Standardizing

How many standard deviations from the mean

The marks on an exam follow a normal distribution with mean $\mu =60$μ = 60 and standard deviation $\sigma =8$σ = 8. A student scores 72. That is 72 − 60 = 12 marks above the mean, and 12 marks is 12 ÷ 8 = 1.5 standard deviations.

This count of standard deviations is the z-score of the value: $z=\frac{x-\mu }{\sigma }$z = (x − μ)/σ. Subtract the mean, then divide by the standard deviation. For the mark of 72, $z=\frac{72-60}{8}=1.5$z = (72 − 60)/8 = 1.5.

A value below the mean has a negative z-score. A mark of 54 gives $z=\frac{54-60}{8}=-\frac{6}{8}=-0.75$z = (54 − 60)/8 = −6/8 = −0.75: three quarters of a standard deviation below the mean. The mean itself has z = 0.

Two moves: shift, then squeeze

Standardizing every value of a normal variable moves its whole curve. Take X ~ $N\left(4,2^2\right)$N(4, 2²). Subtracting the mean, 4, from every value slides the curve 4 to the left, so its center is at 0. Its width has not changed: the standard deviation is still 2.

Dividing every value by 2 then squeezes the curve toward 0 by a factor of 2, so its standard deviation becomes 1. The squeezed curve is twice as tall, since its area is still 1. The result is the standard normal distribution, Z ~ N(0, 1).

The plain curve on the right is $N\left(4,2^2\right)$N(4, 2²). Subtracting 4 slides it to the plain curve centered at 0, which is just as wide. Dividing by 2 squeezes that into the gold curve, the standard normal N(0, 1).

Every normal curve becomes the same curve

Whatever $\mu$μ and $\sigma$σ are, the same two moves turn $N\left(\mu ,{\sigma }^2\right)$N(μ, σ²) into N(0, 1). There is a different normal curve for every mean and standard deviation, but only one standard normal curve, and it is the one that tables are printed for.

The areas carry over unchanged. A mark is less than 72 exactly when its z-score is less than 1.5, so P(X < 72) = P(Z < 1.5). Both are the same share of the marks, and a table of the standard normal gives it: 0.9332, so about 93% of the students scored less than 72.

The exam marks, $N\left(60,8^2\right)$N(60, 8²), with the area to the left of 72 shaded.

The standard normal curve with the area to the left of z = 1.5 shaded. It is the same shape as the curve above, and the shaded area is the same, 0.9332.

A change of units

A z-score measures the same distance from the mean in different units: standard deviations instead of marks. With $\sigma =8$σ = 8, every 4 marks above 60 is half a standard deviation, so 64 is at z = 0.5, 68 at z = 1, 72 at z = 1.5 and 76 at z = 2. A z of 1.5 means one and a half standard deviations above the mean, whatever the original units were.

The marks and their z-scores on two scales pinned together. Each step of 4 marks is half a standard deviation, and 72 sits over z = 1.5.

Comparing across different scales

A student scores 72 in mathematics, where the class mean is 60 and the standard deviation 8, and 80 in English, where the mean is 70 and the standard deviation 10. The English mark is higher, but the z-scores tell a different story: in mathematics $z=\frac{72-60}{8}=1.5$z = (72 − 60)/8 = 1.5, and in English $z=\frac{80-70}{10}=1$z = (80 − 70)/10 = 1.

Compared with the rest of the class, the mathematics result is the stronger one: one and a half standard deviations above the mean, against one.

Going back

Rearranging $z=\frac{x-\mu }{\sigma }$z = (x − μ)/σ gives $x=\mu +z\sigma$x = μ + zσ: multiply the z-score by the standard deviation, then add the mean. It turns a z-score back into a value, and it checks a standardization. For the mark of 72, 60 + 1.5 × 8 = 60 + 12 = 72.

The usual mistakes

Stopping at the gap. 72 − 60 = 12 is the distance in marks; the z-score counts standard deviations, so the gap must still be divided by 8.

Getting the sign the wrong way round. A value below the mean has a negative z-score, and a value above it a positive one: 54 gives −0.75, not 0.75.

Dividing by the variance. In N(60, 64), the 64 is ${\sigma }^2$σ², so $\sigma =8$σ = 8 and $z=\frac{12}{8}=1.5$z = 12/8 = 1.5, not $\frac{12}{64}$12/64.

Dividing before subtracting. $72-\frac{60}{8}=72-7.5=64.5$72 − 60/8 = 72 − 7.5 = 64.5 is not a z-score. The whole gap, 72 − 60, is divided by $\sigma$σ.