Changing data and finding unknowns

Adding the same number to every value

The data \(3,5,6,8,8\) have mean \(6\). Adding \(3\) to every value gives \(6,8,9,11,11\), with mean \(9\). Every value moves \(3\) to the right, and so does the mean, so each distance from the mean is unchanged.

Here both sets have standard deviation \(\sqrt{\dfrac{9+1+0+4+4}5}=\sqrt{3.6}\approx1.90\).

Correcting one value

Changing one value is different: recalculate \(\sum x\) and \(\sum x^2\). Ten values have \(\sum x=120\) and \(\sum x^2=1500\), so the mean is \(12\) and the standard deviation is \(\sqrt{150-144}=\sqrt6\approx2.45\). A value recorded as \(18\) should be \(8\). Then \(\sum x=120-18+8=110\) and \(\sum x^2=1500-18^2+8^2=1240\); the mean is \(11\) and the standard deviation is \(\sqrt{124-121}=\sqrt3\approx1.73\).

Finding unknown values

1. Use the mean: \(\sum x=n\bar x\) (or \(\sum fx=\bar x\sum f\) for a frequency table).
2. Use the standard deviation: \(\sum x^2=n(\sigma^2+\bar x^2)\).
3. Solve, then check by substitution. Frequencies must be whole numbers that are not negative.

The values \(3,7,a,b\) have mean \(6\) and standard deviation \(\sqrt5\). Then \(3+7+a+b=24\), so \(a+b=14\). Also \(9+49+a^2+b^2=4(5+36)=164\), so \(a^2+b^2=106\). With \(b=14-a\): \(a^2+(14-a)^2=106\), \(2a^2-28a+90=0\), \(a^2-14a+45=0\), \((a-5)(a-9)=0\). The two unknown values are \(5\) and \(9\). Check: \(\dfrac{9+49+25+81}4-36=41-36=5\).

Common errors

• Adding a constant does not change the standard deviation; do not add it to the standard deviation as well.
• Correcting one value changes \(\sum x\) and \(\sum x^2\); it is not a shift of every value.
• \(\sum x^2\) is the sum of the squares, not the square of \(\sum x\).