Averages from Frequency Tables

Add the frequencies, not the pet numbers.

Find the value beside the largest frequency.

A frequency of zero means that value did not occur.

Multiply each number of stickers by the number of children receiving it.

Find total goals and total matches separately.

Count the scores, then locate the middle position.

1. There are 2+4+2+1=9 scores, so find the 5th.
2. Score 1 occupies positions 1–2; score 2 occupies positions 3–6. The 5th score is 2.

Find both middle positions for the eight pupils.

1. The middle positions are the 4th and 5th.
2. The cumulative counts are 1, 3, 7 and 8. Both middle pupils have 2 siblings, so the median is 2.

Use value × frequency to find the total score.

1. Total score =1\times2+2\times4+3\times5=25.
2. Total frequency =2+4+5=11.
3. Mean =25\div11=2.2727\ldots, which is 2.3 to 1 decimal place.

Two values tie for the greatest frequency.

1. The greatest frequency is 5. It occurs for values 2 and 3, so the two modes are 2 and 3.

The two middle positions may lie in different rows.

1. There are 10 scores. The 5th score is 2 and the 6th is 3.
2. Median =(2+3)\div2=2.5.

For the mean use both totals; for the range use the ribbon lengths.

Part (a)

1. Total length =2\times4+4\times3+6\times1=26 m.
2. There are 4+3+1=8 pieces.
3. Mean =26\div8=3.25 m.

Part (b)

1. Range =6-2=4 m.

First find the number of tickets, then use the price per ticket.

1. Total tickets =0\times5+1\times8+2\times6+3\times1=23.
2. Money collected =23\times4=£92.

The frequencies must add up to 28.

1. Known frequencies total 7+9+5=21.
2. Missing frequency =28-21=7.

Add the relevant frequencies, then divide by the total number of pupils.

1. At least 3 means 3 or 4 books, so 6+4=10 pupils.
2. Total pupils =3+4+8+6+4=25.
3. Percentage =\frac{10}{25}\times100=40\%.

Find each group’s data total and use only values with positive frequency for the range.

1. Group A has total score 0\times2+1\times6+2\times2=10 and 10 pupils, giving mean 1.
2. Group B has total score 1\times3+2\times4+3\times3=20 and 10 pupils, giving mean 2.
3. A’s range is 2-0=2. B’s range is 3-1=2. B has the greater mean and the ranges are equal.

The frequencies must add up to the number of values. Then use value × frequency to find the total.

Part (a)

1. The known frequencies add to 2+4+2=8.
2. So f=12-8=4.

Part (b)

1. Total =1\times2+2\times4+3\times4+4\times2=2+8+12+8=30.
2. There are 12 values.
3. Mean =30\div12=2.5.

Combine the goal totals and the game counts.

1. Group A: total goals =0\times3+1\times5+2\times2+3\times0=9.
2. Group B: total goals =0\times1+1\times3+2\times4+3\times2=17.
3. Combined total =9+17=26 goals. There are (3+5+2+0)+(1+3+4+2)=20 games.
4. Mean =26\div20=1.3 goals.

Update the frequency of 1, then locate the two new middle positions.

1. The new frequencies are 6,3,4,2,1. There are 6+3+4+2+1=16 scores, so use positions 8 and 9.
2. Score 1 occupies positions 1–6; score 2 occupies positions 7–9.
3. Both middle scores are 2, so the new median is 2.

Replace the three incorrect contributions; the number of results stays the same.

1. Original total =0\times5+1\times4+2\times6+3\times2+4\times3=34.
2. Changing three 0s to 4s adds 3\times4=12, so the correct total is 46.
3. There are still 5+4+6+2+3=20 results.
4. Correct mean =46\div20=2.3.

Use the mean to find the total score of all 15 pupils. How much of that total do the first 10 pupils make?

1. Total of the first 10 scores =0\times1+1\times2+2\times3+3\times4=20.
2. For all 15 pupils the total must be 15\times3=45.
3. The five extra pupils scored 45-20=25 altogether.
4. Each scored 25\div5=5. Check: (20+25)\div15=45\div15=3.