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Prep 2 SS · Mathematics
Probability · The normal or Gaussian distribution

Continuous Density and Distribution Functions

Before you start, make sure you know: Perimeters and areas of polygons (Secondary Year 1)
Before you start, make sure you know: Measures of position (High School Year 1 (semesters 1 and 2))

You get to a Metrobús stop where a bus comes by, without fail, every 5 minutes. You don't know when the last one left, so your wait could be any value between 0 and 5 minutes. What is the probability of waiting less than 2 minutes? You can't count cases one by one: there are infinitely many possible times. The answer will be an area.

The density function

When a random variable $X$ is continuous (time, height, weight), probability isn't split up among individual values. Instead we use a curve $f(x)$, the density function, and the probability of an interval is the area under the curve over that interval.

For the Metrobús stop, $f(x)=k$ between $0$ and $5$. The area is a rectangle with base $5$ and height $k$, and it must equal $1$: $5k=1$, so $k=0.2$.

Uniform density between 0 and 5 minutes with height 0.2; the area between 0 and 2 is shaded 0.2 0 2 5 0.4 f(x) = 0.2 between 0 and 5 minutes

The shaded area measures $2\times0.2=0.4$. Back to our opening question: the probability of waiting less than 2 minutes is $0.4$.

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

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

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

  1. a) Determine the value of the constant $k$ so that $f(x)$ is a valid density function.
  1. b) Graph the function $f(x)$ using the value of $k$ you have obtained.

You can do it on paper and send a photo with your answer.

Solve
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Given the following density function of a continuous random variable $X$, representing the waiting time in minutes in a fairground ride queue before getting on: $$f(x) =…

Given the following density function of a continuous random variable $X$, representing the waiting time in minutes in a fairground ride queue before getting on:

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

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 visitor waits less than $2$ minutes.

Waiting Time: Fairground Ride Queue 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, elsewhere PDF Condition: ∫ f(x) dx = 1 ∴ k = ? ● MOVING FORWARD IN THE FAIRGROUND QUEUE...
Solve
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The battery life of a new wireless headphone model follows a uniform distribution between 4 and 10 hours. Its density function is constant in that interval:…

The battery life of a new wireless headphone model follows a uniform distribution between 4 and 10 hours. Its 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

The distribution function

Calculating a new area every time is slow. The distribution function accumulates the area all at once.

Density curve with the area between a and b shaded, which equals F(b) minus F(a) a b F(b) − F(a) = area between a and b

When $f$ is defined piecewise, $F$ is too: on each piece it is the accumulated area, using whatever shape applies (rectangle, triangle, trapezoid).

Some densities are shaped like a trapezoid. You don't need the trapezoid formula: split it into a rectangle and a triangle, or use the complement.

Trapezoid-shaped density: constant from 0 to 3 and decreasing to 0 at x equal to 7 h 0 3 7 rectangle triangle
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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 density function of this continuous variable is given…

The reaction time, in seconds, of a group of teenagers to a visual stimulus in a video game is being studied. The 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 $k$ so that $f(x)$ is a valid 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 between exactly $0.5$ 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, for an update to download on a console follows a continuous distribution whose density function is in the shape of a trapezium. The…

The waiting time, in minutes, for an update to download on a console follows a continuous distribution whose density function is in the shape of a trapezium. 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 $h$ so that it is a 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:

$f(x) = \begin{cases} k \times x & \text{if } 0 \le x < 2 \\ k \times (4 - x) & \text{if } 2 \le x \le 4 \\ 0 & \text{en el resto} \end{cases}$

  1. a) Determine, with justification, the value of the constant $k$ so that $f(x)$ is a density function of a continuous random variable $X$.
  1. b) Once $k$ has been found, calculate the probability that the variable takes a value in the interval $[1, 3]$, that is, $P(1 \le X \le 3)$.
  1. c) Graph the function $f(x)$ and shade in the drawing the area corresponding to the probability calculated in part b).
Solve

The median and conditional probability

For conditional probability we use $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$. With the Metrobús stop: $P(X>3\mid X>1)=\dfrac{P(X>3)}{P(X>1)}=\dfrac{0.4}{0.8}=0.5$.

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Given the density function $f(x) = \frac{1}{8}(x-1)$ defined for the interval $x \in [1, 5]$ and $f(x)=0$ elsewhere: a) Check that the total area under the curve is…

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

  1. a) Check 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 standard normal curve

Many variables (height, grades, the weight of a product) have a bell-shaped density: the normal $N(\mu,\sigma)$, symmetric about its mean $\mu$. Its area can't be found with simple shapes, so we use a table for the standard normal $Z\sim N(0,1)$ (mean $0$, standard deviation $1$).

The table gives $F(z)=P(Z\le z)$, the accumulated area to the left of $z$. You look up the units digit and the first decimal in the row, and the second decimal in the column.

Standard normal bell curve with the area to the left of z shaded, which is the table value Shaded area: F(z) = P(Z ≤ z) 0 z

Some table values: $P(Z\le0)=0.5$, $P(Z\le1)=0.8413$, $P(Z\le1.5)=0.9332$, $P(Z\le1.96)=0.9750$, $P(Z\le2)=0.9772$.

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

Given a random variable $Z$ following a standard normal distribution $N(0, 1)$, use symmetry properties 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

Intervals and the table in reverse

For an interval, you subtract two table values. For example, $P(0.62\le Z\le1.96)=F(1.96)-F(0.62)$.

You can also use the table backwards: you're given the probability and you look for $z$. You find the value closest to the probability inside the table and read $z$ from its row and column.

  • $P(Z\le k)=0.8413\Rightarrow k=1$.

  • $P(Z\le k)=0.9750\Rightarrow k=1.96$.

  • $P(Z\le k)=0.5\Rightarrow k=0$.

If the given probability is less than $0.5$, then $z$ is negative: look up the complement and change the sign.

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Calculate the probability that the standard normal variable $Z$ takes a value lying in the interval $(0.62 \le Z \le 1.96)$. Remember that to calculate the area of an…

Calculate the probability that the standard normal variable $Z$ takes a value lying 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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Use the $N(0, 1)$ distribution table 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 $N(0, 1)$ distribution table 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

Standardizing any normal

The table is only for $N(0,1)$. For a normal $X\sim N(\mu,\sigma)$ we standardize: we measure how many standard deviations $x$ is from the mean.

If you're asked how many items to expect, multiply the probability by the total: with $500$ orders and $P=0.0668$, we expect $500\times0.0668=33.4$, about $33$.

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The scores on an English placement test for Sixth Form students follow a normal distribution with a mean of $\mu = 75$ and a standard deviation of $\sigma = 8$.…

The scores on an English placement test for Sixth Form students follow a normal distribution with a mean of $\mu = 75$ and a standard deviation of $\sigma = 8$. Standardise the following values of 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 bags of chips produced by a machine follows a distribution $N(150, 5)$ grams. Calculate the probability that a randomly chosen bag weighs less than $160$…

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

σ=5 P(X < 160) = P(Z < 2) X (g) 140 145 150 155 160 165 Z −2 −1 0 1 2 3
Solve
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The heights of 11th-grade students in a city follow a normal distribution with a mean of $170$ cm and a standard deviation of $8$ cm. a) What percentage of students are…

The heights of 11th-grade students in a city follow a normal distribution with a mean of $170$ cm and a standard deviation of $8$ cm.

Normal distribution of height N(170, 8) with the probability regions of the exercise X ~ N(170, 8) μ = 170 cm, σ = 8 cm Height (cm) 154 162 170 μ (mean) 178 180 a) 162 ≤ X ≤ 178 (? %) b) X < 154 P(X < 154) = ? c) X > 180 P(X > 180) = ?
  1. a) What percentage of students are between $162$ and $178$ cm tall?
  2. b) If a sports team selection requires a minimum height of $154$ cm, what is the probability that a randomly chosen student does not meet this requirement?
  3. c) What is the probability that a student is taller than $180$ cm?
Solve

Cutoff values and unknown parameters

There are two types of inverse problems:

  • Cutoff value (for example, "the top $5\,\%$"): you look up $z$ in the table and solve for $x=\mu+z\sigma$.

  • Unknown parameters: each piece of information of the form "$x\,\%$ falls below such-and-such a value" gives a $z$ and therefore an equation $\mu+z\sigma=\text{value}$. With two pieces of information, you solve a system of two equations.

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A courier company has analysed the weight of the packages it sends daily. It has been shown that the weight follows a normal distribution. We know that $15.87 \%$ of the…

A courier company has analysed the weight of the packages it sends daily. It has been shown that the weight follows a normal distribution. We know that $15.87 \%$ of the packages weigh less than $2$ kg and that $2.28 \%$ of the packages weigh more than $8$ kg.

  1. a) Determine the mean and standard deviation of the weight of the packages.
  2. b) What is the probability that a randomly chosen package weighs between $3$ and $6$ kg?
Solve
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The daily usage time of a map app on the cell phones of delivery drivers in a city follows a normal distribution $N(\mu, \sigma)$. It is known that $15.87 \%$ of drivers…

The daily usage time of a map app on the cell phones of delivery drivers in a city follows a normal distribution $N(\mu, \sigma)$. It is known that $15.87 \%$ of 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 driver uses the app for between $50$ and $70$ minutes?

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

In a young musical talent competition, the scores follow a normal distribution with a mean of $72$ points. It is known that to obtain an "Honourable Mention", a score is required such that only $10 \%$ of 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 below $60$ points?

Score σ = ? 60 x₂ 72 μ 84.8 x₁ ? % n = ? 10% N(72, σ) Total: 500 part.
Solve

From the binomial to the normal

If $X\sim B(n,p)$ with $n$ large, calculating $P(X=k)$ one by one is tedious. We can approximate it with a normal distribution.

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

An electronic component manufacturing company claims that $4 \%$ of its touchscreens have some factory micro-defect. If a school purchases 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' continuity correction.

Solve

Summary

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