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Probability · The normal or Gaussian distribution

Continuous Density Functions and the Normal Distribution

You request a ride and the app says the car will arrive "sometime in the next 12 minutes." What is the probability that it shows up at exactly 5 minutes, not a split second earlier or later? And what is the probability that it arrives between 3 and 8 minutes? Continuous probability gives surprising answers to both.

Density functions: area is probability

A continuous random variable can take any value in an interval, like a waiting time, a height, or a battery life. You can't list all the outcomes, so you describe the variable with a curve called a density function $f(x)$. The probability of landing in an interval is the area under the curve over that interval.

Here is why the rules make sense. The total area is $1$ because the variable has to take some value, so the probabilities add up to 100%. The curve can't go negative because there is no such thing as a negative probability.

The simplest density is a uniform distribution: a flat rectangle, because every value in the interval is equally likely.

Uniform density of height 1/12 on 0 to 12 minutes with the area between 3 and 8 minutes shaded 0 3 8 12 1/12 shaded area = 5 · 1/12 = 5/12 x = wait time (minutes)
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The probability density function of a continuous random variable $X$ representing the waiting time (in minutes) for a match to start on a video game server is defined…

The probability density function of a continuous random variable $X$ representing the waiting time (in minutes) for a match to start on a video game server is defined as:

$f(x) = \begin{cases} k & \text{if } 0 \le x \le 5 \\ 0 & \text{otherwise} \end{cases}$

Determine the value of the constant $k$ for $f(x)$ to be a valid probability density function and graph the function.

1 2 3 4 5 k 0 Time x (min) f(x) Area = 1 5 · k = 1 b = 5
Solve
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The battery life of a new model of wireless headphones follows a uniform distribution between 4 and 10 hours. Its probability density function is constant in that…

The battery life of a new model of wireless headphones follows a uniform distribution between 4 and 10 hours. Its probability density function is constant in that interval: $f(x) = \frac{1}{6}$. Calculate the probability that the battery lasts between 5 and 8 hours, interpreting it as the area of a rectangle.

Solve

Finding k and computing areas

Many density functions are piecewise, made of lines. The graph is then a rectangle, a triangle, or a trapezoid, so you only need geometry. A missing constant $k$ is found from the one condition we know: the total area is $1$.

The plan is the same every time. First sketch the graph, then set the total area equal to $1$ and solve for $k$. Finally write each probability as an area, and use the complement ($1 -$ the other part) when that is shorter.

Some densities have a "tent" shape, with two pieces that meet at a peak.

Tent-shaped density on 0 to 6 with peak 1/3 at x = 3 and the area between 2 and 4 shaded 0 2 3 4 6 peak height 1/3 area = 5/9

For a trapezoid, split it into a rectangle plus a triangle, or use $\tfrac{1}{2}(b_1 + b_2)h$ on the part you need.

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The reaction time, in seconds, of a group of teenagers to a visual stimulus in a video game is being studied. The probability density function of this continuous…

The reaction time, in seconds, of a group of teenagers to a visual stimulus in a video game is being studied. The probability density function of this continuous variable is given by the following expression:

$f(x) = \begin{cases} kx & \text{if } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$

Determine:

  1. a) The value of the constant $k$ for $f(x)$ to be a valid probability density function.
  2. b) The probability that a randomly chosen player has a reaction time greater than $1,5$ seconds.
  3. c) The probability that the reaction time is exactly between $0,5$ and $1$ seconds.
0 0,5 1 1,5 2 2k x (s) f(x) f(x) = kx a) Area = 1 c) P(0.5 < x < 1) b) P(x > 1.5)
Solve
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The waiting time, in minutes, until an update is downloaded on a console follows a continuous distribution whose probability density function has the shape of a…

The waiting time, in minutes, until an update is downloaded on a console follows a continuous distribution whose probability density function has the shape of a trapezoid. The function is defined between $0$ and $10$ minutes. Between minutes $0$ and $4$, the function is constant with a value of $h$, and between minutes $4$ and $10$ it decreases linearly until reaching $0$.

  1. a) Calculate the value of $h$ for it to be a probability density function.
  2. b) Calculate the probability that the update takes less than $6$ minutes to download.
  3. c) If a user has been waiting for more than $4$ minutes, what is the probability that the download finishes before minute $8$?
Solve
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Let $f(x)$ be a piecewise function defined as follows:…

Let $f(x)$ be a piecewise function defined as follows:

$f(x) = \begin{cases} k \cdot x & \text{if } 0 \le x < 2 \\ k \cdot (4 - x) & \text{if } 2 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}$

Determine, justifying your answer, the value of the constant $k$ for $f(x)$ to be a probability density function of a continuous random variable $X$. Once $k$ is found, calculate the probability that the variable takes a value in the interval $[1, 3]$ and represent it graphically.

0 1 2 3 4 x k=1/4 2k=1/2 f(x) f(x) P(1 ≤ X ≤ 3) = 0.75
Solve

The cumulative function and the median

Calculating a new area for every question gets slow. The cumulative distribution function collects the area once.

With $F$, any probability is a subtraction:

$$P(a \le X \le b) = F(b) - F(a) \qquad P(X > a) = 1 - F(a)$$

The area up to $b$ includes the area up to $a$, so you subtract the part you don't want.

The median $M$ splits the area into two equal halves, so it is the solution of $F(M) = 0.5$. For a conditional probability with nested events, use $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$. For example, $P(X > 4 \mid X > 3) = \dfrac{P(X > 4)}{P(X > 3)}$, because "$X > 4$" already implies "$X > 3$".

Triangular density on 0 to 6 with the median at about 4.24 splitting the area into two halves of 0.5 area 0.5 area 0.5 M ≈ 4.24
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Given the probability density function $f(x) = \frac{1}{8}(x-1)$ defined for the interval $x \in [1, 5]$ and $f(x)=0$ elsewhere: a) Show that the total area under the…

Given the probability density function $f(x) = \frac{1}{8}(x-1)$ defined for the interval $x \in [1, 5]$ and $f(x)=0$ elsewhere:

  1. a) Show that the total area under the curve is indeed $1$.
  2. b) Find a value $M$ (median) such that $P(X \le M) = 0,5$.
  3. c) Calculate the conditional probability $P(X > 4 | X > 2)$.
1 2 3 4 5 6 0.1 0.2 0.3 0.4 0.5 x f(x) M Area = 0.5 Area = 0.5
Solve

The normal curve and the z-table

The most important density in statistics is the normal distribution, the bell curve. It describes heights, test scores, measurement errors, and many other things. It is written $N(\mu, \sigma)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. The curve is symmetric about $\mu$, so the mean and median are both at the center, and $\sigma$ controls how spread out it is. The total area is $1$.

Normal curve with the region within one standard deviation of the mean shaded μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% about 68% within 1σ, 95% within 2σ, 99.7% within 3σ

There is no easy area formula for the bell curve, so we use a table (or a calculator). To keep one table for all normal curves, we use the standard normal distribution $Z \sim N(0, 1)$, which has mean $0$ and standard deviation $1$. The z-table gives $P(Z \le z)$, the area to the left of $z$.

Selected values of the standard normal table, $P(Z \le z)$ (a full z-table or your calculator has the rest):

$z$ area $z$ area $z$ area
0.00 0.5000 0.85 0.8023 1.50 0.9332
0.45 0.6736 1.00 0.8413 1.52 0.9357
0.48 0.6844 1.12 0.8686 1.62 0.9474
0.50 0.6915 1.15 0.8749 1.78 0.9625
0.62 0.7324 1.19 0.8830 1.96 0.9750
0.67 0.7486 1.20 0.8849 2.00 0.9772
0.75 0.7734 1.28 0.8997 2.05 0.9798
0.81 0.7910 1.33 0.9082 2.38 0.9913
0.84 0.7995 1.42 0.9222 3.00 0.9987

To read a value such as $z = 1.42$, find the row for $1.4$ and the column for $0.02$. Reading the table backward works too. If you are given the area $0.9332$, look for it in the body of the table, and the row and column give $z = 1.50$.

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Given a standard normal distribution $Z \sim N(0, 1)$, use the probability table to find the following values directly: a) $P(Z \le 0,45)$ b) $P(Z \le 1,2)$ c)…

Given a standard normal distribution $Z \sim N(0, 1)$, use the probability table to find the following values directly:

  • a) $P(Z \le 0,45)$

  • b) $P(Z \le 1,2)$

  • c) $P(Z \le 2,38)$

0 k P(Z ≤ k) Z ~ N(0, 1)
Solve
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Use the table of the $N(0, 1)$ distribution in reverse to find the value of the constant $k$ in the following cases: a) $P(Z \le k) = 0,8413$ b) $P(Z \le k) = 0,9750$ c)…

Use the table of the $N(0, 1)$ distribution in reverse to find the value of the constant $k$ in the following cases:

  • a) $P(Z \le k) = 0,8413$

  • b) $P(Z \le k) = 0,9750$

  • c) $P(Z \le k) = 0,5000$

Solve

Symmetry and areas between values

Since the table only shows areas to the left, you combine it with two facts: the total area is $1$, and the curve is symmetric about $0$.

Standard normal curve with both tails shaded, showing that the left tail below negative a equals the right tail above a P(Z < −a) P(Z > a) −a 0 a Symmetry: the two tails have equal area
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For a variable $Z \sim N(0, 1)$, apply the properties of the normal curve (symmetry and total area) to calculate the following probabilities: a) $P(Z > 1,52)$ b)…

For a variable $Z \sim N(0, 1)$, apply the properties of the normal curve (symmetry and total area) to calculate the following probabilities:

  • a) $P(Z > 1,52)$

  • b) $P(Z < -0,84)$

  • c) $P(Z \ge -1,15)$

Solve
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Calculate the probability that the standard normal variable $Z$ takes a value in the interval $(0,62 \le Z \le 1,96)$. Remember that to calculate the area of an interval…

Calculate the probability that the standard normal variable $Z$ takes a value in the interval $(0,62 \le Z \le 1,96)$. Remember that to calculate the area of an interval you must subtract the cumulative areas: $P(a \le Z \le b) = P(Z \le b) - P(Z \le a)$.

Solve
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Given a random variable $Z$ that follows a standard normal distribution $N(0, 1)$, use the properties of symmetry and the total area to calculate the following…

Given a random variable $Z$ that follows a standard normal distribution $N(0, 1)$, use the properties of symmetry and the total area to calculate the following probabilities:

  1. a) $P(Z < 1,42)$
  2. b) $P(Z > -0,75)$
  3. c) $P(1,12 < Z < 2,05)$
  4. d) $P(-1,5 < Z < 1,5)$
  5. e) Find the value of $k$ such that $P(Z < k) = 0,8830$
Solve

Standardizing with z-scores

Real data rarely have mean $0$ and standard deviation $1$. To use the z-table for $X \sim N(\mu, \sigma)$, convert $x$ to a z-score: how many standard deviations $x$ is from the mean.

Subtracting $\mu$ slides the center to $0$, and dividing by $\sigma$ rescales the spread to $1$. The area under the curve doesn't change.

Number line showing scores 59, 67, 75, 83, 91 matched with z-scores -2, -1, 0, 1, 2 when the mean is 75 and sigma is 8 59 67 75 83 91 −2 −1 0 1 2 Each σ = 8 points is one unit of z (here μ = 75) x z
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The scores of an English proficiency exam for high school seniors follow a normal distribution with mean $\mu = 75$ and standard deviation $\sigma = 8$. Standardize the…

The scores of an English proficiency exam for high school seniors follow a normal distribution with mean $\mu = 75$ and standard deviation $\sigma = 8$. Standardize the following values of the variable $X$ to find their corresponding value in $Z$:

  • a) A score of $83$.

  • b) A score of $67$.

  • c) A score of $75$.

Solve
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The weight of the bags of potato chips produced by a machine follows a distribution of $N(150, 5)$ ounces. Calculate the probability that a randomly chosen bag weighs…

The weight of the bags of potato chips produced by a machine follows a distribution of $N(150, 5)$ ounces. Calculate the probability that a randomly chosen bag weighs less than $160$ ounces. To do this, first standardize the variable and look up the result in the table.

σ=5 P(X < 160) = P(Z < 2) X (oz) 140 145 150 155 160 165 Z −2 −1 0 1 2 3
Solve
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The battery life of a specific model of wireless headphones follows a normal distribution with a mean of $30$ hours and a standard deviation of $4$ hours. a) What…

The battery life of a specific model of wireless headphones follows a normal distribution with a mean of $30$ hours and a standard deviation of $4$ hours.

  1. a) What percentage of headphones have a battery that lasts between $26$ and $34$ hours?
  2. b) If the manufacturer guarantees that the headphones last at least $22$ hours, what is the probability of finding headphones that do not meet the guarantee?
  3. c) What is the probability that the battery lasts more than $35$ hours?
22 26 30 34 35 μ-2σ μ-σ μ μ+σ a) 26 ≤ x ≤ 34 b) x < 22 c) x > 35 N(30, 4) μ=30, σ=4
Solve

Percentages, counts, and cutoff values

Once you can standardize, three kinds of questions follow.

How many? Multiply the group size by the probability. In a group of $n$ people, you expect about $n \cdot P$ to land in the interval.

What interval? A symmetric interval around the mean that holds $95\%$ of the data is $\mu \pm 1.96\sigma$, because $P(-1.96 < Z < 1.96) = 2(0.9750) - 1 = 0.95$.

What cutoff? Work backward in two steps. First find $z$ from the area to the left, using the table in reverse. Then convert back with

$$x = \mu + z\sigma$$ Normal curve with the top 10 percent shaded to the right of z equal to 1.28 top 10% μ x = μ + 1.28σ The area to the left of the cutoff is 0.90

For the top $5\%$ you need an area to the left of $0.95$. That falls halfway between $z = 1.64$ and $z = 1.65$, so use $z \approx 1.645$. For the top $10\%$ (area $0.90$), use $z \approx 1.28$.

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The scores obtained by the students of a school in the college entrance exam follow a normal distribution $N(6,8 ; 1,2)$. a) If accessing a highly demanded major…

The scores obtained by the students of a school in the college entrance exam follow a normal distribution $N(6,8 ; 1,2)$.

  1. a) If accessing a highly demanded major requires a score higher than $8,5$, what percentage of the school's students would have the option to enter?
  2. b) The school decides to award an honorable mention to the $5 \%$ of students with the best scores. From what score is this mention received?
  3. c) What is the probability that a student has a score lower than $5$?
Solve
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In a study on the monthly leisure expenditure of young people in a city, it has been determined that it follows a normal distribution with a mean of $\$80…

In a study on the monthly leisure expenditure of young people in a city, it has been determined that it follows a normal distribution with a mean of $\$80$ and a standard deviation of $\$15$.

  1. a) Calculate the interval centered on the mean that comprises $95 \%$ of the young people.
  2. b) If we select a young person at random, what is the probability that their monthly expenditure does not deviate from the mean by more than $\$10$?
  3. c) If there are $5.000$ young people in the city, how many of them are estimated to spend more than $\$110$ a month?
Solve
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A farmer classifies the oranges from his harvest by their diameter. The diameter follows a normal distribution $N(8 ; 0,6)$ in inches. a) Oranges with a diameter of less…

A farmer classifies the oranges from his harvest by their diameter. The diameter follows a normal distribution $N(8 ; 0,6)$ in inches.

  1. a) Oranges with a diameter of less than $7$ inches are destined for juice manufacturing. What probability does an orange have of ending up in the juice factory?
  2. b) "Extra" oranges are those whose diameter is between $8,5$ and $9,5$ inches. What percentage of the harvest is labeled as "Extra"?
  3. c) If the farmer wants to separate the largest $10 \%$ of the oranges for a luxury export, what must be the minimum diameter of these oranges?
Solve

When μ or σ is unknown

Sometimes the problem tells you a percentage and a value, and you must find $\sigma$ or $\mu$. The z-score formula still holds, but now $z$ comes from the table and the unknown is what you solve for: $z = \dfrac{x - \mu}{\sigma}$. Convert the percentage to a $z$ first, then solve the one equation.

Number line showing the mean 30 minutes at z equal 0, 35 at z equal 1, and 40 at z equal 2, with sigma equal 5 minutes 30 (μ) 35 40 z = 0 z = 1 z = 2 10 = 2σ gives σ = 5 minutes

The same idea finds a mean you can adjust. If a soda machine has $\sigma = 2$ oz and you want only $2.28\%$ of cups to have less than $12$ oz, then $z = -2$ and $12 = \mu - 2(2)$. So $\mu = 12 + 4 = 16$ oz. In general, $\mu = x - z\sigma$.

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In a young musical talent contest, the scores follow a normal distribution with a mean of $72$ points. It is known that to obtain an "Honorable Mention," a score is…

In a young musical talent contest, the scores follow a normal distribution with a mean of $72$ points. It is known that to obtain an "Honorable Mention," a score is required such that only $10 \%$ of the participants achieve it. If the minimum score to receive this mention is $84,8$ points, what is the standard deviation of the scores? If $500$ participants enter, how many are expected to have a score lower than $60$ points?

Score σ = ? 60 x₂ 72 μ 84.8 x₁ ? % n = ? 10% N(72, σ) Total: 500 part.
Solve
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It has been verified that the weight of the $500$-ounce coffee bags leaving a packaging machine follows a normal distribution $N(505, 10)$. Quality control rejects those…

It has been verified that the weight of the $500$-ounce coffee bags leaving a packaging machine follows a normal distribution $N(505, 10)$. Quality control rejects those bags that weigh less than $490$ ounces or more than $525$ ounces.

  1. a) What percentage of bags is rejected by the factory?
  2. b) If the manager decides to adjust the machine so that the mean weight is $\mu$, but keeping the standard deviation $\sigma = 10$, what value should $\mu$ have so that the risk of a bag weighing less than $500$ ounces is only $2,5 \%$?
Solve
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On a streaming platform, the time users watch a show before abandoning it follows a normal distribution. It is known that the mean time is $45$ minutes. If $25 \%$ of…

On a streaming platform, the time users watch a show before abandoning it follows a normal distribution. It is known that the mean time is $45$ minutes. If $25 \%$ of users watch for more than $60$ minutes, calculate the standard deviation. Determine the interval symmetrical with respect to the mean $(45 - d, 45 + d)$ that contains $95 \%$ of the users' viewing times.

Solve

Finding both μ and σ

When both $\mu$ and $\sigma$ are unknown, you need two pieces of information. Each one gives an equation $x = \mu + z\sigma$, and you solve the pair of linear equations with elimination (or substitution).

Number line with 20 hours at z equal minus 1, 30 hours at z equal 0, and 50 hours at z equal 2 20 30 50 z = −1 z = 0 z = 2 1σ 2σ
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A courier company has analyzed the weight of the parcels it sends daily. It has been verified that the weight follows a normal distribution. We know that $15,87 \%$ of…

A courier company has analyzed the weight of the parcels it sends daily. It has been verified that the weight follows a normal distribution. We know that $15,87 \%$ of the parcels weigh less than $2$ pounds and that $2,28 \%$ of the parcels weigh more than $8$ pounds.

  1. a) Determine the mean and standard deviation of the weight of the parcels.
  2. b) What is the probability that a randomly chosen parcel weighs between $3$ and $6$ pounds?
Solve
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The daily usage time of a map application on the cell phones of delivery drivers in a city follows a normal distribution $N(\mu, \sigma)$. It is known that $15,87 \%$ of…

The daily usage time of a map application on the cell phones of delivery drivers in a city follows a normal distribution $N(\mu, \sigma)$. It is known that $15,87 \%$ of the delivery drivers use it for less than $40$ minutes a day, while $2,28 \%$ use it for more than $80$ minutes. Calculate the parameters $\mu$ and $\sigma$ of this distribution. What is the probability that a randomly chosen delivery driver uses the application between $50$ and $70$ minutes?

Solve
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The battery life of a model of wireless headphones follows a normal distribution with a mean of $30$ hours. If it is known that the probability of a battery lasting more…

The battery life of a model of wireless headphones follows a normal distribution with a mean of $30$ hours. If it is known that the probability of a battery lasting more than $34$ hours is the same as the probability of it lasting less than $26$ hours, and both are $0,1587$, calculate the variance of the distribution. If a user needs the headphones to work during a $33$-hour trip without recharging, what is the probability that the battery runs out before reaching their destination?

Solve

Normal approximation to the binomial

A binomial variable counts successes in $n$ independent trials, each with success probability $p$. Its mean is $np$ and its standard deviation is $\sqrt{np(1-p)}$. For a large $n$, adding up binomial probabilities is painful, but the bar graph looks like a bell curve. So we can use $N\!\left(np,\ \sqrt{np(1-p)}\right)$ instead.

A binomial count is a whole number, but the normal curve is continuous. The bar for $X = 12$ really covers the stretch from $11.5$ to $12.5$. The continuity correction means you widen each endpoint by $0.5$ so that you include the whole bar.

Bar graph of a discrete distribution where the bar for 12 spans 11.5 to 12.5 10 11 12 13 14 11.5 12.5 The bar for X = 12 covers 11.5 to 12.5

Some problems call this Yates's correction. It is the same continuity correction.

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An electronic component manufacturing company claims that $4 \%$ of its touch screens have some factory micro-defect. If a school acquires a batch of $250$ screens for…

An electronic component manufacturing company claims that $4 \%$ of its touch screens have some factory micro-defect. If a school acquires a batch of $250$ screens for its classrooms, calculate the probability that:

  1. a) There are exactly $12$ defective screens.
  2. b) There are between $8$ and $15$ defective screens (inclusive).

Note: Justify whether it is possible to approximate the binomial distribution to the normal distribution and apply Yates's continuity correction.

Solve

Back to the start: the chance that the car arrives at exactly 5 minutes is $0$, because a single point has no area. The chance it arrives between 3 and 8 minutes is the area of a rectangle, $\tfrac{5}{12} \approx 0.417$.

Summary

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