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Worked solution

Find a mean by symmetry and by summation

The discrete random variable XX has distribution

xx−2-2−1-1012
P(X=x)P(X=x)0.30.10.200.10.3

Find the mean of XX.

CJC · Lecture · Example 8

Method 1: pair the symmetric masses

The probabilities match at −2-2 and 22 and at −1-1 and 11, so the distribution is symmetric about x=0x=0. Therefore the mean is 00. The Teacher note recommends this example to explain symmetry: equal masses at opposite distances give equal and opposite contributions.

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Method 2: compute every weighted contribution

Retain the source working table:

xx−2-2−1-1012
P(X=x)P(X=x)0.30.10.200.10.3
xP(X=x)xP(X=x)−0.6-0.6−0.1-0.100.10.6

E(X)=−0.6+(−0.1)+0+0.1+0.6=0E(X)=-0.6+(-0.1)+0+0.1+0.6=0.

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Common mistakes

The mean is the centre of symmetry here because paired values have equal probabilities. It is not found by selecting the largest probability (there are two modes, −2 and 2).

Read the hint