Statistics

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high importance~3 Q in Tier 128 formulas⚡ 10 shortcuts5 subtopics

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Mean, weighted mean and combined mean

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⏱ 4 min read🧩 5 question types🎯 16 practice Q

The idea in one minute

The average (mean) spreads a total equally: average = total ÷ count.

Almost every exam question hides the total. Rebuild it first with total = average $\times$ count, and the question becomes plain addition and subtraction.

The one habit that solves most questions

An average is just a total shared equally.

\text{average} = \frac{\text{total}}{\text{count}}, \qquad \text{total} = \text{average} \times \text{count}

Five bats score 40, 55, 35, 60, 50. Total = 240, average = $240 \div 5 = 48$ .

Rule: Always write the total first: average × count. Nine of ten average questions are solved by that one line.

Finding a missing value

The average fixes the total. The total reveals the missing entry.

The average of five numbers is 27, so the total is $5 \times 27 = 135$ . Four of them are 24, 31, 19 and 28, adding to 102. The fifth number is $135 - 102 = 33$ .

Example: A cricketer's average over 8 innings is 45. Runs in the 9th innings that lift the average to 50: $9 \times 50 - 8 \times 45 = 450 - 360 = 90$ .

Someone joins or leaves

When a new member joins a group, the total grows by that member's value and the count grows by 1.

24 students average 35 kg, so the class weighs $24 \times 35 = 840$ kg. A teacher of weight $w$ joins and the average becomes 35.5 kg:

840 + w = 25 \times 35.5 = 887.5 \Rightarrow w = 47.5 \text{ kg}

Leaving works the same way: subtract the member, then divide by the smaller count.

Tip: Check the direction of the pull. If the average rose, the newcomer is above the new average — here 47.5 > 35.5.

Combining two groups

Two classes with different sizes cannot be averaged by adding the two averages and halving.

\text{combined average} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}

Class A: 30 students, mean 62. Class B: 20 students, mean 58. Combined $= \dfrac{30 \times 62 + 20 \times 58}{50} = \dfrac{3020}{50} = 60.4$ .

Watch: The combined average must sit between 62 and 58, nearer the bigger class. 60.4 does; 60 (the plain midpoint) is the planted option.

A wrong entry is corrected

A number was misread. Undo the wrong entry and add the right one.

The average of 30 numbers is 20, so the recorded total is 600. The entry 31 should have been 13. Corrected total $= 600 - 31 + 13 = 582$ , and the correct average $= 582 \div 30 = 19.4$ .

Tip: Only the difference between the wrong and right entries matters: $-31 + 13 = -18$ , and $-18 \div 30 = -0.6$ , so $20 - 0.6 = 19.4$ .

Averages by deviations

Adding big numbers is slow. Pick a round base near the values and track the differences.

Values 44, 46, 43, 47 around 45: deviations $-1, +1, -2, +2$ add to 0. The average is exactly 45.

\text{average} = \text{base} + \frac{\text{sum of deviations}}{\text{count}}

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Mean and a missing value

How to spot it:

The average of a list is given with all values but one; the missing value is asked.

x = n\bar{x} - \sum \text{known}

Method

Rebuild the total: average $\times$ count.
Add the known values.
Subtract that sum from the total.

Why it works:

The average pins the total, and the total pins the missing entry.

Try this

The average of five numbers is 27. Four of the numbers are 24, 31, 19 and 28. The fifth number is:

Show solution

Total $= 5 \times 27 = 135$ .
Known sum $= 24 + 31 + 19 + 28 = 102$ .
Fifth number $= 135 - 102 = 33$ .

Answer

Type 2very common3 practice Q

A member joins or leaves

How to spot it:

A teacher joins a class, a player retires, one reading is dropped; the average shifts.

w = (n+1)\bar{x}_{new} - n\bar{x}_{old}

Method

Old total $=$ old average $\times$ old count.
New total $=$ new average $\times$ new count.
The new member equals the difference.

Why it works:

The change in the total can only come from the person who joined or left.

Try this

The average weight of 24 students is 35 kg. When a teacher joins them, the average weight becomes 35.5 kg. The weight of the teacher is:

Show solution

Old total $= 24 \times 35 = 840$ kg.
New total $= 25 \times 35.5 = 887.5$ kg.
Teacher $= 887.5 - 840 = 47.5$ kg.

Answer

47.5 kg

Type 3common2 practice Q

Combined average of two groups

How to spot it:

Two sections, teams or batches with their own sizes and averages; one overall average asked.

\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}

Method

Multiply each group's size by its average.
Add the two products.
Divide by the total size.

Why it works:

Totals add; counts add; the average is their ratio.

Try this

Section A has 30 students with mean marks 62. Section B has 20 students with mean marks 58. The mean marks of all 50 students is:

Show solution

Total marks $= 30 \times 62 + 20 \times 58 = 1860 + 1160 = 3020$ .
Total students $= 50$ .
Mean $= 3020 \div 50 = 60.4$ .

Answer

60.4

Type 4common2 practice Q

Correcting a misread value

How to spot it:

A value was copied wrongly; the correct average is asked.

\bar{x}_{new} = \frac{n\bar{x}_{old} - \text{wrong} + \text{right}}{n}

Method

Write the wrong total: average $\times$ count.
Remove the wrong entry, add the correct one.
Divide by the count.

Why it works:

Only the difference between the two entries changes the total.

Try this

The average of 30 numbers was computed as 20. Later it was found that 31 had been misread as 13. The correct average is:

Show solution

Wrong total $= 30 \times 20 = 600$ .
Corrected total $= 600 - 31 + 13 = 582$ .
Correct average $= 582 \div 30 = 19.4$ .

Answer

19.4

Type 5common2 practice Q

Raising the average

How to spot it:

How much must the next score/inning earn to reach a target average.

x = (n+1)\bar{x}_{target} - n\bar{x}_{old}

Method

Total needed at the target average with one extra count.
Subtract the current total.
The difference is the required score.

Why it works:

The next value must cover its own share plus the lift on all earlier values.

Try this

A batsman's average in 8 innings is 45 runs. How many runs in the 9th innings will raise his average to 50?

Show solution

Needed total $= 9 \times 50 = 450$ .
Current total $= 8 \times 45 = 360$ .
Required runs $= 450 - 360 = 90$ .

Answer

90 runs

Formula sheet

Average

\bar{x} = \frac{\text{sum of values}}{\text{count}}

Total

\text{total} = \bar{x} \times n

Combined average

\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2}

Missing value

x = n\bar{x} - \sum(\text{known values})

New member

w = (n+1)\bar{x}_{new} - n\bar{x}_{old}

Use the same pattern for a leaving member with n-1.

Shortcuts that save time

⚡ Deviations from a round base

Pick a round number near the values. Add the small differences and divide by the count. The base can be anything.

Example

Find the average of 44, 46, 43 and 47.

Show solution

Base 45: deviations $-1, +1, -2, +2$ .
Sum of deviations $= 0$ .
Average $= 45 + 0 \div 4 = 45$ .

Answer

⚡ Watch where the average is pulled

The combined average always lies between the two group averages, closer to the bigger group. Use this to reject impossible options in seconds.

Example

Group P has 60 people averaging 40, group Q has 40 people averaging 60. The combined average is:

Show solution

Between 40 and 60, nearer 40 (P is bigger).
$\dfrac{60 \times 40 + 40 \times 60}{100} = \dfrac{4800}{100} = 48$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Averaging two group averages without weighting.

Multiply each average by its group size first; only sizes equal allow a plain midpoint.

Mistake 02

Dividing the old total by the new count after someone joins.

Add the newcomer's value to the total, then divide by the new count.

Mistake 03

Subtracting the correct entry when fixing a misread.

Remove the wrong entry and add the correct one.

Mistake 04

Forgetting to convert units before averaging (cm and m mixed).

Bring every value to one unit first.

Mistake 05

Solving for a missing value without writing the total.

Total $= n\bar{x}$ first; the missing value is total minus the known sum.

Quick revision

Read this the night before the exam.

average $=$ total $\div$ count; total $=$ average $\times$ count.
Missing value $= n\bar{x} -$ sum of known values.
New member: $w = (n+1)\bar{x}_{new} - n\bar{x}_{old}$ .
Combined: $\dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2}$ , between the two, near the bigger.
Misread entry: swap wrong for right, then divide by the count.
Deviations: base plus (sum of deviations ÷ count).

Practice: 16 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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