Introduction to Statistics · Chapter 5 · The Normal Distribution

Using the Normal Distribution

Turn a shaded region under the normal curve into a probability — then run the calculation backwards to find the value sitting at a given percentile.

bookSHelf · Introduction to Statistics · §5.2 · a self-paced section

Using the Normal Distribution · bookSHelf Intro Stats§5.2

Learning objectives — by the end of this section you will be able to

Objectives

Read a shaded area as a probability, and convert between the area to the left of a value and the area to its right §5.2.1
Use technology to find $P(x < a)$ , $P(x > a)$ , and $P(a < x < b)$ for any normal distribution §5.2.2–3
Find the value $k$ that sits at a given percentile, and state what it means in the original units §5.2.4
Translate “at least”, “at most”, “the bottom quartile”, and “the middle 50%” into the areas a calculator needs §5.2.5–6

Using the Normal Distribution · bookSHelf Intro Stats§5.2

§5.2.1 — probability is area

Reading a shaded region

For a continuous variable, probability is area under the curve. The shaded region is the area to the left of $x$ — exactly $P(X < x)$ .

$P(X > x) = 1 - P(X < x)$

A single point has zero width, so zero area — “less than” and “at most” mean the same thing here, unlike the discrete distributions of the last chapter.

Figure 5.2.1: Probability is area — the shade sweeps in from the tail and stops at $x$ .

Using the Normal Distribution · bookSHelf Intro Stats§5.2

Insight Note — one cut, two answers

Whichever piece you get, subtract from 1

Drawing the line through $x$ slices the curve into exactly two pieces. Whichever piece the calculator hands you, subtracting from 1 gets you the other — you never have to compute a right tail directly.

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Your turn — flip a left tail

Try It Now 5.2.1

Try It Now 5.2.1

If the area to the left of $x$ is 0.012, what is the area to the right?

The two areas fill the whole curve, so they add to 1.

\text{area to the right} = 1 - 0.012 = 0.988

Answer: the area to the right of $x$ is 0.988.

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§5.2.1 — worked example

Worked Example 5.2.1

Worked Example 5.2.1 — Flipping a Left Tail into a Right Tail

If the area to the left is 0.0228, what is the area to the right?

Step 1. Name what you were given — 0.0228 is $P(X < x)$ .

Step 2. Subtract from the total area under any normal curve, which is 1.

1 - 0.0228 = 0.9772

Answer: the area to the right is 0.9772.

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§5.2.2 — naming the boundary

Definition 5.2.1: Percentile

Definition 5.2.1 — Percentile of a Normal Distribution

For a normal random variable $X$ and a number $p$ with $0 < p < 100$ , the $p^{\text{th}}$ percentile is the value $k$ for which

P(X < k) = \frac{p}{100}.

$p$ percent of the values are at or below $k$ , and the remaining $100-p$ percent are at or above.

Definition 5.2.1: A percentile is a place on the axis — name the area and the boundary follows.

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§5.2.2 — the same point, a different name

Definition 5.2.2 — Critical Value

A value $k$ located on the $x$ -axis that separates a distribution into a specified area to its left and the remaining area to its right is called a critical value.

Percentile names the point by the share of data below it; critical value names it by the job it does of splitting the curve — the same point on the axis, two vocabularies.

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Context Pause — no calculator in the room

Appendix H has the tables

You can do all of this without technology using the probability tables in Appendix H. The tables carry their own instructions for how to read them, and they are what statisticians used for most of the twentieth century.

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§5.2.2 — deciding which command you need

normalcdf or invNorm?

normalcdf takes a range of $x$ -values and returns the area — value in, probability out.
invNorm takes an area and returns the $x$ -value — probability in, value out.
Ends with a unit — points, hours, centimeters? That is a percentile: run invNorm.
Ends with “probability” or “percent”? Run normalcdf.

Sketch the curve first: shade what the problem describes, then check whether the unknown is the shaded amount or the boundary of the shading.

Using the Normal Distribution · bookSHelf Intro Stats§5.2

Your turn — golf scores

Try It Now 5.2.2

Try It Now 5.2.2

Golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65.

$X \sim N(68, 3)$ . A score in, a probability out — normalcdf.

P(x < 65) = \text{normalcdf}(-10^{99}, 65, 68, 3) = 0.1587

Answer: about a 16% chance.

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§5.2.2 — worked example

Worked Example 5.2.2: Final Exam Scores

Worked Example 5.2.2 — final exam scores, $X \sim N(63, 5)$

Four questions on one distribution — two probabilities, two percentiles.

$P(x > 65) = \text{normalcdf}(65, 10^{99}, 63, 5) = 0.3446$
$P(x < 85) = \text{normalcdf}(0, 85, 63, 5) \approx 1$ — about 100%
90th percentile: $k = \text{invNorm}(0.90, 63, 5) = 69.4$
70th percentile: $k = \text{invNorm}(0.70, 63, 5) = 65.6$

Same mean and standard deviation throughout — only the direction of the question changes.

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Try it in rāSHio — running both directions on one curve

Example 5.2.2 runs both directions on one distribution, and rāSHio runs them the same two ways. Choose Distributions → Normal, set mean 63 and standard deviation 5, then pick Right with cutoff 65 for part a’s 0.3446, and switch to Inverse (Left) with probability 0.90 for part c’s $k = 69.4$ .

Using the Normal Distribution · bookSHelf Intro Stats§5.2

Figure 5.2.4

Reading a tail area in rāSHio

Figure 5.2.4: Reading a tail area in rāSHio — Distributions → Normal, with the mean and standard deviation set by hand.

Using the Normal Distribution · bookSHelf Intro Stats§5.2

§5.2.3 — shading a strip instead of a tail

Between questions and quartiles

A “between” question shades a strip in the middle — normalcdf already takes a lower and an upper bound, so supply both.
Read the sentence first: which number is the lower edge, which is the upper? A swapped pair returns a negative area with no warning.
Quartile language translates to a percentile: bottom quartile → 0.25, median → 0.50, third quartile → 0.75.
Watch for “maximum” — it is the tell that a quartile question wants a value (invNorm), not a probability.

Using the Normal Distribution · bookSHelf Intro Stats§5.2

Context Pause — the strip is the answer

How tall is the pile of density?

A between-question asks how much of the population sits inside a window. “How many households use the computer between 1.8 and 2.75 hours” is the same question as “how tall is the pile of the density between those two marks”.

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Your turn — a between question

Try It Now 5.2.3

Try It Now 5.2.3

Golf scores: $X \sim N(68, 3)$ . Find the probability that a golfer scored between 66 and 70.

P(66 < x < 70) = \text{normalcdf}(66, 70, 68, 3) = 0.4950

Answer: just under half the team’s scores land in that four-point window.

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§5.2.3 — worked example

Worked Example 5.2.3: Time Spent on a Home Computer

Worked Example 5.2.3 — entertainment time, $X \sim N(2, 0.5)$ hours

a. Find $P(1.8 < x < 2.75)$ . b. Find the maximum number of hours per day for the bottom quartile of households.

a. $P(1.8 < x < 2.75) = \text{normalcdf}(1.8, 2.75, 2, 0.5) = 0.5886$

b. The bottom quartile’s maximum is the 25th percentile: $k = \text{invNorm}(0.25, 2, 0.5) = 1.66$ hours.

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Try it in rāSHio — a between, then an inverse

The strip in part a is its own mode. In rāSHio choose Distributions → Normal, set mean 2 and standard deviation 0.5, pick Between with bounds 1.8 and 2.75 to get 0.5886, then switch to Inverse (Left) with probability 0.25 for part b’s bottom-quartile 1.66 hours.

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§5.2.4 — a percentile is only half an answer

Write the interpretation sentence

“The 80th percentile is 48.6” tells a reader almost nothing on its own. “80% of smartphone users in this age range are 48.6 years old or younger” tells them everything.

The template: [percent] of the [population] are [value] [units] or [less / more] — fill it in every time a problem asks you to interpret a percentile.

The units come from the mean and standard deviation you were given — never a bare number, never a probability.

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Insight Note — percentile is a rank, not a score

Your position in line, not your grade

Being at the 80th percentile does not mean you got 80% of something right. It means 80% of the group is at or below you. The percentile is your position in the line, not your grade.

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Your turn — smartphone ages

Try It Now 5.2.4

Try It Now 5.2.4

Smartphone user ages (13 to 55+) approximately follow $X \sim N(36.9, 13.9)$ . a. Find the 30th percentile and interpret it. b. Find $P(x < 27)$ .

a. $k = \text{invNorm}(0.30, 36.9, 13.9) = 29.6$ — 30% of users are 29.6 years old or younger.

b. $P(x < 27) = \text{normalcdf}(-10^{99}, 27, 36.9, 13.9) = 0.2382$

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§5.2.4 — worked example

Worked Example 5.2.4: Ages of Smartphone Users

Worked Example 5.2.4 — $X \sim N(36.9, 13.9)$ years

a. Between 23 and 64.7 years old. b. At most 50.8 years old. c. The 80th percentile, interpreted.

a. $\text{normalcdf}(23, 64.7, 36.9, 13.9) = 0.8186$

b. “At most” is a left tail: $\text{normalcdf}(-10^{99}, 50.8, 36.9, 13.9) = 0.8413$

c. $\text{invNorm}(0.80, 36.9, 13.9) = 48.6$ — 80% of users are 48.6 years old or younger.

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§5.2.5 — two more phrasings

The IQR, and “at least”

$IQR = Q_3 - Q_1$ — get $Q_1$ and $Q_3$ from invNorm at 0.25 and 0.75, then subtract.
“At least $k$ ” means $x \geq k$ — the area to the right of $k$ . invNorm only accepts an area to the left.
So subtract first: an area of 0.40 on the right becomes $1 - 0.40 = 0.60$ before you call invNorm.
Skipping the subtraction fails quietly — you get a plausible-looking wrong age, with nothing on the screen to warn you.

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Insight Note — the IQR is a width, not a location

A spread, not a place on the axis

$Q_1$ and $Q_3$ are places on the axis; the distance between them is a spread. Two distributions can share an IQR while sitting nowhere near each other.

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Your turn — exam score quartiles

Try It Now 5.2.5

Try It Now 5.2.5

Exam scores: $\mu = 81$ points, $\sigma = 15$ points. a. Find $Q_1$ and $Q_3$ . b. The middle 50% of scores fall between what two values?

Q_1 = \text{invNorm}(0.25, 81, 15) = 70.9, \qquad Q_3 = \text{invNorm}(0.75, 81, 15) = 91.1

Answer: the middle 50% of the exam scores are between 70.9 and 91.1 points.

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§5.2.5 — worked example

Worked Example 5.2.5: The IQR and an At-Least Question

Worked Example 5.2.5 — $X \sim N(36.9, 13.9)$ years

a. Calculate the IQR. b. Forty percent of smartphone users are at least what age?

a. $Q_3 = 46.2754$ , $Q_1 = 27.5246$ , so $IQR = Q_3 - Q_1 = 18.8$ years.

b. $P(x \geq k) = 0.40$ is a right-hand area, so the left-hand area is $1 - 0.40 = 0.60$ : $\text{invNorm}(0.60, 36.9, 13.9) = 40.4$ years.

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Try it in rāSHio — where the subtraction bites

In rāSHio choose Distributions → Normal, set mean 36.9 and standard deviation 13.9, and run Inverse (Left) at 0.25 and 0.75 for the quartiles (IQR 18.8), then at 0.60 — not 0.40 — for part b’s 40.4 years. Ask it for 0.40 too: the 33.4 it hands back is the below-the-mean answer this subsection warns you to catch.

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§5.2.6 — the middle P%

Splitting the leftover between two tails

If the middle $P$ % sits in the center of the curve, $100 - P$ percent is left over for the two tails.
The normal distribution is symmetric, so those tails are equal — each carries half of what is left.
Lower boundary $k_1$ is the percentile at the tail area; upper boundary $k_2$ is the percentile at the tail area plus the middle chunk.

Middle 20%: $1 - 0.20 = 0.80$ outside, so each tail is 0.40. $k_1$ is the 40th percentile and $k_2$ is the 60th, since $0.40 + 0.20 = 0.60$ .

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Your turn — mandarin orange diameters

Try It Now 5.2.6

Try It Now 5.2.6

Orange diameters: $X \sim N(5.85, 0.24)$ cm. a. The middle 40% falls between what two values? b. Find the 16th percentile.

a. Outside the middle 40% is 0.60, so each tail is 0.30: $k_1 = \text{invNorm}(0.30, 5.85, 0.24) = 5.72$ cm, $k_2 = \text{invNorm}(0.70, 5.85, 0.24) = 5.98$ cm.

b. $\text{invNorm}(0.16, 5.85, 0.24) = 5.61$ cm — 16% of oranges are 5.61 cm or smaller.

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§5.2.6 — worked example

Worked Example 5.2.6: Diameters of Mandarin Oranges

Worked Example 5.2.6 — $X \sim N(5.85, 0.24)$ cm

a. Find $P(x > 6.0)$ . b. The middle 20% falls between what two values? c. Find the 90th percentile.

a. $\text{normalcdf}(6, 10^{99}, 5.85, 0.24) = 0.2660$

b. Each tail carries 0.40: $k_1 = \text{invNorm}(0.40, 5.85, 0.24) = 5.79$ cm, $k_2 = \text{invNorm}(0.60, 5.85, 0.24) = 5.91$ cm.

c. $\text{invNorm}(0.90, 5.85, 0.24) = 6.16$ cm — ninety percent of oranges are at most 6.16 cm.

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Try it in rāSHio — two inverse lookups, not one

Part b’s middle 20% is two inverse lookups, not one. In rāSHio choose Distributions → Normal, set mean 5.85 and standard deviation 0.24, then run Inverse (Left) at 0.40 and again at 0.60 to get the 5.79 cm and 5.91 cm edges — the two percentiles that leave 40% in each tail.

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Key Terms

Four terms from this section

percentile — the value $k$ with a stated percent of the distribution at or below it: $P(X < k) = p/100$ .

critical value — a value on the $x$ -axis that splits a distribution into a specified area to its left and the rest to its right.

normalcdf — lower bound, upper bound, mean, and standard deviation in; the area between the bounds out.

invNorm — area to the left, mean, and standard deviation in; the value $k$ at that percentile out.

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The headline result

Value in, probability out — or the reverse

normalcdf and invNorm are inverses on the same curve: one turns bounds into an area, the other turns an area into a boundary.

$\text{normalcdf}(\text{lower}, \text{upper}, \mu, \sigma) \rightarrow$ a probability, and $\text{invNorm}(\text{area left}, \mu, \sigma) \rightarrow$ a value on the $x$ -axis. Every problem in this section is deciding which direction you need.

† “At least $k$ ” hands you the area on the right. invNorm only takes the area on the left — subtract from 1 before you type. Skipping it is the single most common error in this section, and it fails quietly.

5.2

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§5.2 — conclusions

What §5.2 leaves you with

The core idea

Sketch the region the question describes. If the unknown is the shaded amount, run normalcdf; if the unknown is the boundary, run invNorm. Quartiles, percentiles, and critical values are all the same machinery wearing different vocabulary.

Watch for

“At least” and “at most” can hand you an area on the wrong side — convert to a left-area before invNorm. And a swapped normalcdf bound returns a negative number with no warning, so read the sentence before you type.

Next: §5.3 — Normal Distribution: Lap Times. Back to start.