Wize University Calculus 2 Textbook > Applications of Integrals

Probability

Surface Area of RevolutionPrevious Section

0:00 / 0:00

Random Variables, PDFs, CDFs     
A random variable is a variable whose values depend on the outcomes of a random experiment – we don’t know for sure the probability of a certain outcome occuring, but we can determine the probability for the occurrence of any particular outcome

A continuous random variable can only take on any values within a certain range
Example: height of a randomly selected student at UBC
Example: time it takes for a randomly chosen student to write   their MATH exam
Probability Density Function (PDF) p(x)\color{orange}p\left(x\right)p(x)
Suppose that we have a random variable XXX that can take on any value x∈[a,b]x\in\left[a,b\right]x∈[a,b]
Then the criteria for the probability density function are
p(x)≥0p(x)\ge0p(x)≥0 for all xxx
∫abp(x)dx=1\int_a^bp\left(x\right)dx=1∫ab​p(x)dx=1 (total area under the curve is 1)
→\to→Probability that X takes on a value between a1a_1a1​ and a2a_2a2​ is ∫a1a2p(x)dx\int_{a_1}^{a_2}p\left(x\right)dx∫a1​a2​​p(x)dx

Wize Tip
If the PDF is defined on the interval (−∞,∞)\left(-\infty,\infty\right)(−∞,∞) then ∫−∞∞p(x)dx=1\int_{-\infty}^{\infty}p\left(x\right)dx=1∫−∞∞​p(x)dx=1

Cumulative Distribution Function   (CDF) F(x)\color{orange}F\left(x\right)F(x)
The cumulative function represents the probability that the random variable takes on a value in the range (a,x)(a,x)(a,x)
F(x)=∫axp(s)dsF(x)=\int_a^xp\left(s\right)dsF(x)=∫ax​p(s)ds
F(x)F\left(x\right)F(x) is an increasing, continuous function
By the Fundamental Theorem of Calculus I, F′(x)=p(x)F'\left(x\right)=p\left(x\right)F′(x)=p(x)
By the Fundamental Theorem of Calculus II, F(a2)−F(a1)=∫a1a2p(x)dxF\left(a_2\right)-F\left(a_1\right)=\int_{a_1}^{a_2}p\left(x\right)dxF(a2​)−F(a1​)=∫a1​a2​​p(x)dx
Important properties: F(b)=1F\left(b\right)=1F(b)=1 and F(a)=0F\left(a\right)=0F(a)=0

Wize Tip
If the PDF is defined on the interval (−∞,∞)\left(-\infty,\infty\right)(−∞,∞), then F(−∞)=0F\left(-\infty\right)=0F(−∞)=0 and F(∞)=1F\left(\infty\right)=1F(∞)=1

Practice: PDF  
Given the probability density function p(t)=k(1−t2)p\left(t\right)=k\left(1-t^2\right)p(t)=k(1−t2) for 0≤t≤10\le t\le10≤t≤1,
a. determine the value of kkk such that p(t)p\left(t\right)p(t) is a proper probability density function.
b. what is the probability that t<0.5t<0.5t<0.5?
c. find the cumulative distribution function F(t)F\left(t\right)F(t).

a. k=23k=\frac{2}{3}k=32​
b. P<0.5=34P_{<0.5}=\frac{3}{4}P<0.5​=43​
c. F(t)=2t−2t33F\left(t\right)=\frac{2t-2t^3}{3}F(t)=32t−2t3​

a. k=23k=\frac{2}{3}k=32​
b. P<0.5=114P_{<0.5}=\frac{11}{4}P<0.5​=411​
c. F(t)=2t−t33F\left(t\right)=\frac{2t-t^3}{3}F(t)=32t−t3​

a. k=32k=\frac{3}{2}k=23​
b. P<0.5=1116P_{<0.5}=\frac{11}{16}P<0.5​=1611​
c. F(t)=3t−t32F\left(t\right)=\frac{3t-t^3}{2}F(t)=23t−t3​

a. k=32k=\frac{3}{2}k=23​
b. P<0.5=34P_{<0.5}=\frac{3}{4}P<0.5​=43​
c. F(t)=3t−t32F\left(t\right)=\frac{3t-t^3}{2}F(t)=23t−t3​

None of the above

I don't know

Practice: CDF
Find a, b, and k such that this is a valid cumulative distribution function
F(x)={ax<0k arctan⁡x0≤x<1bx≥1F\left(x\right)=
\begin{cases}
a&x<0\\
k\ \arctan x&0\le x<1\\
b&x\ge1
\end{cases}F(x)=⎩⎨⎧​ak arctanxb​x<00≤x<1x≥1​

a=0, b=0, k=4πa=0,\ b=0,\ k=\frac{4}{\pi}a=0, b=0, k=π4​

a=0, b=1, k=4πa=0,\ b=1,\ k=\frac{4}{\pi}a=0, b=1, k=π4​

a=0, b=0, k=π4a=0,\ b=0,\ k=\frac{\pi}{4}a=0, b=0, k=4π​

a=0, b=1, k=π4a=0,\ b=1,\ k=\frac{\pi}{4}a=0, b=1, k=4π​

a=−1, b=1, k=π4a=-1,\ b=1,\ k=\frac{\pi}{4}a=−1, b=1, k=4π​

I don't know

0:00 / 0:00

Mean, Median, and Variance
Suppose we have a probability density function p(x)p\left(x\right)p(x) defined for a random variable on the interval a≤x≤ba\le x\le ba≤x≤b

Mean (Expected value)
μ=x‾=∫abx p(x)dx\displaystyle \mu=\overline{x}=\int_a^bx\ p\left(x\right)dxμ=x=∫ab​x p(x)dx
*This is like the weighted average

Median
The median xmedx_{med}xmed​ is a value in the interval a≤xmed≤ba\le x_{med}\le ba≤xmed​≤b such that 
∫axmedp(x)dx=∫xmedbp(x)dx=12\displaystyle \int_a^{x_{med}}p\left(x\right)dx=\int_{x_{med}}^bp\left(x\right)dx=\frac{1}{2}∫axmed​​p(x)dx=∫xmed​b​p(x)dx=21​
(i.e. F(xmed)=12F\left(x_{med}\right)=\frac{1}{2}F(xmed​)=21​)

Wize Tip
In a symmetric distribution, the mean and median will be the same, but will likely be different in a non-symmetric distribution.

Variance
The variance VVV is the "average square distance between the mean and the whole distribution"
V=∫ab(x−μ)2p(x)dx\displaystyle V=\int_a^b(x-\mu)^2p(x)dxV=∫ab​(x−μ)2p(x)dx

Standard deviation
The standard deviation σ\sigmaσ is the "average distance between the mean and the whole distribution". It is just the square root of the variance
σ=V\sigma=\sqrt Vσ=V​
Moments

Practice
The flight altitude xxx of a certain bird (measured from the ground) is given by the probability density function p(x)=5e−3xp(x)=5e^{-3x}p(x)=5e−3x for x≥0x\ge0x≥0.
a) What is the average altitute HavgH_{avg}Havg​?
b) What is the probability that a bird's flight altitude is higher than 3?

a) 49\frac{4}{9}94​
b) 53e9\frac{5}{3e^9}3e95​

a) 49\frac{4}{9}94​
b) 53e5−53e9\frac{5}{3e^5}-\frac{5}{3e^9}3e55​−3e95​

a) 59\frac{5}{9}95​
b) 53e9\frac{5}{3e^9}3e95​

a) 59\frac{5}{9}95​
b) 53e5−53e9\frac{5}{3e^5}-\frac{5}{3e^9}3e55​−3e95​

None of the above

I don't know

Practice: Median
Which of the following PDFs has the largest median?

A

B

C

I don't know

Extra Practice

The probability density function for the number of times a child screams during a roller coaster is given by
f(x)={16π(1−cos⁡(2x)),0≤x≤6π0,otherwisef(x) = \left \{ \begin{array}{ll} 
\displaystyle \frac{1}{6 \pi}(1 - \cos(2x)), & 0\leq x \leq 6 \pi \\\\
0, & \text{otherwise}
\end{array}
\right .f(x)=⎩⎨⎧​6π1​(1−cos(2x)),0,​0≤x≤6πotherwise​
Find the mean number of screams over the course of the ride.

Let XXX be a random variable with values that range within [0,2][0, 2][0,2] with pdf
p(x)=38x2p(x) = \frac{3}{8}x^2p(x)=83​x2
Compute the mean, and standard deviation of this distribution.

Let XXX be a continuous random variable with a probability density function
f(x)={ce−4x,0≤x0,otherwisef(x) = \left \{ 
\begin{array}{ll}
ce^{-4x}, & 0 \leq x \\
0, & \text{otherwise}
\end{array}
\right .f(x)={ce−4x,0,​0≤xotherwise​
First, determine the value of ccc that makes the function a PDF. Then determine P(9≤X≤11)P(9 \leq X \leq 11)P(9≤X≤11) .

Wize University Calculus 2 Textbook > Applications of Integrals

Probability

Popular Courses

Random Variables, PDFs, CDFs

Practice: PDF

Practice: CDF

Mean, Median, and Variance

Practice

Practice: Median

Extra Practice