A bell appeared at the end of the previous lesson: as \( n \) grew, the coupons' binomial smoothed out into a symmetric, bell-shaped silhouette. That silhouette is the normal distribution (or Gaussian), the most important template in all of statistics — and an old acquaintance: it is the "bell-shaped" form we already described when studying histograms in Module 2, and the reason behind the 68-95-99.7 empirical rule that we accepted back then without justification. In this lesson we turn it into a calculation tool: you will learn what its two parameters \( \mu \) and \( \sigma \) mean, how to standardize any value (reuniting with the z-scores of 02-03), how to read the Z table to obtain probabilities and percentiles, and how to apply it all to real NovaMarket decisions: what is the probability that a store exceeds €50,000 in sales in one day? How much cash should be prepared to cover 95% of days? We will close with something just as valuable: how to suspect that a variable is not normal before using the template.

1. Why the normal shows up everywhere
2. The probability density function and the parameters \( \mu \) and \( \sigma \)
3. Properties and the 68-95-99.7 empirical rule
4. Standardization: the standard normal and the z-score
5. How to read the Z table
6. Computing probabilities: the stores' daily sales
7. Percentiles: planning cash for 95% of days
8. Informal normality check (and a counterexample: delivery times)

The daily sales of a NovaMarket store are the sum of thousands of individual receipts. A person's height is the sum of thousands of small genetic and environmental effects. The error of a scale is the sum of many tiny perturbations. The pattern repeats: when a quantity results from adding many small, independent contributions, its distribution tends to be normal, almost regardless of what each individual contribution looks like. This astonishing fact has a name — the central limit theorem — and we devote the whole of lesson 04-04 to it; for now the practical consequence is enough: a great many "sum-like or average-like" business variables are approximately normal, which is why this template is the first one an analyst tries.

At NovaMarket, the perfect candidate is daily sales per store: in Module 2 we saw that its histogram was symmetric and bell-shaped, with mean €43,983 and standard deviation €6,200. In this lesson we will model it as

\[ V \sim N(43{,}983;\ 6{,}200) \]

which reads "\( V \) follows a normal distribution with mean 43,983 and standard deviation 6,200".

The normal is a continuous distribution, so — as we saw in 03-03 — it is described by a probability density function, and probabilities are areas under the curve. Its density has a celebrated formula:

\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} , e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]

Rest easy: we will never calculate with this formula directly (the areas under it have no closed-form expression; that is what the tables are for). What matters is what its two parameters say:

Each pair \( (\mu, \sigma) \) defines a different normal: the chain's daily sales are \( N(43{,}983;\ 6{,}200) \), but a small store like Cuenca might be \( N(21{,}000;\ 2{,}900) \) — same shape, different center and width. The normal family is a template with two dials, just as the binomial was with \( n \) and \( p \).

Every normal, whatever its \( \mu \) and \( \sigma \), satisfies:

• Perfect symmetry around \( \mu \): hence mean = median = mode (recall from Module 2 that in symmetric distributions all three coincide).
• Unimodal: a single peak, at \( \mu \), with tails that decay smoothly on both sides without ever touching the axis.
• The total area under the curve is 1 (it is a probability distribution).
• \( P(V = x) = 0 \) for any exact value: like every continuous variable, only intervals carry probability. Practical consequence: \( P(V > 50{,}000) \) and \( P(V \geq 50{,}000) \) are equal.

And the star property, which in Module 2 we presented as the "empirical rule" and can now state as an exact property of the normal:

The immediate business reading: a day with €60,000 in sales at an average store sits at almost \( 3\sigma \) — extremely rare if everything is normal, so it deserves investigation (a public holiday? a promotion? a recording error?). It is the same logic as the outliers of 02-03, now with exact probabilities behind it.

There are infinitely many normals, but we only need one table, thanks to a trick: any normal can be converted into the standard normal \( Z \sim N(0;\ 1) \) (mean 0, standard deviation 1) through the transformation

\[ z = \frac{x - \mu}{\sigma} \]

This formula is exactly the z-score from Module 2: "how many standard deviations from the mean is this value?". Back then we used it to compare values on different scales; now it also gives us probabilities: if \( V \) is normal, the standardized value \( z \) is an observation from the standard normal, and its probability is looked up in the table.

An immediate example: a €50,000 sales day is equivalent to

\[ z = \frac{50{,}000 - 43{,}983}{6{,}200} = \frac{6{,}017}{6{,}200} \approx 0.97 \]

that is, 0.97 standard deviations above the mean. Every question about \( V \) translates in this way into a question about \( Z \).

The Z table (or standard normal table) records the cumulative distribution function of \( Z \): for each value \( z \), it gives \( \Phi(z) = P(Z \leq z) \), the area to the left of \( z \). It is organized like this: the rows give the units and tenths of \( z \), and the columns, the hundredth. An excerpt:

To look up \( \Phi(0.97) \): row "0.9", column ".07" → 0.8340. That is, \( P(Z \leq 0.97) = 0.8340 \).

With the area to the left and two rules, everything can be solved:

• Area to the right (complement): \( P(Z > z) = 1 - \Phi(z) \).
• Negative values (symmetry): \( \Phi(-z) = 1 - \Phi(z) \). For example, \( \Phi(-0.97) = 1 - 0.8340 = 0.1660 \). (Many printed tables include only the positive half for this very reason.)
• Between two values: \( P(a < Z < b) = \Phi(b) - \Phi(a) \) — the area up to \( b \) minus the area up to \( a \).

Case 1 — "What is the probability that a store exceeds €50,000 today?" Marta needs it to size staffing reinforcements. A three-step method that will always be the same:

1. Standardize: \( z = \dfrac{50{,}000 - 43{,}983}{6{,}200} \approx 0.97 \).
2. Table: \( \Phi(0.97) = 0.8340 \).
3. Translate the question: we want the area to the right: \( P(V > 50{,}000) = 1 - 0.8340 = 0.1660 \).

On roughly 16.6% of days, an average store exceeds €50,000. Across 42 stores, we would expect about \( 42 \times 0.166 \approx 7 \) stores to top that figure on any given day.

Case 2 — "And of it falling below €35,000?" (the threshold that triggers a store review):

1. \( z = \dfrac{35{,}000 - 43{,}983}{6{,}200} = \dfrac{-8{,}983}{6{,}200} \approx -1.45 \).
2. By symmetry: \( \Phi(-1.45) = 1 - \Phi(1.45) = 1 - 0.9265 = 0.0735 \).
3. The question already asked for the area to the left: \( P(V < 35{,}000) \approx 0.0735 \), 7.4% of days.

Careful with the interpretation: the fact that the threshold trips on 7.4% of days with the business running perfectly normally means that, in a chain of 42 stores, every day about 3 stores will sit below €35,000 with nothing wrong at all. If the review protocol does not account for this, the team will end up chasing ghosts.

Case 3 — "Probability of a 'typical' day, between €40,000 and €50,000?"

1. \( z_1 = \dfrac{40{,}000 - 43{,}983}{6{,}200} \approx -0.64 \); \( z_2 \approx 0.97 \).
2. \( \Phi(-0.64) = 1 - 0.7389 = 0.2611 \); \( \Phi(0.97) = 0.8340 \).
3. \( P(40{,}000 < V < 50{,}000) = 0.8340 - 0.2611 = 0.5729 \approx 57.3,% \).

Sometimes the question runs the other way: not "what probability does this value have?" but "what value leaves a given probability below it?". It is the percentile problem from Module 2, now solved with the model.

Case: cash and staffing for the 95th percentile. Operations wants to plan treasury and shifts to cover "every day except the strongest 5%". We look for the value \( v_{95} \) such that \( P(V \leq v_{95}) = 0.95 \).

1. Table in reverse: we search inside the table for the area 0.9500 and read off which \( z \) it corresponds to. It sits exactly between \( \Phi(1.64) = 0.9495 \) and \( \Phi(1.65) = 0.9505 \); we take the midpoint value \( z_{95} = 1.645 \) (a classic worth memorizing).
2. De-standardize (solve the \( z \) formula for \( x \)):

\[ x = \mu + z \cdot \sigma = 43{,}983 + 1.645 \times 6{,}200 = 43{,}983 + 10{,}199 = \text{€}54{,}182 \]

By preparing operations for €54,182 in daily sales, an average store is covered on 95% of days. The same method works downward: the 1st percentile (for severe-drop alerts) uses \( z = -2.33 \) (because \( \Phi(2.33) = 0.9901 \)) and gives \( 43{,}983 - 2.33 \times 6{,}200 \approx \text{€}29{,}537 \).

Other values of \( z \) you will use again and again (keep them handy):

The 1.96 in particular will be the undisputed star of Module 5.

The normal is a template, not an obligation: before using it you must check that it fits. The informal check, with tools you already master from Module 2:

1. Histogram: is it roughly symmetric, unimodal and bell-shaped? Very long tails on one side, two peaks or a sharp cutoff edge rule out the normal.
2. Mean vs median: in a normal they coincide. If they clearly drift apart (remember: the mean gets dragged by the tail), there is skewness.
3. Empirical rule: do ~68% of the data really fall within \( \bar{x} \pm s \) and ~95% within \( \bar{x} \pm 2s \)? If not, the template does not describe your data.
4. Impossible-value test: if \( \mu - 2\sigma \) or \( \mu - 3\sigma \) lands on absurd values (negative for amounts or times), the normal cannot be a good description.

Two counterexamples from NovaMarket itself:

• The receipt amount (mean €32.40, \( s = \text{€}21.50 \)). The impossible-value test fails spectacularly: \( 32.40 - 2 \times 21.50 = -\text{€}10.60 \). A normal with those parameters would assign appreciable probability to negative receipts (standardizing 0: \( z = -1.51 \), meaning 6.5% of "impossible" receipts). The real histogram, as we saw in Module 2, is right-skewed: many small receipts and a tail of big shopping carts. Do not use the normal here.
• E-commerce delivery times. If they were normal, mean and median would coincide at 25.5 h, and from \( F(48) = 0.899 \) we would deduce \( z = 1.27 \) and therefore \( \sigma = (48 - 25.5)/1.27 \approx 17.7 \) h. But then \( P(T < 0) \) would correspond to \( z = -1.44 \): 7.5% of deliveries in negative time! Delivery times are right-skewed (mostly fast, with a tail of delays); we will meet their natural template in the next lesson.

There is a specialized plot for this check, the QQ-plot (it compares the quantiles of your data against those of a normal: if the result is a straight line, you have normality). Knowing it exists is enough for us; with the histogram, mean-vs-median and the empirical rule you can get very far. And a preview that will resolve a pending paradox: even though the individual receipt is not normal, the mean of many receipts will be — that is the magic of the central limit theorem.

• Forgetting what the table gives. The standard Z table gives the area to the left. For "greater than", subtract from 1; for "between", subtract two readings. Always sketch a bell and shade the requested area before touching the table: it is the number-one antidote against sign errors.
• Getting the direction wrong with negative z. \( \Phi(-1.45) \) is not \( -0.9265 \) or \( 0.9265 \): it is \( 1 - 0.9265 = 0.0735 \). Probabilities are never negative or greater than 1.
• Standardizing backwards (\( \frac{\mu - x}{\sigma} \)) flips the sign of z and turns a 16% into an 84%. The subtraction is always "value minus mean".
• Applying the normal without checking the shape. With skewed variables (amounts, times, incomes) the normal gives badly wrong probabilities in the tails — exactly where decisions are made. Always run the "impossible-value test".
• Confusing a day's probability with a store count. "On 7.4% of days a store drops below €35,000" and "today about 3 of the 42 stores will be below it" are two readings of the same number; in reports, give both so management can calibrate.
• Tip: memorize the 68-95-99.7 trio and the z-values in the percentile table. They let you judge in your head, in a meeting, without a table, whether a figure is routine or extraordinary.

Exercise 1. With \( V \sim N(43{,}983;\ 6{,}200) \), compute the probability that tomorrow's sales at Zaragoza-Centro fall between €38,000 and €45,000. (Round the z-values to two decimals; \( \Phi(0.16) = 0.5636 \).)

Exercise 2. The committee wants a "red alert" that, by pure chance, trips only on the 1% of days with the lowest sales. Below what daily sales figure should the threshold be set? And what if a "yellow alert" were wanted for the lowest 10%? (\( z_{90} = 1.282 \).)

Exercise 3. A colleague proposes computing "the probability that a receipt exceeds €60" as \( P(Z > \frac{60 - 32.40}{21.50}) = P(Z > 1.28) \approx 0.10 \). Explain why the result is not reliable and which checks would have exposed it. With which Module 2 tool would you estimate that probability honestly?

Solution 1. \( z_1 = \frac{38{,}000 - 43{,}983}{6{,}200} \approx -0.97 \) and \( z_2 = \frac{45{,}000 - 43{,}983}{6{,}200} \approx 0.16 \). \( \Phi(-0.97) = 1 - 0.8340 = 0.1660 \); \( \Phi(0.16) = 0.5636 \). \( P = 0.5636 - 0.1660 = 0.3976 \approx 39.8,% \). Common mistake: subtracting the z-values (\( 0.16 - (-0.97) \)) instead of the areas — z-values are not probabilities.

Solution 2. 1st percentile: \( z = -2.33 \), so \( x = 43{,}983 - 2.33 \times 6{,}200 = 43{,}983 - 14{,}446 = \text{€}29{,}537 \). Yellow alert (10th percentile): \( z = -1.282 \), \( x = 43{,}983 - 1.282 \times 6{,}200 \approx 43{,}983 - 7{,}948 = \text{€}36{,}035 \). Common mistake: using a positive \( z \) and getting thresholds above the mean; for low percentiles, z is negative and is subtracted.

Solution 3. The receipt amount is not normal: its histogram is right-skewed (Module 2) and it fails the impossible-value test (\( \mu - 2\sigma = -\text{€}10.60 \), negative receipts). A normal with \( \mu = 32.40 \) and \( \sigma = 21.50 \) underestimates the real right tail, so that 10% is unreliable exactly where it matters. The honest approach: use the empirical relative frequency — count in Module 2's frequency table/histogram what proportion of real receipts exceeds €60 (the frequentist approach of 03-01). Underlying lesson: the template is no substitute for looking at the data.

You now command the queen of distributions: the normal is fully defined by its center \( \mu \) and its width \( \sigma \); the 68-95-99.7 rule gives its skeleton; and the standardize → table → translate method turns any probability or percentile question into two operations and one lookup. Just as important: you can detect when the bell does not fit — receipt amounts and delivery times have proved it. Precisely for those unruly cases (counts of rare events, waiting times, attempts until a success) statistics keeps more templates in the drawer, each with its typical situation of use. In the next lesson, Other Important Distributions, we complete the toolbox with the Poisson, geometric, hypergeometric, uniform and exponential distributions.