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Algebra 2 · 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 (Algebra 1)

Aaliyah scored 1350 on the SAT. Suppose the scores of all test takers form a bell-shaped pattern with a mean of 1,050 and a standard deviation of 200. What percent of test takers did she beat? Heights, battery life, reaction times and test scores all pile up in the same symmetric bell shape. That shape is the normal distribution, and by the end of this page you will answer questions like Aaliyah's in three lines.

Continuous variables and density curves

A continuous random variable can take any value in an interval: the minutes you wait for a food truck, the ounces in a bottle, the inches in a person's height. We can't list every outcome and give each one a probability, so we use a curve instead.

A flat density curve of height 1/5 between 0 and 5 minutes, with the area from 0 to 2 shaded. f(x) = 1/5 P(X < 2) = 0.4 0 2 5 x (min)
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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 normal curve

The most important density curve is the bell-shaped normal curve.

Two bell curves with the same mean: a tall narrow one with a small standard deviation and a short wide one with a large standard deviation. μ small σ: narrow, tall large σ: wide, short

Drag the sliders and watch what changes. The mean slides the curve left and right. The standard deviation squeezes or stretches it. The area stays equal to $1$.

A normal curve with the regions within 1, 2 and 3 standard deviations of the mean marked as about 68%, 95% and 99.7% of the area. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% 95% 99.7%

The interval in the next exercise is exactly one standard deviation on each side of the mean, so the empirical rule gives you a first estimate. You will get the exact value two sections from now.

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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

The standard normal and the z-table

Every context has its own $\mu$ and $\sigma$, and nobody prints a table for each one. The solution is to use one reference curve: the standard normal $Z \sim N(0, 1)$, with mean $0$ and standard deviation $1$. Its areas are tabulated, and your calculator knows them too.

The z-table gives the area to the left of $z$, which is $P(Z < z)$. Here is a mini table with the values you will need on this page. Use the full table from your class or your calculator for anything else.

$z$ $P(Z $z$ $P(Z $z$ $P(Z $z$ $P(Z
0.00 0.5000 0.25 0.5987 0.38 0.6480 0.50 0.6915
0.53 0.7019 0.67 0.7486 0.80 0.7881 0.82 0.7939
0.84 0.7995 1.00 0.8413 1.04 0.8508 1.20 0.8849
1.24 0.8925 1.28 0.8997 1.30 0.9032 1.45 0.9265
1.50 0.9332 1.60 0.9452 1.65 0.9505 1.72 0.9573
1.96 0.9750 2.00 0.9772 2.01 0.9778 2.33 0.9901
2.50 0.9938
The standard normal curve with the area to the left of z shaded; this area is what the z-table lists. area = P(Z < z) 0 z
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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
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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

Areas between two z-values

To find the area between two values, take the big left area and cut off the small left area.

A standard normal curve with the area between z equal to a and z equal to b shaded, with the formula as a difference of two left areas. a b P(a < Z < b) = P(Z < b) − P(Z < a)

The symmetric formula also settles the empirical rule exactly: $P(-1 < Z < 1) = 2(0.8413) - 1 = 0.6826$ and $P(-2 < Z < 2) = 2(0.9772) - 1 = 0.9544$. Those are the "$68\%$" and "$95\%$".

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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
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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

Standardizing: z-scores

Now we connect any normal variable to the standard one.

Diagram showing that the formula z equals (x minus mean) over standard deviation turns any normal variable into the standard normal. X ~ N(μ, σ) Z ~ N(0, 1) z = (x − μ)/σ A z-value says how many standard deviations x is above the mean (z > 0) or below the mean (z < 0).

Why does this work? Subtracting $\mu$ slides the center to $0$. Dividing by $\sigma$ rescales the width so the standard deviation becomes $1$. The shape of the curve is not changed, so areas are preserved.

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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
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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

Normal probabilities in real situations

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

A bell curve for food truck wait times with mean 8 minutes and the area between 6 and 11 minutes shaded. Food truck wait: X ~ N(8, 2) min Find P(6 < X < 11) 6 8 11 z = −1 z = 1.5
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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
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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

Working backward: from an area to $z$ and $x$

Sometimes you know the probability and you want the cutoff. For instance, "How high must you score to be in the top $10\%$?" Now you read the table backward: find the area in the body of the table, then read $z$ from the margins. On a calculator use invNorm(area, μ, σ).

A normal curve with the top 10 percent tail shaded, starting at z equal to 1.28, which is x equal to 628 when the mean is 500 and the standard deviation is 100. top 10% 500 x = 628 z ≈ 1.28
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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
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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
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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

Finding the mean or the standard deviation

Sometimes the missing piece is $\mu$ or $\sigma$ itself. The idea is the same as before: turn the given probability into a $z$-value, then solve a linear equation $z = \frac{x - \mu}{\sigma}$ for the unknown. If both $\mu$ and $\sigma$ are unknown you get two equations, which you solve by elimination.

A normal curve with the lowest 20 percent shaded below x equal to 60 and the highest 10 percent shaded above x equal to 90, used to find the mean and standard deviation. 20% 10% 60 μ ? 90 z = −0.84 z = 1.28
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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
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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
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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

Symmetric intervals around the mean

Quality control often asks for an interval centered at the mean that holds a given percent of the data. Because the curve is symmetric, the two tails are equal.

A normal curve with the central 80 percent shaded between z equal to minus 1.28 and z equal to 1.28, and a tail of 10 percent on each side. central 80% 10% 10% μ − 1.28σ μ μ + 1.28σ
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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
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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
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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

Multi-step cutoffs and conditional probability

The hardest problems combine the tools above. Three patterns show up again and again.

A normal curve split into three regions at the 10th and 80th percentiles: the bottom 10 percent, the middle 70 percent and the top 20 percent. bottom 10% middle 70% top 20% 10th percentile 80th percentile
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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
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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
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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

Normal approximation to the binomial

A binomial variable $X \sim B(n, p)$ counts successes in $n$ independent trials, each with success probability $p$. Its mean is $np$ and its standard deviation is $\sqrt{np(1-p)}$. Adding up dozens of binomial terms by hand is painful, but when $n$ is large the histogram looks like a bell curve, and we can use the normal table instead.

A histogram of a binomial distribution with many trials closely matched by a smooth bell curve. B(n, p) with large n ≈ bell curve of N(np, √(npq)), q = 1 − p

The binomial takes only whole numbers, but the normal curve is continuous. A bar at $k$ of the histogram covers the interval from $k - 0.5$ to $k + 0.5$. The continuity correction makes the normal area match the bars.

Binomial Normal $X'$
$P(X = k)$ $P(k - 0.5 < X' < k + 0.5)$
$P(X \le k)$ $P(X' < k + 0.5)$
$P(X < k)$ $P(X' < k - 0.5)$
$P(X \ge k)$ $P(X' > k - 0.5)$
$P(X > k)$ $P(X' > k + 0.5)$
$P(a \le X \le b)$ $P(a - 0.5 < X' < b + 0.5)$

Some courses skip the continuity correction. These exercises use it, so always include it here.

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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
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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: Aaliyah's score is $1.5$ standard deviations above the mean, so $z = 1.5$ and $P(Z < 1.5) = 0.9332$. She beat about $93\%$ of test takers.

Summary

See all the exercises in this section