2. Corrected count below 20 is 5+10=15. Percentage =15÷20×100=75\%.

Distinguish a class boundary from a value inside a class.

Part (a)

1. The first two classes contain 5+7=12 ribbons, all shorter than 30 cm.

Part (b)

1. Some of the 7 ribbons in 20≤l<30 could be shorter than 25 cm, but their individual lengths are not given. The exact number cannot be found.

The missing counts are not equal; use the stated relationship.

Part (a)

1. Known frequencies total 15+3=18. The two missing frequencies total 36-18=18.
2. They are in ratio 1:2, so the frequency for 0 siblings is 18÷3=6.

Part (b)

1. The frequency for 2 siblings is twice 6, so it is 12. Check: 6+15+12+3=36.