IEE380 Cheatsheet 1 - cheat sheet for exam 1 Probability and Statistics for Engineering Problem Solving (Arizona State University)

IEE380 Cheatsheet 1 - cheat sheet for
exam 1


Probability and Statistics for
Engineering Problem Solving (Arizona
State University)

, RK IEE380 Probability and Statistics Cheat Sheet 1
1. Binomial Distribution 11. T-Distribution & Residuals 13. Working with CDFs (Cumula-
Identify: Fixed n, success probability p, count x. tive Distribution Functions)
T-distribution: Used when population standard
n x n−x
deviation is unknown and sample size is small. Key Rule: If you’re given a CDF, you do not
PMF: P (X = ·x) = x p (1 − Defined by degrees of freedom df = n − 1.
Mean: µ = n p need to integrate. Just evaluate F (x2) − F
p)
Variance: σ2 = n ·p · (1 − p) Calculator: (x1) to get P (x1 < X < x2). Only integrate if
Calculator: • tcdf(lower, upper, df): probability area you’re given a PDF.
• binompdf(n, p, x) for P (X = x) Example Problem: The CDF for a door hinge
• invT(p, df ): t-score for left-tail area p length X (in mm) is:
• binomcdf(n, p, x) for P (X ≤ x) Example: Find P (T ≤ 1.2) with df = ,
, 0 x < 74.5
4:
tcdf(-1E99, 1.2, 4) ⇒ 0.8518
1
2. Poisson Distribution Residual Definition: F (x) (x − 74.5) 74.5 ≤ x ≤ 77.5
= ,3
,
Residual = yobserved — ypredicted 1 x > 77.5

Identify: Random events over time/space, rate What is the probability that the hinge is between
λ. Residuals should follow a normal distribution with 74.1 and 75.2 mm?
x −λ
xe mean 0.
PMF: P (X = x) = λ ! Solution:
Mean/Variance: µ = σ2 =√λ
P (74.1 < X < 75.2) = F (75.2) − F (74.1)
Standard Deviation: σ = λ 12. Poisson: Expected Value Over
Calculator: Interval F (74.1) = 0 (since x < 74.5)

• poissonpdf(λ, x) for P (X = x) 1 0.7
0.2333
Expected Value: For Poisson rate λ over t units, F (75.2) = (75.2 − 74.5) ≈
• poissoncdf(λ, x) for P (X ≤ x) expected value is: = 3 3
So the final probability is: 0.2333
E[X] = λ · t Common Probability Calculations:
3. Continuous Uniform Distribution Example: If a textbook has 0.04 errors/page and
• (a) P (X ≤ 1) = P (X = 1)
204 pages: • (b) P (X > 1) = P (X = 2) + P (X = 3)
Identify: Equal likelihood over [a, b].
• (c) P (2 < X < 8) = P (X = 3) since the
PDF: f (x) = b−a1
,a≤x≤b E[X] = 0.04 · 204 = 8.16 errors
support is only {1,2,3}
Mean: µ = a+b
2
2
Do NOT use poissonpdf for expected val- • (d) P (X ≤ 3 or X > 3) = 1, a universal
Variance: σ2 = (b−a)
12
ues. probability rule.
Steps: Key Properties:
• P (x1 ≤ X ≤ x2) = f (x) (x2 − x1) 7. Bayes’ Theorem (Conditional • The total probability must sum to 1.
• P (X = x) = 0 for continuous
Probability) • The probabilities of disjoint events add up.
distributions. Formula: • If the PMF function does not sum to 1, it
x1−a is not a valid PMF.
• P (X ≥ x1) = 1 − F (x1) = 1 − b−a . P (B | A) P (A)
P (A | B) = P (B)
10. Basic Probability Rules
Steps:
4. Exponential Distribution Σ
• P (B) = (over all ways A can happen) 1. Independence: Events A and B are indepen-
Identify: Waiting times; mean life is µ = 1
, so
λ
rate is λ = 1 . dent if:
PDF: f (x) = mean
λe−λx, x > 0 P (A ∩ B) = P (A)P (B)
8. Custom PDFs and CDFs
CDF: F (x) = 1 − e −λx Otherwise, they are dependent.
Mean/Variance: µ = σ2 = λ1 2. Conditional Probability: The probability
Mean: of A given B is:
Calculator: ∫ b
µ= x f (x)
• P (X > k) = e −λk
dx P (A ∩ B)
P (A| B) = P (B)
Variance: a
• P (a ≤ X ≤ b) = F (b) − F
(a)
ln(P ) ∫ b If P (A |B) = P (A), then A and B are indepen-
• Solve for t: t = − σ2 = x2 f (x) dx— µ2
λ dent.
a
3. Complement Rule: The probability of Ac,

5. Exponential + Binomial Combi- PDF/CDF Relationship: the complement of A, is:
nation • f (x) = dxd F (x) P (Ac) = 1 − P (A)
, x
• F (x) = f (t) dt
Identify: If multiple independent trials exist, use a 4. Law of Total Probability: If B 1, B 2, ..., Bn
**Exponential** for individual waiting times, are mutually exclusive and exhaustive, then:
then **Binomial** for the total count. 9. Probability Mass Functions Σn
Steps: (PMFs) P (A) = P (A|Bi)P (Bi)
i=1
• Compute p = P (X < k) = 1 − e −λk
.
Definition: A probability mass function (PMF) 5. Bayes’ Theorem: Given events A and B:
• Let Y ~ Binomial(n, p) to model occur- assigns probabilities to discrete values of a
rences. random variable X, satisfying:
Σk−1 n i | B) = P (B|A)P (A)
P (A
Σ P (B)
• P (Y ≥ k) = 1 − i=0 p (1 − p)n−i. P (X = x) = 1, P (X = x) ≥ 0 for all x
i
x
6. Multiplication Rule: The probability of the
intersection of dependent events:
Example PMF: