The normal family and standardising

One shape, two parameters

The normal distribution is the bell-shaped, symmetric density that describes an enormous range of measured quantities — and, from Module 8 on, describes sample means whatever the data look like. It is a family: every member has the same shape, located and scaled by two parameters. The mean $\mu$ says where the peak sits; the standard deviation $\sigma$ says how wide the bell is. The curve is symmetric about $\mu$, and about 68 % of its area lies within one $\sigma$ of $\mu$, 95 % within two, 99.7 % within three — the empirical rule of Module 3, which was really a statement about this curve.

Notation. This course writes

$$X\sim N(\mu ,{\sigma }^2)$$

with the variance as the second argument. Northfield Credit's invoice amounts have mean \$120 and standard deviation \$15, so $X\sim N(120,15^2)=N(120,225)$. Read the notation before the number: 225 is ${\sigma }^2$, and the SD you standardise with is $\sqrt{225}=15$. IBS1 writes $N(\mu ,\sigma )$ with the SD, and R's pnorm takes the SD (L7.6); the course's choice is the handouts', and it is the one on the paper.

Standardising

The standard normal is the member with $\mu =0$ and $\sigma =1$, written $Z$. The cumulative table (L1.3, L7.3) is printed for $Z$ alone, so every normal question is converted into a $Z$ question by measuring the distance from the mean in standard deviations:

$$z=\frac{x-\mu }{\sigma }$$

$z$ is a z-score: how many SDs above ($z>0$) or below ($z<0$) the mean $x$ lies. An invoice of \$138 at Northfield is

$$z=\frac{138-120}{15}=\frac{18}{15}=1.20$$

standard deviations above the mean. A \$99 invoice is $z=(99-120)/15=-1.40$ — below the mean, and the sign carries that information. Since $X\sim N(120,15^2)$, $Z=(X-120)/15\sim N(0,1)$: subtracting $\mu$ and dividing by $\sigma$ are exactly Module 6's linear transformation with $a=1/\sigma$ and $b=-\mu /\sigma$, which is why the result has mean 0 and variance 1.

Un-standardising

The inverse problems of L7.4 run the other way: the table hands you a $z$ and you need the dollar amount. Solve the z-score formula for $x$:

$$x=\mu +z\sigma$$

An invoice 0.8 SDs below the mean is $x = 120 + (-0.8)(15) = 120 - 12 = \$108$. Two SDs above is $120 + 2(15) = \$150$. The two formulas are the same equation read in opposite directions; every normal question in the course is one of them followed by a table lookup, or a table lookup followed by one of them.

Why the same curve

Standardising works because the normal family is closed under linear transformation: any $N(\mu ,{\sigma }^2)$ is $N(0,1)$ stretched by $\sigma$ and shifted by $\mu$. So one table serves every normal, and — the fact Module 8 rests on — a sample mean of normal data, being a linear combination, is itself normal.