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Statistics & correlation · Lesson 1 of 3

Mean and standard deviation by the assumed mean

The assumed mean method for the mean, variance and standard deviation of lists and grouped tables, and finding missing values from a given mean or standard deviation.

18 minYou should already know: Dispersion & cumulative frequency Statistics: data & averages
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In General Maths you found the mean, variance and standard deviation of lists and frequency tables (see measures of spread). Further Maths questions often add “using an assumed mean of …”: a way to keep the numbers small.

The assumed mean

Choose a value AA near the middle of the data, and work with the deviations d=x−Ad = x - A instead of the numbers themselves. The mean moves back by AA:

xd = x − 5548−753−2550594627Σd = 2, so mean = 55 + 2 ÷ 5 = 55.4small deviations are easier to add than the numbers
Deviations from an assumed meanMean = A + Σd ÷ n
xˉ=A+∑fd∑fσ=∑fd2∑f−(∑fd∑f)2\begin{gathered} \bar x = A + \frac{\sum fd}{\sum f} \\ \sigma = \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2} \end{gathered}

Subtracting AA from every value doesn’t change the spread, so the standard deviation of the dd‘s is the standard deviation of the xx‘s. For grouped data, xx is the class mark (the middle of the class).

Choose an assumed meanPick any class mark as A
classxfd = x − 22fdfd²
10–14124−10−40400
15–19179−5−45225
20–242212000
25–29277535175
30–343231030300
total35−201100
21.429mean = 22 + −20 ÷ 355.577standard deviation
With A = 22: Σfd = −20 and Σfd² = 1100. Mean = 22 + −20 ÷ 35 = 21.429; standard deviation = √(1100 ÷ 35 − (−20 ÷ 35)²) = 5.577. Try another A: the answers don't change.

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q6

Age (years) 12–14 15–17 18–20 21–23 24–26
Frequency 6 10 3 2 1

The table shows the distribution of ages of 22 students in a school. Using an assumed mean of 19, calculate, correct to three significant figures, the:

mean age;

standard deviation of the distribution.

  1. The table

    • Class marks: 13, 16, 19, 22, 25{13,\ 16,\ 19,\ 22,\ 25}; frequencies 6, 10, 3, 2, 1{6,\ 10,\ 3,\ 2,\ 1}.
    • d=x−19{d = x - 19}: −6, −3, 0, 3, 6{-6,\ -3,\ 0,\ 3,\ 6}.
    • fd{fd}: −36, −30, 0, 6, 6{-36,\ -30,\ 0,\ 6,\ 6}, so ∑fd=−54{\sum fd = -54}.
    • fd2{fd^2}: 216, 90, 0, 18, 36{216,\ 90,\ 0,\ 18,\ 36}, so ∑fd2=360{\sum fd^2 = 360}.

    Think first. Class marks 13, 16, 19, 22, 25. What is d = x − 19 for each?

  2. The mean

    • xˉ=19+−5422=19−2.4545=16.5{\bar x = 19 + \frac{-54}{22} = 19 - 2.4545 = 16.5} (3 s.f.).

    Think first. A + Σfd ÷ Σf, with Σf = 22.

  3. The standard deviation

    • ∑fd2∑f=36022=16.3636{\frac{\sum fd^2}{\sum f} = \frac{360}{22} = 16.3636}.
    • (∑fd∑f)2=2.45452=6.0248{\left(\frac{\sum fd}{\sum f}\right)^2 = 2.4545^2 = 6.0248}.
    • σ=16.3636−6.0248{\sigma = \sqrt{16.3636 - 6.0248}}.
    • =10.3388≈3.22{= \sqrt{10.3388} \approx 3.22}.

More: the assumed mean

Mean deviation

The mean deviation is the average distance from the mean, ignoring signs: ∑f∣x−xˉ∣∑f\frac{\sum f|x - \bar x|}{\sum f}. Find the mean first, then each distance.

More: mean deviation

Missing values from the mean or standard deviation

Turn each given fact into an equation:

  • a mean gives the total: ∑x=nxˉ\sum x = n\bar x;
  • a standard deviation gives the total of squares: ∑x2=n(σ2+xˉ2)\sum x^2 = n(\sigma^2 + \bar x^2).

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q12

The mean and the standard deviation of marks scored by eight people in an interview are 7 and 3\sqrt3 respectively. If six of the marks are 6, 7, 9, 5, 5 and 6, find the two remaining marks.

  1. Use the mean

    • The total of all 8 marks is 8×7=56{8 \times 7 = 56}.
    • The six known marks add to 38{38}, so the two missing ones add to a+b=18{a + b = 18}.

    Think first. 8 marks with mean 7. What is their total?

  2. Use the standard deviation

    • 3=∑x28−49{3 = \frac{\sum x^2}{8} - 49}, so ∑x2=8×52=416{\sum x^2 = 8 \times 52 = 416}.
    • The six known squares add to 252{252}, so a2+b2=164{a^2 + b^2 = 164}.

    Think first. σ² = Σx²/8 − 7². What is Σx²?

  3. Solve

    • a2+(18−a)2=164{a^2 + (18 - a)^2 = 164}, so 2a2−36a+160=0{2a^2 - 36a + 160 = 0}.
    • Divide by 2: a2−18a+80=0{a^2 - 18a + 80 = 0}, so (a−8)(a−10)=0{(a - 8)(a - 10) = 0}.
    • The two marks are 8{8} and 10{10}.

    Think first. Substitute b = 18 − a.

More: missing values

Your turn

WAEC 2016 · Paper 2 · Q6

The mean of the numbers 1,4,k,(k+4)1, 4, k, (k + 4) and 1111 is (k+1)(k + 1). Calculate the:

  1. (a)

    value of kk;

  2. (b)

    standard deviation.

Worked solution (try it first)

(a)

  1. The five numbers add to 2k+202k + 20, so their mean is 2k+205=k+1\dfrac{2k + 20}{5} = k + 1.
  2. Multiply by 5: 2k+20=5k+52k + 20 = 5k + 5, so 3k=153k = 15 and k=5k = 5.

(b)

  1. The numbers are 1,4,5,9,111, 4, 5, 9, 11, with mean 6.
  2. The deviations are −5,−2,−1,3,5-5, -2, -1, 3, 5.
  3. ∑d2=25+4+1+9+25=64\sum d^2 = 25 + 4 + 1 + 9 + 25 = 64, so σ=645\sigma = \sqrt{\dfrac{64}{5}}
    =85= \dfrac{8}{\sqrt5}
    =855= \dfrac{8\sqrt5}{5}
    ≈3.578\approx 3.578.

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