Mean, Median, Mode and Range

Count how often each value occurs.

Subtract the smallest value from the largest.

The list is ordered. Find the value with two values on each side.

Order the numbers before finding the middle.

Add all three values and divide by three.

Find the two middle values and average them.

1. The two middle values are 5 and 7.
2. Median =(5+7)\div2=6.

The pupil who read no books still counts as one pupil.

1. Total books =0+2+3+7=12.
2. Mean =12\div4=3 books.

Look for a tie for the highest frequency.

1. 4 and 6 each occur twice. Every other value occurs once, so the modes are 4 and 6.

A mode needs to occur more often than the other values.

1. No value repeats. There is no mode; 6.5 would be the median, not the mode.

Find the distance from the lowest temperature to the highest.

1. The highest is 5 °C and the lowest is -3 °C.
2. Range =5-(-3)=8 °C.

Put the times in order for the median; add them for the mean.

Part (a)

1. In order: 17,\ 18,\ 19,\ 21,\ 25. The middle time is 19 seconds.

Part (b)

1. Total =18+21+17+19+25=100.
2. Mean =100\div5=20 seconds.

Use the mean to find the total of all six scores.

1. Required total =6\times8=48.
2. Known total =5+7+8+9+10=39.
3. Missing score =48-39=9.

Recover the original total before adding the fifth bag.

1. Original total mass =4\times7=28 kg.
2. New total =28+12=40 kg.
3. There are now five bags. New mean =40\div5=8 kg.

Use the median for a typical time and the range for consistency.

1. B has the smaller median: 21<24, so its typical journey is shorter.
2. A has the smaller range: 5<11, so its journey times are more consistent.

For six values, the median is the mean of the third and fourth values.

1. The middle values are x and 10. Their total is 9\times2=18.
2. So x=18-10=8. This fits between 6 and 10.

Find each group total before combining the groups.

1. First total =4\times12=48. Second total =6\times7=42.
2. Combined total =48+42=90 for 4+6=10 pupils.
3. Combined mean =90\div10=9. Averaging 12 and 7 would wrongly give the groups equal weight.

Use the old mean to find the old total, then subtract the removed number.

1. Original total =5\times14=70.
2. Remaining total =70-22=48.
3. New mean =48\div4=12.

Find the two averages, then consider the unusually long task.

Part (a)

1. Total =6+7+8+9+30=60.
2. Mean =60\div5=12 minutes.

Part (b)

1. The ordered list has middle value 8. Four of the five times are between 6 and 9 minutes.
2. The unusually long task makes the mean 12 minutes. The median of 8 minutes better represents a typical task in this set.

The range gives the largest number. The mean gives the total.

1. The smallest number is 2, so y=2+9=11.
2. The total must be 5\times5=25.
3. Then x=25-(2+4+4+11)=4. The ordered list is 2,4,4,4,11, so the missing values are 4 and 11.

Recover the incorrect total, replace the copied value, then divide again.

1. The incorrect total was 16\times8=128.
2. Replace 29 with 13: correct total =128-29+13=112.
3. The number of values stays 8. Correct mean =112\div8=14.