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Sums of observations: X1 + X2 + ... + Xn versus nX

Edexcel IAL Mathematics · WST03 Statistics 3

The most common S3 error is confusing the total of n separate observations with n times one observation. They have the same mean but very different variances.

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n independent observations

If X1, X2, ..., Xn are independent observations of X ~ N(μ, σ²), then

X1 + X2 + ... + Xn ~ N(nμ, nσ²).

Each observation brings its own variance σ², so the variances add n times.

n times one observation

nX ~ N(nμ, n²σ²), because Var(nX) = n^2Var(X).

This describes one value scaled up, for example one plank cut and its length multiplied by 6. It is much more variable than the sum of 6 separate planks, because in a sum high and low values partly cancel.

Apples, A ~ N(150, 10²) g	Mean	Variance
Bag of 6 apples: A1 + ... + A6	900	600
One apple, mass × 6: 6A	900	3600
Mean of 6 apples: (A1 + ... + A6)/6	150	100/6

Deciding which to use

Ask: are there n different items, each with its own random value? Then use a sum. Is one random value being multiplied? Then use nX. Examiners report confusion between nσ² and n²σ² (for example 3 × 0.3 against 3² × 0.3); write out the variance calculation each time.

Totals can include several types: 6 apples in a box B ~ N(50, 4²) gives a total ~ N(900 + 50, 600 + 16).

Worked example

Apples have mass A ~ N(150, 10²) grams and a box has mass B ~ N(50, 4²) grams. Find the probability that a box of 6 apples weighs more than 1000 g.

Total T = A1 + ... + A6 + B.
E(T) = 6 × 150 + 50 = 950; Var(T) = 6 × 100 + 16 = 616.
z = 50/sqrt(616) = 2.0146
P(Z > 2.01) = 1 - 0.9778

Answer: P = 0.0222 (tables)

Key idea: Sum of n separate observations: variance nσ². One observation times n: variance n²σ².

Check yourself

Apples have mass A ~ N(150, 10²) g. Find the variance of the total mass of 6 randomly chosen apples.
Show answer
600. 6 × 10² = 600.
Apples have mass A ~ N(150, 10²) g. One apple is weighed and its mass is multiplied by 6. Find the variance of this quantity.
Show answer
3600. Var(6A) = 6² × 100 = 3600.
Apples have mass A ~ N(150, 10²) g. A bag holds 6 randomly chosen apples. Find the probability that the bag's apples weigh less than 870 g in total. Give your answer to 4 decimal places.
Show answer
0.1103. T ~ N(900, 600); z = -30/sqrt(600) = -1.2247; P = 0.1112 from tables.
Apples have mass A ~ N(150, 10²) g. Find P(6A > 920), where 6A is six times the mass of one apple. Give your answer to 4 decimal places.
Show answer
0.3694. 6A ~ N(900, 3600); z = 20/60 = 0.3333; P = 0.3707.
Apples have mass A ~ N(150, 10²) g and boxes B ~ N(50, 4²) g, all independent. Find the probability that a box with 6 apples in it weighs more than 1000 g. Give your answer to 4 decimal places.
Show answer
0.0220. Total ~ N(950, 616); z = 2.0146; P = 0.0222.
Apples have mass A ~ N(150, 10²) g. Find the probability that the mean mass of 6 randomly chosen apples is less than 145 g. Give your answer to 4 decimal places.
Show answer
0.1103. Mean ~ N(150, 100/6); z = -5/sqrt(100/6) = -1.2247; P = 0.1112.
A carpenter cuts one length L ~ N(2, 0.01) metres and uses it as a template for 4 identical pieces laid end to end. Which describes the total length?
- 4L ~ N(8, 0.16)
- L1 + L2 + L3 + L4 ~ N(8, 0.04)
- 4L ~ N(8, 0.04)
- 4L ~ N(2, 0.16)
Show answer
4L ~ N(8, 0.16). One random length is copied 4 times, so it is 4L with variance 16 × 0.01 = 0.16.

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Last updated 8 November 2026. Spotted a mistake? Tell us.