B.3 Normal Distribution (Pinkie Length)

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

collect a real measurement from 30 people and summarize it with a histogram, a mean, and a standard deviation;
fit a normal distribution to your own data by using the sample mean and sample standard deviation as $\mu$ and $\sigma$;
compute the median, quartiles, IQR, percentiles, and a tail probability twice — once from the data and once from the fitted normal curve — and say why the two columns disagree.

B.3.1 Stats Lab: Normal Distribution (Pinkie Length)

Definition B.3.1: Empirical Distribution

The distribution you get by tallying data you actually collected — the frequencies, relative frequencies, and summary statistics of a real sample. An empirical distribution is measured, so it changes every time the experiment is rerun.

Its counterpart is the theoretical distribution, which is a curve chosen to describe the experiment. Once its parameters are fixed, the theoretical distribution stops moving; rerunning the experiment cannot change it.

This section is a lab, not a reading. In §5.1 and §5.2 you were handed a mean and a standard deviation and asked to compute probabilities from them. Here nobody hands you anything. You are going to measure 30 pinky fingers, work out the mean and standard deviation of what you measured, use those two numbers to name a normal distribution, and then check whether that curve actually describes the fingers you measured.

Class Time:

Names:

You will end up with two columns of numbers describing the same thing: an empirical column computed straight from your 30 measurements, and a theoretical column computed from the normal curve you fitted to them. The two columns will not match exactly. The size and the direction of the mismatch is the lesson — a normal distribution is a smooth curve with no gaps and no ends, and your data is 30 values rounded to the nearest half centimeter, so a certain amount of disagreement is built in before anyone measures anything.

Write down what you actually did, including anything that went sideways. If two people used different rulers, if one person measured to the knuckle and another to the crease, if you surveyed 29 people and rounded a 30th from memory — record it. A lab report that hides its own irregularities cannot be checked by anyone, and being checkable is most of what makes a result worth anything.

Try It Now B.3.1

Rowan is running the lab with their girlfriend Blake, who is in the same evening section. They produce two sets of numbers: the mean, quartiles, and percentiles of their 30 measurements, and the mean, quartiles, and percentiles of the normal curve they fit to them. Which of the two would come out differently if a different class ran this same lab tomorrow, and what does that tell you about which one gets called "the right answer"?

Solution

Step 1 — ask what each set is built from. The first set is a tally of 30 fingers that a particular group of people brought to a particular classroom. It is a measurement.

Step 2 — ask what the second set is built from. The theoretical values come from a formula applied to a mean and a standard deviation. Once those two parameters are named, every other theoretical number — median, quartiles, percentiles, tail probabilities — follows from the formula and nothing else.

Answer: both would change, and that is what makes this lab different from the uniform-distribution lab in §A.3. There the theoretical distribution was fixed in advance, so only the data moved. Here the theoretical curve is fitted to your sample, so a different class gets a different $\bar{x}$ and $s$ and therefore a different curve. Neither column is "the right answer." What you are testing is narrower and more honest: given that you chose a normal curve with these two parameters, does that curve describe these 30 measurements well?

B.3.2 Collect the Data

Measure the length of your pinky finger, in centimeters.

Randomly survey 30 adults for their pinky finger lengths. Round the lengths to the nearest 0.5 cm and record them in Table B.3.1.

**Table B.3.1** — Your 30 pinky finger lengths, in centimeters, rounded to the nearest 0.5 cm.
1–5	6–10	11–15	16–20	21–25	26–30

Try it in rāSHio

Type your 30 measurements straight into rāSHio instead of writing them out twice — File → Delimited List… accepts them separated by commas, spaces, or one per line, so you can enter them in the order you measured them. Everything below reads from that one column, so getting it in cleanly once saves re-keying it for the histogram, the summary statistics, and the sort.

Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil, and scale the axes. Put the finger length on the horizontal axis and label it Length of Finger, and put the frequency on the vertical axis and label it Frequency.

Try it in rāSHio

Once the column is in, rāSHio’s Graph → Histogram draws your five or six intervals for you: set the bin start below your smallest measurement and the bin width to whatever cuts your range into that many pieces. Draw the pencil version first — choosing the intervals yourself is the point of step 2 — then use rāSHio to check you counted each bar right.

Calculate the following.

a. $\bar{x} =$ _________

b. $s =$ _________

Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences describing the general shape of the curve. Keep it simple: does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end?

Rounding to 0.5 cm is doing something to your data

Real pinky lengths take any value a ruler can resolve. Rounding forces all 30 into a handful of allowed values, which is what makes a histogram drawable — and it is also why the smooth curve you draw next can only ever be an approximation of a staircase.

Since every measurement lands on a half centimeter, your 30 values will pile up on a small number of distinct lengths, and a histogram with five or six intervals will mostly be counting how many people share each of those lengths. That is convenient for drawing and inconvenient for pretending the data is continuous. Keep it in mind, because it is the single biggest reason your two columns of numbers will disagree later. Choose your intervals before you look at the counts, not after. It is tempting to slide the boundaries around until the picture looks bell-shaped, and a histogram built that way tells you about your own preferences rather than about the fingers you measured. Pick a starting boundary below your smallest value, pick a width that divides the range into five or six pieces, and draw whatever comes out. If the result looks lopsided, that is a finding, not a mistake to be corrected. The whole point of drawing the smooth curve through the bar tops is to decide, by eye, whether a normal curve is a defensible model here — and you cannot make that call honestly if you have already tuned the picture toward the answer you want.

Throughout the rest of this lab, one worked group's data is used to show every calculation. That group surveyed 30 adults and got the counts in Table B.3.2. Use it to check that you can reproduce each step; your own numbers will differ.

**Table B.3.2** — One group's 30 pinky lengths, tallied.
Length (cm)	Frequency
4.5	1
5.0	2
5.5	5
6.0	8
6.5	7
7.0	4
7.5	2
8.0	1

Try It Now B.3.2

Using the group's data in Table B.3.2, find the sample mean $\bar{x}$.

Solution

Step 1 — total the measurements. Multiply each length by how many people had it, then add.

$$ \sum x = 1(4.5) + 2(5.0) + 5(5.5) + 8(6.0) + 7(6.5) + 4(7.0) + 2(7.5) + 1(8.0) = 186.5 $$

Step 2 — divide by the count.

$$ \bar{x} = \frac{186.5}{30} \approx 6.22 \text{ cm} $$

Answer: $\bar{x} \approx 6.22$ cm. Notice that the mean is not one of the allowed measurements — nobody in the room has a 6.22 cm pinky, because the ruler was only read to the nearest half centimeter. That is normal and it is not an error. The mean is a balance point for the whole group, not a prediction about any one person.

Try It Now B.3.3

Tayen takes the same 30 measurements and works out the sample standard deviation $s$. Do it with them: the sum of the values is 186.5 and the sum of their squares is 1,177.75.

Solution

Step 1 — subtract off the part explained by the mean. With $n = 30$ and $\bar{x} \approx 6.2167$,

$$ s^2 = \frac{\sum x^2 - n\bar{x}^2}{n - 1} = \frac{1177.75 - 30(6.2167)^2}{29} \approx \frac{1177.75 - 1159.41}{29} \approx \frac{18.34}{29} \approx 0.6325 $$

Step 2 — take the square root.

$$ s = \sqrt{0.6325} \approx 0.7953 \approx 0.80 \text{ cm} $$

Answer: Tayen gets $s \approx 0.80$ cm. Divide by $n - 1$, not by $n$ — these 30 adults are a sample, not every adult there is. Carry the unrounded $\bar{x}$ through the squaring step: rounding it to 6.22 first shifts $n\bar{x}^2$ by about a tenth, which is small here but is exactly the kind of slippage that grows when the numbers are bigger.

Try it in rāSHio

rāSHio’s Stats → Summary Statistics returns the mean, standard deviation, median, and both quartiles for your column in one step. Use it to CHECK the $\bar{x}$ and $s$ you just worked out, not to replace them — if the tool and your arithmetic disagree you have either mis-keyed a measurement or dropped a term out of the sum of squares, and both are worth finding now.

Figure B.3.1 — Checking your hand-computed $\bar{x}$ and $s$ in rāSHio: Stats → Summary Statistics. The walkthrough uses its own demo column; the steps are the ones you run on your 30 measurements.

B.3.3 Analyze the Distribution

Definition B.3.2: Normal Distribution

A continuous random variable $X$ whose probability density function is

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

where $\mu$ is the mean of the distribution and $\sigma$ is its standard deviation. The notation is $X \sim N(\mu, \sigma)$. When $\mu = 0$ and $\sigma = 1$, the variable is called the standard normal distribution, usually written $Z \sim N(0, 1)$.

Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution of the data you collected?

$X \sim$ _________(_________, _________)

How does the histogram help you arrive at the approximate distribution?

Two dials, and that is the whole instrument

A normal distribution knows nothing about pinky fingers. Give it a center and a spread and it will hand back every percentile and probability you ask for — which is exactly why fitting one to your data is so cheap, and why you should check the fit rather than assume it.

Filling in that blank is two separate decisions, and the lab asks for both. The histogram answers the first one: it tells you which family of curves is worth trying. A single hump near the middle with the bars falling away on both sides is the shape a normal curve has, so a normal curve is a reasonable candidate. A picture that runs flat across, or piles up at one end and trails off the other, would be telling you to reach for a different family — uniform or exponential, both of which you met in Chapter 5.

Your $\bar{x}$ and $s$ answer the second decision: which normal curve. Every normal distribution has the same bell shape, and $\mu$ and $\sigma$ are the only two dials — $\mu$ slides the whole curve left or right, $\sigma$ makes it wider and flatter or narrower and taller. So once you have decided on the family from the picture, you set $\mu = \bar{x}$ and $\sigma = s$ and the curve is fully determined. This is the step where the theoretical column stops being independent of your data, and it is worth being clear-eyed about: you are not testing whether pinky lengths are normal in general, you are testing how well the best-fitting normal curve tracks the sample it was fitted to.

Try It Now B.3.4

Ethan, who tallied the counts in Table B.3.2 for his group, found $\bar{x} \approx 6.22$ cm and $s \approx 0.80$ cm. Write his approximate theoretical distribution, and say what would have to be true of his histogram for that choice to be defensible.

Solution

Step 1 — set the parameters from the sample. Use $\mu = \bar{x} = 6.22$ and $\sigma = s = 0.80$.

Step 2 — write it in the standard notation.

$$ X \sim N(6.22,\ 0.80) $$

where $X$ is the length in centimeters of a randomly chosen adult's pinky finger.

Answer: $X \sim N(6.22, 0.80)$. For that to be defensible his histogram needs one hump, roughly centered near 6.22, falling away on both sides at about the same rate — which is what the counts in Table B.3.2 do: 1, 2, 5, 8, 7, 4, 2, 1. It rises to a peak at 6.0 and comes back down, slightly heavier on the right. Notice that the peak of the data sits at 6.0 while the fitted mean is 6.22. That gap is a small warning that the fit is not perfect, and it comes back in the Discussion Questions.

B.3.4 Describe the Data

Definition B.3.3: Interquartile Range

The distance between the first and third quartiles of a data set:

$$ IQR = Q_3 - Q_1 $$

It measures the width of the middle half of the data. Because it ignores everything below $Q_1$ and above $Q_3$, a single unusually large or small value cannot move it.

Using the data you collected, complete the following statements. Order the data first — every statistic below except the probability is read off a position in the sorted list.

$IQR =$ _________

The 15th percentile is _________.

The 85th percentile is _________.

Median is _________.

What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm?

Explain the meaning of the 85th percentile of this data.

Everything in this list is counted, not computed from a formula, and the counting rule is the one from §2.3: to locate the $k$th percentile of $n$ ordered values, find the index $i = \frac{k}{100}(n + 1)$. If $i$ lands on a whole number, that position holds the percentile. If it lands between two positions, average the two values on either side. The quartiles are the 25th and 75th percentiles, so they use the same rule, and the median is the 50th.

The one item in the list that is not counted is the probability question, and the wording is deliberately awkward: it asks for a theoretical probability while you are still in the empirical column. Answer it from the data anyway — count how many of your 30 measurements are strictly greater than 6.5 and divide by 30 — and hold that number, because you will compute the genuinely theoretical version of it in the next section and the two will not agree. Watch the word "more than" carefully while you count. A measurement of exactly 6.5 is not more than 6.5, so it does not go in the numerator. In a rounded data set that distinction is worth real probability: with lengths recorded to the nearest half centimeter, a whole block of people sit exactly on 6.5, and including or excluding them changes the answer by a lot more than it would in data recorded to three decimals.

Try It Now B.3.5

Marisol is working from the same 30 measurements in Table B.3.2. Help her find the median, $Q_1$, $Q_3$, and the IQR.

Solution

Step 1 — write the running totals. In order, the cumulative counts are 4.5 cm through position 1, 5.0 cm through 3, 5.5 cm through 8, 6.0 cm through 16, 6.5 cm through 23, 7.0 cm through 27, 7.5 cm through 29, and 8.0 cm at 30.

Step 2 — locate the median.

$$ i = \frac{50}{100}(30 + 1) = 15.5 $$

Not a whole number, so average positions 15 and 16. Both fall in the 6.0 cm block, so the median is 6.0 cm.

Step 3 — locate $Q_1$.

$$ i = \frac{25}{100}(31) = 7.75 $$

Average positions 7 and 8. Both are 5.5 cm, so $Q_1 = 5.5$ cm.

Step 4 — locate $Q_3$.

$$ i = \frac{75}{100}(31) = 23.25 $$

Average positions 23 and 24. Position 23 is the last 6.5 cm and position 24 is the first 7.0 cm, so $Q_3 = \frac{6.5 + 7.0}{2} = 6.75$ cm.

Step 5 — take the difference.

$$ IQR = Q_3 - Q_1 = 6.75 - 5.5 = 1.25 \text{ cm} $$

Answer: Marisol gets median $= 6.0$ cm, $Q_1 = 5.5$ cm, $Q_3 = 6.75$ cm, $IQR = 1.25$ cm. The middle half of her group spans 1.25 cm — a bit over a centimeter separates a typical short pinky from a typical long one, which is a smaller spread than most people guess before they measure.

Try it in rāSHio

Every statistic in this list is read off a position, so the column has to be in order first: Transform → Sort in rāSHio puts your 30 measurements in ascending order in one step. Then count the positions by hand — the index $i = \frac{k}{100}(n+1)$ tells you where to look, and walking to that spot yourself is what makes a percentile feel like a location in the data rather than a formula.

Figure B.3.2 — Ordering a column in rāSHio: Transform → Sort. The walkthrough sorts its own demo column; the steps are the ones you run on your 30 measurements.

Try It Now B.3.6

Claire takes the same 30 measurements. Find the 15th and 85th percentiles with her, and explain in a sentence what the 85th percentile means here.

Solution

Step 1 — locate the 15th percentile.

$$ i = \frac{15}{100}(31) = 4.65 $$

Average positions 4 and 5. Both sit in the 5.5 cm block, so the 15th percentile is 5.5 cm.

Step 2 — locate the 85th percentile.

$$ i = \frac{85}{100}(31) = 26.35 $$

Average positions 26 and 27. Both are 7.0 cm, so the 85th percentile is 7.0 cm.

Step 3 — say what it means. A percentile splits the ordered data at a stated share.

Answer: Claire finds the 15th percentile at 5.5 cm and the 85th at 7.0 cm. The 85th percentile means about 85% of the adults she surveyed have a pinky finger 7.0 cm or shorter, and about 15% have one longer than that. Say it about people, not about numbers — "85% of the values are below 7.0" is true but tells the reader nothing they can picture, and the whole reason to report a percentile instead of a mean is that a percentile is a statement about where an individual stands in a group.

B.3.5 Theoretical Distribution

Definition B.3.4: z-score

The linear transformation

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

applied to a value $x$ of a normal random variable $X \sim N(\mu, \sigma)$. The result says how many standard deviations $x$ sits above (positive) or below (negative) the mean. Applying the transformation to the whole distribution turns it into the standard normal, $Z \sim N(0, 1)$, which is what makes distributions with different units comparable.

Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation.

$IQR =$ _________

The 15th percentile is _________.

The 85th percentile is _________.

Median is _________.

What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm?

Explain the meaning of the 85th percentile of this data.

Every number in this column comes out of the two parameters and the standard normal table, and none of them looks at your sorted list. The pattern is always the same: convert the percentile you want into a $z$-score using the standard normal, then undo the transformation with $x = \mu + z\sigma$ to get back into centimeters. Going the other direction — from a length to a probability — you compute the $z$-score of the length first and then read the area beyond it.

Two of the six answers you can write down without any table at all. The median of a normal distribution equals its mean, because the curve is symmetric about $\mu$, so exactly half the area sits on each side. And the meaning of the 85th percentile is the same sentence you wrote in the empirical column, with one word changed: it is the length such that 85% of the distribution falls at or below it, rather than 85% of your 30 measurements. Keep the parallel structure when you write the two explanations, because setting them side by side is what makes the comparison in the Discussion Questions readable.

Try It Now B.3.7

Tou has moved on to the theoretical column. For $X \sim N(6.22, 0.80)$, find the theoretical median, $Q_1$, $Q_3$, and IQR as he would. Use $z = \pm 0.6745$ for the quartiles.

Solution

Step 1 — the median. A normal distribution is symmetric about its mean, so the median is $\mu = 6.22$ cm.

Step 2 — undo the transformation at $z = -0.6745$.

$$ Q_1 = \mu + z\sigma = 6.22 + (-0.6745)(0.80) \approx 6.22 - 0.54 = 5.68 \text{ cm} $$

Step 3 — do the same at $z = +0.6745$.

$$ Q_3 = 6.22 + (0.6745)(0.80) \approx 6.22 + 0.54 = 6.76 \text{ cm} $$

Step 4 — subtract.

$$ IQR = 6.76 - 5.68 = 1.08 \text{ cm} $$

Answer: Tou gets median $= 6.22$ cm, $Q_1 \approx 5.68$ cm, $Q_3 \approx 6.76$ cm, $IQR \approx 1.08$ cm. His quartiles came out symmetric about the mean — 0.54 cm on each side — and they had to, because the curve is symmetric and $z = \pm 0.6745$ is a matched pair. Your empirical quartiles almost certainly were not symmetric about your $\bar{x}$, and that asymmetry is one of the concrete things the Discussion Questions want you to comment on.

Try It Now B.3.8

For the same distribution, find the theoretical 15th and 85th percentiles. Use $z = \pm 1.0364$.

Solution

Step 1 — the 15th percentile. It sits below the mean, so the $z$-score is negative.

$$ x = \mu + z\sigma = 6.22 + (-1.0364)(0.80) \approx 6.22 - 0.83 = 5.39 \text{ cm} $$

Step 2 — the 85th percentile. By symmetry it sits the same distance above the mean.

$$ x = 6.22 + (1.0364)(0.80) \approx 6.22 + 0.83 = 7.05 \text{ cm} $$

Answer: the theoretical 15th percentile is about 5.39 cm and the 85th is about 7.05 cm. Hold these against the empirical 5.5 cm and 7.0 cm from Try It Now B.3.6. The curve reaches further out on both ends than the data did — it predicts a shorter 15th percentile and a longer 85th — which is the beginning of the answer to the discussion question, and is exactly what a smooth curve with genuine tails does when it is laid over 30 values rounded onto seven or eight allowed lengths.

Try It Now B.3.9

Yesenia and her wife measured last, and both came in above 6.5 cm. For $X \sim N(6.22, 0.80)$, find the theoretical probability that a randomly chosen pinky length is more than 6.5 cm.

Solution

Step 1 — standardize the boundary.

$$ z = \frac{x - \mu}{\sigma} = \frac{6.5 - 6.22}{0.80} = \frac{0.28}{0.80} = 0.35 $$

Step 2 — read the area to the left. From the standard normal table, the area to the left of $z = 0.35$ is about 0.6368.

Step 3 — take the complement, because the question asks for "more than."

$$ P(X > 6.5) = 1 - 0.6368 = 0.3632 $$

Answer: about 0.3632, or roughly 36% — so a pair landing where Yesenia and her wife did is not unusual at all. On a calculator this is normalcdf(6.5, 1E99, 6.22, 0.80). Note that $P(X > 6.5)$ and $P(X \geq 6.5)$ are the same number here: the normal distribution is continuous, so the probability of landing on any single exact value is zero. That is precisely the assumption your rounded data violates, since seven of the group's 30 people landed on exactly 6.5.

Try it in rāSHio

rāSHio’s Distributions → Normal does both directions of this section in one panel: enter $\mu = 6.22$ and $\sigma = 0.80$, then ask for the area above 6.5 to get the probability, or ask for the value at 0.15 and 0.85 to get the two percentiles. Same two operations as the $z$-table, without the rounding you pick up converting to $z$ and back.

Figure B.3.3 — Reading a normal distribution in rāSHio: Distributions → Normal. The walkthrough uses its own demo values; for this section you enter $\mu = 6.22$ and $\sigma = 0.80$.

B.3.6 Discussion Questions

Do the data you collected give a close approximation to the theoretical distribution? In complete sentences, and comparing the results in the sections titled Describe the Data and Theoretical Distribution, explain why or why not.

Answer it statistic by statistic rather than with one overall verdict. For each pair — median, IQR, 15th percentile, 85th percentile, and the probability — state your empirical value, state the theoretical value, and say how far apart they are in centimeters. A comparison that reports gaps in the units of the data can be checked by a reader; "pretty close" cannot.

Try It Now B.3.10

Work the discussion question all the way through for the group's data. Line up all five pairs, measure each gap, and say what pattern the gaps show and what causes it.

Solution

Step 1 — line the pairs up.

**Table B.3.3** — The group's empirical and theoretical values, side by side.
Statistic	Empirical	Theoretical	Gap
Median	6.00 cm	6.22 cm	0.22 cm
IQR	1.25 cm	1.08 cm	0.17 cm
15th percentile	5.50 cm	5.39 cm	0.11 cm
85th percentile	7.00 cm	7.05 cm	0.05 cm
$P(X > 6.5)$	0.2333	0.3632	0.1299

Step 2 — judge the four length statistics. Every gap is 0.22 cm or less, on measurements whose rounding to the nearest half centimeter can shift a value by 0.25 cm all by itself. Four out of five are within half a rounding step, which is about as close as this data is capable of coming.

Step 3 — find the pattern. The empirical median sits below the fitted mean, and the empirical IQR is wider than the theoretical one. The first says the data is slightly heavier on the right: the peak sits at 6.0 cm with 14 measurements above it and only 8 below. The second says the sample's middle half is wider than the fitted curve predicts. A normal curve cannot be lopsided, so fitting one moves the center up toward the long side and narrows the middle half relative to what the data actually did.

Step 4 — deal with the probability. This is the one real disagreement, at 0.13. It is not a lopsidedness problem — it is the rounding. Seven of the 30 people measured exactly 6.5 cm, and "more than 6.5" excludes all seven from the empirical count. The continuous curve puts no weight on any single value, so it splits that block across the boundary instead of dropping it.

Answer: yes, the approximation is close for the four length statistics — all within 0.22 cm — and clearly off for the tail probability, by 0.13. The two failures have different causes and should be reported separately: the small length gaps come from a slightly right-heavy sample being described by a symmetric curve, while the large probability gap comes from using a continuous model on data rounded onto half centimeters. Reporting the probability gap as evidence that "pinky lengths are not normal" would be the wrong conclusion. It is evidence that a continuous model and a rounded measurement disagree at a boundary where a lot of the data is stacked, which is a fact about the ruler rather than about fingers.

Key Terms

empirical distribution — the distribution of data actually collected, summarized by frequencies and sample statistics; a measured quantity that changes each time the experiment is rerun.

interquartile range (IQR) — the width of the middle half of a data set, $Q_3 - Q_1$.

normal distribution — a continuous random variable with density $f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$, written $X \sim N(\mu, \sigma)$, where $\mu$ is the mean and $\sigma$ the standard deviation.

standard normal distribution — the normal distribution with $\mu = 0$ and $\sigma = 1$, written $Z \sim N(0, 1)$.

theoretical distribution — the curve chosen to model an experiment; here, the normal curve fitted to the sample by setting $\mu = \bar{x}$ and $\sigma = s$.

z-score — the transformation $z = \frac{x - \mu}{\sigma}$, giving the number of standard deviations a value lies above or below the mean.