Back to homeMathematics · Integrated Math III · Change subject
Start for free
Math III · Mathematics
Probability · The normal or Gaussian distribution

The Normal Distribution

Before you start, make sure you know: Linear equations (Math 8 (Pre-Algebra))
Before you start, make sure you know: Intervals, half-lines and neighborhoods (Integrated Math II)

Jayden scored 84 on a unit test where the class mean was 72 and the standard deviation was 8. His cousin Emma scored 640 on a practice-test section where scores average 500 with a standard deviation of 100. Whose score is more impressive? You can't compare 84 and 640 directly, so you need a common yardstick. By the end of this page you'll have one.

The Bell Curve

Heights of 11th graders, the time a food truck takes to make an order, scores on a standardized test: when you measure many individuals, the values often pile up around an average and thin out symmetrically on both sides. That shape is the bell curve.

First, a reminder of how continuous variables work. A density curve $f(x)$ describes a continuous random variable $X$. The probability that $X$ lands in an interval is the area under the curve over that interval, and the total area is always $1$. A single point has no width, so $P(X = a) = 0$. That is why $P(X \le a)$ and $P(X < a)$ are the same number.

Practice★☆☆Not started
Given the following probability density function of a continuous random variable $X$, which represents the waiting time in minutes for a video game server to assign a…

Given the following probability density function of a continuous random variable $X$, which represents the waiting time in minutes for a video game server to assign a match:

$$f(x) = \begin{cases} k & \text{if } 0 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}$$

Calculate the value of the constant $k$ so that the area under the function is equal to $1$ and graphically find the probability that a player waits less than $2$ minutes.

Waiting Time: Video Game Server Probability Density Function (Continuous Variable X) P(X < 2) = ? Total Area = 1 Time in minutes (x) Density f(x) 0 1 2 3 4 5 k = ? f(x) = k, 0 ≤ x ≤ 4 f(x) = 0, otherwise PDF Condition: ∫ f(x) dx = 1 ∴ k = ? ● SEARCHING FOR MATCH ON SERVER...
Solve

The most important density curve in statistics is the normal curve.

Use the sliders to see what each parameter does. The mean $\mu$ slides the bell left and right. The standard deviation $\sigma$ makes it taller and narrower, or lower and wider.

Bell curve with the 68-95-99.7 rule marked at one, two and three standard deviations from the mean. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ ≈ 68% ≈ 95% ≈ 99.7%

Reading the Z-Table: Left and Right Tails

Every normal distribution can be converted to one reference curve: the standard normal $Z \sim N(0, 1)$, with mean $0$ and standard deviation $1$. Its areas are tabulated.

The standard normal table gives the area to the left of $z$, that is, $P(Z < z)$. To read it, split $z$ into its first decimal (the row) and its second decimal (the column). For $z = 1.15$, go to row $1.1$ and column $0.05$. On a TI-84 the same number is normalcdf(-1E99, 1.15).

Here are a few values for the examples on this page.

z 0 0.50 0.67 0.84 0.88 0.98 1.00 1.15 1.28 1.50 1.645 1.96 2.00 2.58
P(Z < z) 0.5000 0.6915 0.7486 0.7995 0.8106 0.8365 0.8413 0.8749 0.8997 0.9332 0.9500 0.9750 0.9772 0.9951
Standard normal curve with the area to the left of z shaded and the remaining right tail labeled as one minus that area. P(Z < z) 1 − P(Z < z) 0 z The table gives the shaded area, to the LEFT of z
Practice★☆☆Not started
Using the standard normal distribution table $N(0, 1)$, obtain the following direct probabilities: a) $P(Z < 1,45)$ b) $P(Z \le 0,38)$ c) $P(Z < 2,01)$

Using the standard normal distribution table $N(0, 1)$, obtain the following direct probabilities:

  1. a) $P(Z < 1,45)$
  2. b) $P(Z \le 0,38)$
  3. c) $P(Z < 2,01)$
0 z P(Z < z) N(0, 1)
Solve
Practice★☆☆Not started
In a standard normal distribution $Z \sim N(0, 1)$, take advantage of the symmetry properties of the bell curve to calculate the following probabilities: a)…

In a standard normal distribution $Z \sim N(0, 1)$, take advantage of the symmetry properties of the bell curve to calculate the following probabilities:

  1. a) $P(Z > 0,82)$
  2. b) $P(Z \ge 1,96)$
  3. c) $P(Z > 2,5)$

Remember that $P(Z > k) = 1 - P(Z \le k)$.

Solve

Probabilities Between Two Z-Values

Areas between two values are a difference of two left areas: take everything to the left of the upper value and remove everything to the left of the lower value.

Standard normal curve with the area between two values a and b shaded, equal to the left area up to b minus the left area up to a. a b P(a < Z < b) = P(Z < b) − P(Z < a)
Practice★☆☆Not started
Find the probability that a standard normal variable $Z$ is found in the following intervals: a) $P(0,5 < Z < 1,5)$ b) $P(1 \le Z \le 2,33)$ Illustrate part a) by…

Find the probability that a standard normal variable $Z$ is found in the following intervals:

  1. a) $P(0,5 < Z < 1,5)$
  2. b) $P(1 \le Z \le 2,33)$

Illustrate part a) by drawing a sketch of the normal curve and shading the corresponding area.

−2 −1 0 1 2 0,5 1,5 Z N(0, 1) P(0.5 < Z < 1.5)
Solve
Practice★★☆Not started
Given a random variable $Z$ that follows a standard normal distribution $N(0, 1)$, use the probability table to calculate: a) $P(Z < 1,45)$ b) $P(Z > 0,82)$ c)…

Given a random variable $Z$ that follows a standard normal distribution $N(0, 1)$, use the probability table to calculate:

  1. a) $P(Z < 1,45)$
  2. b) $P(Z > 0,82)$
  3. c) $P(Z \le -1,24)$
  4. d) $P(-0,5 < Z < 1,72)$
−3 −2 −1 0 1 2 3 −0,5 1,72 z
Solve

Z-Scores: Standardizing

A table for every possible $\mu$ and $\sigma$ would be endless. Instead, we standardize: convert $X$ to the number of standard deviations it sits from the mean.

A positive $z$ means $x$ is above the mean, and a negative $z$ means it is below. The number tells you how many standard deviations away.

Bell curve for scores with mean 72 and standard deviation 8, showing the x-axis values above the matching z-scores and a score of 84 marked at z equal to 1.5. 48 56 64 72 80 88 96 −3 −2 −1 0 1 2 3 x z x = 84 → z = 1.5
Practice★☆☆Not started
Let $X$ be a random variable that follows a normal distribution with mean $\mu = 50$ and standard deviation $\sigma = 10$. Carry out the standardization process to…

Let $X$ be a random variable that follows a normal distribution with mean $\mu = 50$ and standard deviation $\sigma = 10$. Carry out the standardization process to calculate the probability $P(X < 65)$.

Clearly indicate the value of the transformed variable $Z = \frac{X - \mu}{\sigma}$ before looking it up in the table.

Solve
Practice★☆☆Not started
The scores of an Applied Mathematics exam follow a normal distribution with a mean of $6,2$ points and a standard deviation of $1,5$ points. If we choose a student at…

The scores of an Applied Mathematics exam follow a normal distribution with a mean of $6,2$ points and a standard deviation of $1,5$ points. If we choose a student at random, calculate the probability that their score is higher than $7$.

Solve
Practice★☆☆Not started
The time it takes a courier company to deliver a parcel in a given city follows a normal distribution $N(30, 5)$, where time is measured in minutes. Calculate the…

The time it takes a courier company to deliver a parcel in a given city follows a normal distribution $N(30, 5)$, where time is measured in minutes. Calculate the probability that a parcel takes between $25$ and $35$ minutes to be delivered.

σ=5 30 25 35 Time (min) N(30, 5) P(25≤X≤35)
Solve

Normal Probabilities in Real Situations

When a problem gives you $\mu$ and $\sigma$, the routine is always the same.

Bell curve of order times with mean 7 minutes and standard deviation 2 minutes, with the area above 10 minutes shaded. 1 3 5 7 9 11 13 10 P(X > 10) ≈ 0.0668 min
Practice★★☆Not started
The time that teenagers at a high school spend playing video games daily follows a normal distribution with a mean of $\mu = 110$ minutes and a standard deviation of…

The time that teenagers at a high school spend playing video games daily follows a normal distribution with a mean of $\mu = 110$ minutes and a standard deviation of $\sigma = 25$ minutes. If we select a student at random, calculate the probability that they:

  1. a) Play for more than two hours a day.
  2. b) Play for between 90 and 130 minutes.
  3. c) Play for less than 60 minutes.
VIDEO GAME TIME DISTRIBUTION Continuous Random Variable: X ~ N(110, 25) Mean (μ) = 110 min | Std. Dev. (σ) = 25 min f(x) = 1 σ√(2π) e -(x-μ)² / 2σ² a) More than 2 hours (X > 120 min) 110 120 P(X > 120) = ? b) Between 90 and 130 min (90 < X < 130) 90 130 110 P(90 < X < 130) = ? c) Less than 60 min (X < 60 min) 110 60 P(X < 60) = ?
Solve
Practice★★☆Not started
The scores obtained by 11th grade students in an Applied Mathematics for Social Sciences exam follow a distribution $N(6,4; 1,5)$. a) What percentage of students has…

The scores obtained by 11th grade students in an Applied Mathematics for Social Sciences exam follow a distribution $N(6,4; 1,5)$.

  1. a) What percentage of students has obtained a score higher than 7?
  2. b) If at least a 5 is needed to pass, what is the probability that a student fails?
  3. c) In a class of 30 students, how many are expected to have a score between 5 and 8?
Solve

Using the Table in Reverse

Sometimes you know the area and need the $z$-value. Now you search for the area in the body of the table and read $z$ from the row and column edges. On a TI-84, use invNorm(area).

Some problems call the unknown value the abscissa $k$: that just means its horizontal-axis position, the $z$-value.

Standard normal curve with a left tail of area 0.2514 shaded to the left of an unknown negative value k. area 0.2514 k = ? 0 Area in the body of the table → z on the edges
Practice★☆☆Not started
Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of the abscissa $k$ in the following cases: a) $P(Z < k) = 0,9452$ b)…

Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of the abscissa $k$ in the following cases:

  1. a) $P(Z < k) = 0,9452$
  2. b) $P(Z \le k) = 0,9750$
  3. c) $P(Z < k) = 0,5000$
Solve
Practice★★☆Not started
In the standard normal distribution $Z \sim N(0, 1)$, find the value of the abscissa $k$ in each of the following cases: a) $P(Z < k) = 0,9418$ b) $P(Z > k) = 0,1075$ c)…

In the standard normal distribution $Z \sim N(0, 1)$, find the value of the abscissa $k$ in each of the following cases:

  1. a) $P(Z < k) = 0,9418$
  2. b) $P(Z > k) = 0,1075$
  3. c) $P(Z < k) = 0,2236$
  4. d) $P(-k < Z < k) = 0,95$
Solve

Symmetric Intervals

"The middle $90\%$" or "values within some distance of the mean" describe an interval $-k < Z < k$ centered at $0$.

Standard normal curve with the middle area between minus k and k shaded and two equal tails on either side. middle: p (1−p)/2 (1−p)/2 −k k
Practice★★☆Not started
In a soft drink bottling plant, the content of the cans follows a normal distribution with a mean of $33$ fl oz and a standard deviation of $0,5$ fl oz. a) Calculate the…

In a soft drink bottling plant, the content of the cans follows a normal distribution with a mean of $33$ fl oz and a standard deviation of $0,5$ fl oz.

  1. a) Calculate the probability that a can contains less than $32,2$ fl oz.
  2. b) Determine the interval of values, symmetrical with respect to the mean, in which $90 \%$ of the content of the cans is found.
Can Content Distribution Modelo Normal: X ~ N(33; 0.5) a) Probability that it contains < 32.2 fl oz 33 32,2 P = ? b) Central 90% interval 33 ? ? 90%
Solve
Practice★★★Not started
Let $Z$ be a random variable that follows a standard normal distribution $N(0, 1)$. Calculate the value of the positive constant $a$ such that the following equality…

Let $Z$ be a random variable that follows a standard normal distribution $N(0, 1)$. Calculate the value of the positive constant $a$ such that the following equality holds:

$$P(|Z| \le a) = 3 \cdot P(Z > a)$$

Reason your answer graphically using the symmetry properties of the bell curve.

0 -a a 3A A A N(0, 1) P(Z < -a) = A P(|Z| ≤ a) = 3A P(Z > a) = A
Solve

Percentiles: From a Probability to a Value

"What score do you need to be in the top $10\%$?" asks for the opposite of reading a probability: you are given an area and must find the value $x$.

Bell curve of scores with mean 500 and the top 10 percent tail shaded to the right of an unknown cutoff score. top 10% μ = 500 x = ? left area 0.90 → z₀ ≈ 1.28 → x = μ + 1.28σ

Conditional probabilities

Sometimes you are told something about the value before asking the question: "Given that the order took more than $7$ minutes, what is the probability it took more than $10$?" Use $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$.

Practice★★☆Not started
To access a highly demanded specialized vocational program, a test is taken whose score follows a distribution $N(72, 12)$. If there are only spots for the $25 \%$ of…

To access a highly demanded specialized vocational program, a test is taken whose score follows a distribution $N(72, 12)$. If there are only spots for the $25 \%$ of candidates with the best scores:

  1. a) What is the minimum score required to obtain a spot?
  2. b) If a student has obtained $80$ points, are they in the selected group? Reason your answer by calculating their percentile.
Solve
Practice★★★Not started
The reaction time of players in a professional esports tournament follows a normal distribution $N(180, 25)$, where time is measured in milliseconds. The tournament…

The reaction time of players in a professional esports tournament follows a normal distribution $N(180, 25)$, where time is measured in milliseconds. The tournament organizers want to classify the players into three categories according to their speed:

  • Elite category: the $5\%$ of the fastest players.

  • Amateur category: the $75\%$ of players with intermediate times.

  • Beginner category: the rest of the players (the slowest).

Calculate the reaction times that delimit each of these categories.

t₁ μ = 180 ms t₂ Elite 5% Amateur 75% Beginner 20% N(180, 25)
Solve
Practice★★★Not started
A courier company uses a fleet of electric vans whose battery range is normally distributed with a mean of $210$ miles and a standard deviation of $35$ miles. The…

A courier company uses a fleet of electric vans whose battery range is normally distributed with a mean of $210$ miles and a standard deviation of $35$ miles. The logistics manager decides to renew the batteries of those vans that are in the $12$ percentile of lowest range to avoid incidents on the routes.

  1. a) Determine the maximum range below which the batteries will be renewed.
  2. b) If a van is selected at random and it is known that its range is above the mean, what is the probability that its range exceeds $260$ miles?
Solve

Finding an Unknown Mean or Standard Deviation

Sometimes the missing piece is $\mu$ or $\sigma$. You still standardize, but now you solve for the parameter. Use a probability to find $z_0$ with reverse lookup, then solve the linear equation.

Bell curve with a small right tail of 2.28 percent shaded beyond 12.4 ounces, which is 2 standard deviations above the unknown mean. 2.28% μ = ? 12.4 12.4 lies 2 standard deviations above μ
Practice★★☆Not started
It is known that the weight of newborns in a hospital follows a normal distribution. If $15,87 \%$ of the babies weigh less than $2,8$ pounds and the population standard…

It is known that the weight of newborns in a hospital follows a normal distribution. If $15,87 \%$ of the babies weigh less than $2,8$ pounds and the population standard deviation is $0,4$ pounds, calculate the mean weight of the newborns in that hospital.

Solve
Practice★★☆Not started
A study on the battery life of a new smartphone model indicates that its lifespan follows a normal distribution with a mean of $32$ hours. If it has been found that only…

A study on the battery life of a new smartphone model indicates that its lifespan follows a normal distribution with a mean of $32$ hours. If it has been found that only $2,28 \%$ of the phones have a battery that lasts less than $26$ hours:

  1. a) Find the standard deviation of this distribution.
  2. b) What is the probability that a randomly chosen phone lasts more than $35$ hours?
μ=32h 26h 35h 2,28% P(X>35)=? 2σ 1σ
Solve
Practice★★★Not started
A factory produces bags of coffee whose theoretical weight is $500$ ounces. Due to the machinery, the actual weight follows a normal distribution with a standard…

A factory produces bags of coffee whose theoretical weight is $500$ ounces. Due to the machinery, the actual weight follows a normal distribution with a standard deviation of $12$ ounces. According to consumer regulations, any bag weighing less than $485$ ounces or more than $525$ ounces is considered "out of weight."

Calculate what the mean weight ($\mu$) to which the filling machine must be adjusted should be so that the probability of a bag being "out of weight" by excess (more than $525$ oz) is exactly $2,5\%$. With this new mean, what percentage of bags will weigh less than the legal minimum of $485$ oz?

Solve

Two Unknowns: μ and σ Together

If a problem gives two probability facts and you don't know $\mu$ or $\sigma$, each fact yields one equation $x = \mu + z_0\sigma$. You then have a system of two linear equations, which you can solve by elimination.

Bell curve with a 20 percent left tail shaded below 60 and a 10 percent right tail shaded above 90, used to set up two equations for the mean and standard deviation. 20% 10% 60 90 z = −0.84 z = 1.28
Practice★★★Not started
In a prestigious music academy, it has been observed that the scores obtained by applicants in the music theory test follow a normal distribution $N(\mu, \sigma)$. It is…

In a prestigious music academy, it has been observed that the scores obtained by applicants in the music theory test follow a normal distribution $N(\mu, \sigma)$. It is known that $15\%$ of the applicants obtained a score higher than $82$ points, while only $10\%$ of the students failed to exceed the minimum score of $45$ points.

Calculate the value of the mean ($\mu$) and the standard deviation ($\sigma$) of this distribution of scores.

σ 45 μ 82 10% 15%
Solve
Practice★★★Not started
The volume of soft drink that a vending machine pours into $330$ fl oz cups follows a normal distribution $N(\mu, \sigma)$. It has been found that $95\%$ of the cups…

The volume of soft drink that a vending machine pours into $330$ fl oz cups follows a normal distribution $N(\mu, \sigma)$. It has been found that $95\%$ of the cups contain an amount of soft drink between $318$ fl oz and $338$ fl oz.

Determine the value of the mean ($\mu$) and the standard deviation ($\sigma$) of the volume of soft drink poured by the machine, assuming that the given interval is symmetrical with respect to the mean.

μ 318 338 95% d d
Solve

The Normal Approximation to the Binomial

A social media post goes viral with probability $0.004$. Out of $2{,}500$ posts, how likely are exactly $12$ viral ones? Computing $\binom{2500}{12}(0.004)^{12}(0.996)^{2488}$ is painful, and adding up many such terms is worse. When $n$ is large, the binomial histogram looks like a bell, so we can use a normal curve instead.

Histogram of a binomial distribution with a normal bell curve drawn over its bars. Bars: binomial Curve: normal B(n, p) with large n ≈ N(np, √(npq))

The binomial counts whole numbers, but the normal curve is continuous. The bar for the integer $k$ covers the interval from $k - 0.5$ to $k + 0.5$, so we move $0.5$ to include or exclude the whole bar. This is called the continuity correction.

A single binomial bar for the integer k covering the interval from k minus 0.5 to k plus 0.5 on a continuous axis. The bar for k becomes the interval [k − 0.5, k + 0.5] k k−0.5 k+0.5
Binomial question Normal question (with correction)
$P(X = k)$ $P(k - 0.5 < X' < k + 0.5)$
$P(X \le k)$ $P(X' < k + 0.5)$
$P(X < k)$ $P(X' < k - 0.5)$
$P(X \ge k)$ $P(X' > k - 0.5)$
$P(X > k)$ $P(X' > k + 0.5)$
$P(a \le X \le b)$ $P(a - 0.5 < X' < b + 0.5)$

Here $X'$ is the normal variable with $\mu = np$ and $\sigma = \sqrt{npq}$.

Practice★★★Not started
On a microblogging social network, the probability of a published message going "viral" (more than $10.000$ shares) is $0,004$. A data analyst studies a random sample of…

On a microblogging social network, the probability of a published message going "viral" (more than $10.000$ shares) is $0,004$. A data analyst studies a random sample of $2.500$ messages published in one day.

Using the normal approximation to the binomial distribution (with the corresponding continuity correction), calculate the probability that in this sample:

  1. a) There are exactly $12$ viral messages.
  2. b) There are between $8$ and $15$ viral messages (inclusive).
Solve
Practice★★★Not started
It is known that the weight of newborn babies in a region follows a normal distribution with a mean of $3,25$ pounds and a standard deviation of $0,6$ pounds. A private…

It is known that the weight of newborn babies in a region follows a normal distribution with a mean of $3,25$ pounds and a standard deviation of $0,6$ pounds. A private clinic considers a newborn to have "low weight" if their weight is less than $2,5$ pounds.

  1. a) If $150$ babies are born in that region in a month, what is the probability that more than $35$ of them have low weight? (Use the normal approximation).
  2. b) What should the maximum weight be so that only the $5\%$ of babies with the highest weight in the region are considered in a "special monitoring" category?
Solve

Back to the start. Jayden's z-score of $1.5$ beats Emma's $1.4$, so relative to their groups Jayden did slightly better. Standardizing turned two scores on different scales into the same yardstick.

Summary

See all the exercises in this section