Lesson 6 of 35 · Quantitative Reasoning: arithmetic

Statistics for Quant: averages, medians and spread

Work with sums instead of averages, find medians, handle weighted and evenly spaced sets, and judge standard deviation without computing it.

About 20 minutes

In this lesson

  • Solve mean questions through the sum, including values added or removed.
  • Find a weighted average, a median and the mean of an evenly spaced set quickly.
  • Predict how changes to a set affect its standard deviation without calculating it.
  • Find the largest or smallest possible value under a fixed mean and median.

The idea

Think in sums

The mean of a list is its sum divided by the number of values, so

sum=mean×number of values\text{sum} = \text{mean} \times \text{number of values}

Almost every averages question becomes easy once you switch from means to sums. If five numbers average 18, they add to 90. If a sixth number lifts the average to 20, the six add to 120, so the new number is 120−90=30120 - 90 = 30. A second way to see the same thing: the new number must equal the new mean (20) plus enough to lift each of the five old numbers by 2, that is 20+5×2=3020 + 5 \times 2 = 30.

Weighted averages

When groups of different sizes are combined, the overall mean is a weighted average:

mean=n1m1+n2m2n1+n2\text{mean} = \frac{n_1 m_1 + n_2 m_2}{n_1 + n_2}

It lies between the two group means, closer to the larger group. The balance method from mixtures applies: the distances from the overall mean to the group means are in the inverse ratio of the group sizes. Averaging the two means directly is right only when the groups are the same size.

Median, mode and range

The median is the middle value once the list is sorted. With an even number of values it is the mean of the two middle values. The median ignores how far the extreme values are from the middle, so one huge value moves the mean but not the median. The mode is the most frequent value (a list can have several). The range is the largest value minus the smallest.

Evenly spaced sets

In a list with a constant gap between consecutive values (consecutive integers, consecutive multiples of 5, and so on), the mean equals the median, and both equal the average of the first and last values:

mean=median=first+last2\text{mean} = \text{median} = \frac{\text{first} + \text{last}}{2}

The number of values is last−firstgap+1\frac{\text{last} - \text{first}}{\text{gap}} + 1; forgetting the +1+1 is the usual off-by-one slip. The sum is the number of values times the mean, so the integers from 11 to 40 add to 30×25.5=76530 \times 25.5 = 765.

Standard deviation without computing it

Standard deviation measures how far the values typically lie from the mean. You will rarely need to compute it; you need to know how it reacts:

Change to every value Effect on the standard deviation
add or subtract the same constant none: the values move together
multiply by kk multiplied by ∣k∣\lvert k \rvert
add a new value equal to the mean decreases (unless it is already 0)
add a new value far from the mean increases

To compare two sets, compare the distances of their values from their own means. {1,5,9}\{1, 5, 9\} and {101,105,109}\{101, 105, 109\} have the same standard deviation; {1,9,17}\{1, 9, 17\} has twice as much as either.

Extremes under constraints

"What is the greatest possible value of the largest number?" asks you to make every other number as small as the conditions allow, and then let the sum do the rest. Read every condition: whether the numbers must be different, positive, integers, and where the median sits.

Worked examples

Example 1: Finding a removed value

The mean of eight numbers is 15. When one number is removed, the mean of the remaining seven is 14. What number was removed?

  1. Switch to sums: the eight numbers add to 8×15=1208 \times 15 = 120 and the seven add to 7×14=987 \times 14 = 98.

  2. The removed number is 120−98=22120 - 98 = 22.

  3. Check another way: removing a value above the mean pulls the mean down. Each of the seven remaining values "gave" 1 to cover the drop, so the removed value was 15+7×1=2215 + 7 \times 1 = 22.

Answer

The removed number is 22. Answering 1, the change in the mean, is the quick misreading.

Example 2: Mean of two classes

A class of 20 students has a mean score of 72, and a class of 30 students has a mean score of 82. What is the mean score of all 50 students?

  1. Total of scores: 20×72+30×82=1440+2460=390020 \times 72 + 30 \times 82 = 1440 + 2460 = 3900.

  2. Mean: 390050=78\frac{3900}{50} = 78.

  3. Check with the balance: 78 is 6 above 72 and 4 below 82, and 6:4=3:26 : 4 = 3 : 2, the inverse of the class sizes 20:3020 : 30.

Answer

The mean is 78. The simple average of the class means, 77, ignores that the larger class has the higher mean.

Example 3: Consecutive even integers

The sum of nine consecutive even integers is 234. What is the largest of them?

  1. The set is evenly spaced, so its mean equals its median: 2349=26\frac{234}{9} = 26, the fifth (middle) number.

  2. The largest is four steps of 2 above the middle: 26+8=3426 + 8 = 34.

  3. Check: the integers run from 18 to 34, and 18+342×9=26×9=234\frac{18 + 34}{2} \times 9 = 26 \times 9 = 234.

Answer

The largest is 34. Stepping up by 1 instead of 2 gives 30, the slip of treating even integers as consecutive integers.

Common traps

  • Averaging averages

    The mean of two groups is the simple average of their means only when the groups are the same size. Otherwise go back to sums, or use the balance, and the larger group pulls the result its way.

  • A median from an unsorted list

    The median is the middle of the sorted list. Questions often list values out of order, or add new values, so the old middle is no longer the middle. Sort first, every time.

  • Shifting a set and expecting the spread to change

    Adding the same number to every value moves the mean but leaves the standard deviation unchanged. Multiplying every value by a number scales the standard deviation by the same factor (ignoring sign).

  • Ignoring "different" or "positive"

    When you minimise the other values to make one value as large as possible, the conditions decide how small they may be. Different positive integers can be no smaller than 1, 2, 3 and so on; values above the median must exceed it if the numbers are different.

Check yourself

Answer each question before you open its answer.

Question 1: A sixth number lifts the mean

Problem Solving

The mean of five numbers is 18. When a sixth number is added, the mean of the six numbers is 20. What is the sixth number?

  1. A.20
  2. B.22
  3. C.30
  4. D.32
  5. E.120
Show the answer

Answer: C. 30

Working

Sum of five: 5×18=905 \times 18 = 90. Sum of six: 6×20=1206 \times 20 = 120. The sixth number is 120−90=30120 - 90 = 30. Check: it equals the new mean plus 2 for each of the five old numbers, 20+5×2=3020 + 5 \times 2 = 30.

Why the other options are wrong

  • A. 20 is the new mean, not the added number; adding 20 would lift the mean only to 1106≈18.3\frac{110}{6} \approx 18.3.
  • B. This adds the rise of 2 only once; the new number must supply 2 extra for each of the five original numbers.
  • D. This adds 2 for six numbers; only the five original numbers needed lifting.
  • E. 120 is the sum of all six numbers, not the sixth number.

What carries over

Turn every mean into a sum before adding or removing values.

Question 2: Largest value under a mean and median

Problem Solving

A list of five different positive integers has a mean of 12 and a median of 14. What is the greatest possible value of the largest integer in the list?

  1. A.28
  2. B.29
  3. C.30
  4. D.43
  5. E.46
Show the answer

Answer: A. 28

Working

The five integers add to 5×12=605 \times 12 = 60. In order, a<b<14<d<ea < b < 14 < d < e. To make ee as large as possible, make the others as small as allowed: a=1a = 1, b=2b = 2, d=15d = 15. Then e=60−1−2−14−15=28e = 60 - 1 - 2 - 14 - 15 = 28. Check: {1,2,14,15,28}\{1, 2, 14, 15, 28\} has sum 60, mean 12 and median 14.

Why the other options are wrong

  • B. This sets the fourth number equal to the median, 14, but the integers must be different, so it has to be at least 15.
  • C. This repeats values (1,1,14,141, 1, 14, 14), ignoring that the integers are different.
  • D. This leaves out the fourth number, which must be larger than 14 and also counts towards the sum.
  • E. This subtracts only the median from the sum of 60, as if the other three numbers could be zero.

What carries over

To maximise one value, minimise all the others within every stated condition.

Question 3: Which set is less spread out?

Problem Solving

Set SS is {10,14,18,22,26}\{10, 14, 18, 22, 26\}. Which of the following sets has a smaller standard deviation than SS?

  1. A.{15,19,23,27,31}\{15, 19, 23, 27, 31\}
  2. B.{20,28,36,44,52}\{20, 28, 36, 44, 52\}
  3. C.{6,14,18,22,30}\{6, 14, 18, 22, 30\}
  4. D.{10,14,18,18,22,26}\{10, 14, 18, 18, 22, 26\}
  5. E.{10,14,18,22,26,34}\{10, 14, 18, 22, 26, 34\}
Show the answer

Answer: D. {10,14,18,18,22,26}\{10, 14, 18, 18, 22, 26\}

Working

SS has mean 18, and its squared distances from the mean are 64,16,0,16,6464, 16, 0, 16, 64, a total of 160, so its variance is 1605=32\frac{160}{5} = 32. Adding another 18, the mean itself, keeps the mean at 18 and the total at 160 but spreads it over six values: 1606≈26.7<32\frac{160}{6} \approx 26.7 < 32. The standard deviation therefore falls. Each other option keeps or increases the spread.

Why the other options are wrong

  • A. Adding 5 to every value moves the set without changing its spread; the standard deviation is the same.
  • B. Doubling every value doubles every distance from the mean, so the standard deviation doubles.
  • C. The two outer values moved further from the mean of 18, so the spread increased.
  • E. 34 lies far from the mean, so adding it increases the spread.

What carries over

Adding a value at the mean lowers the standard deviation; shifting all values leaves it alone.

Key points

  • Sum = mean × number of values: switch to sums first.

  • A weighted mean lies closer to the larger group.

  • Sort before finding a median; in an evenly spaced set, mean = median = (first + last) ÷ 2.

  • Shifting values leaves the standard deviation alone; scaling scales it; a value at the mean lowers it.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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