📚 MATH 130 Practice Final

Interactive Walkthrough: Questions 1 – 8

🧠 Question 1: True/False Intuition Check

Click each card to reveal the answer and explanation!

Statement 1

If X is a random variable, then E[aX + b] = aE[X] + b

🔄 Click to reveal

✅ TRUE

Concept: Linearity of Expectation

Intuition: Expectation is an "average"

Scaling (aX): Double values → double average
Shifting (+b): Add 5 to all → average +5

Statement 2

The probability that a continuous random variable equals zero is always zero.

🔄 Click to reveal

✅ TRUE

Concept: Continuous Probability

Intuition: Probability = Area under curve

A single point has no width → no area → P = 0

⚠️ Trap: P=0 doesn't mean "impossible"!

Statement 3

The PDF of a continuous random variable must always be ≤ 1.

🔄 Click to reveal

❌ FALSE

Concept: PDF vs. Probability

Key: PDF = density, not probability!

Counter-example: Uniform on [0, 0.1]
Width = 0.1, Area must = 1
Height = 1/0.1 = 10 > 1 ✓

Statement 4

A standard normal distribution has mean 1 and variance 1.

🔄 Click to reveal

❌ FALSE

Concept: Standard Normal Parameters

Correction: Standard Normal Z has:

μ = 0, σ² = 1

💡 "Standard" = centered at zero, unit variance

📊 Visual Demo: Linearity of Expectation

See how scaling and shifting affects the distribution:

Scale factor (a):

a = 1

Shift (b):

b = 0

E[aX + b] =

Original E[X] = 5 (shown in blue) | Transformed E[aX+b] (shown in purple)

📊 Visual Demo: PDF Can Be > 1

Adjust the interval width to see how PDF height changes:

Uniform distribution on [0, width]:

Narrow (0.1) width = 0.5 Wide (2.0)

PDF Height (density):

2.0

Area (must = 1):

1.0 ✓

📈 Question 2: Finding PDF Parameters

The Problem:

f(x) = a + bx on (0, 2), and E[X] = 5/4. Find a, b and Var(X).

🎯 Strategy

We have 2 unknowns (a and b), so we need 2 equations!

Equation 1: Probability Axiom

Total area under PDF must equal 1

∫₀² f(x)dx = 1

Equation 2: Expectation Definition

Given E[X] = 5/4

∫₀² x·f(x)dx = 5/4

💡 Visual Intuition

f(x) = a + bx is a linear function. We need to find the right a and b!

🔧 Solve the System

Our two equations:

(1) 2a + 2b = 1

(2) 2a + (8/3)b = 5/4

From (1): solve for 2a

2a = 1 - 2b

Substitute into (2):

(1 - 2b) + (8/3)b = 5/4
1 + (2/3)b = 5/4
(2/3)b = 1/4
b = 3/8

Back-substitute for a:

2a = 1 - 2(3/8) = 1 - 3/4 = 1/4
a = 1/8

✅ Solution!

a = 1/8

= 0.125

b = 3/8

= 0.375

Final PDF:

f(x) = 1/8 + (3/8)x

📊 Part (b): Compute Var(X)

Variance Formula:

Var(X) = E[X²] - (E[X])²

We know: E[X] = 5/4

Need: E[X²]

Calculate E[X²]:

E[X²] = ∫₀² x²·f(x) dx

= ∫₀² x²(1/8 + 3x/8) dx

= ∫₀² (x²/8 + 3x³/8) dx

= [x³/24 + 3x⁴/32]₀²

= 8/24 + 48/32

= 1/3 + 3/2 = 11/6

Final calculation:

Var(X) = E[X²] - (E[X])²

= 11/6 - (5/4)²

= 11/6 - 25/16

= 88/48 - 75/48

= 13/48

✅ Final Answer!

Var(X) = 13/48

≈ 0.2708

📋 Summary

1/8

3/8

Var(X)

13/48

🔬 Interactive PDF Explorer

Experiment with different values of a and b to see how they affect the PDF:

Parameter a:

a = 0.125

Parameter b:

b = 0.375

Total Area

1.00

✓ Valid PDF

E[X]

1.25

Target: 1.25

📞 Question 3: Conditional Probability with Poisson

The Problem:

A call center receives an average of 3 wrong calls per day (X ~ Poi(3)).

Given that 1 wrong call has already been received today (X ≥ 1), what is the probability of receiving at least 2 more wrong calls (Total X ≥ 3)?

🎯 What We're Being Asked

Key Insight

We want: P(Total ≥ 3 | Total ≥ 1)

This is a conditional probability question!

Breaking it down:

📌 X = total wrong calls today
📌 X ~ Poisson(λ = 3)
📌 Already know X ≥ 1 (condition)
📌 Want P(X ≥ 3 | X ≥ 1)

📊 Visual: Poisson(3) Distribution

λ = 3 means we expect about 3 calls on average

📊 Poisson PMF Refresher

Poisson PMF Formula:

P(X = k) = e^-λ · λ^k / k!

With λ = 3:

P(X = 0) = e^-3 · 3⁰/0! = e^-3

P(X = 1) = e^-3 · 3¹/1! = 3e^-3

P(X = 2) = e^-3 · 3²/2! = 4.5e^-3

📋 PMF Values Table

k	P(X=k) formula	≈ value
0	e^-3	0.0498
1	3e^-3	0.1494
2	4.5e^-3	0.2240
Sum (0,1,2)	8.5e^-3	0.4232

🧮 Interactive Poisson Calculator

Explore how λ affects the conditional probability:

λ (average rate):

λ = 3

P(X ≥ 3 | X ≥ 1)

60.67%

P(X ≥ 3) unconditional

57.68%

📉 Question 4: Gamma/Exponential PDF

The Problem:

Given f(x) = Cx·e^-x/2 for x > 0

(a) Find the constant C

(b) Find the CDF F(x)

🔍 Pattern Recognition

This is a Gamma Distribution!

The general form is: f(x) ∝ x^α-1e^-λx

Identify Parameters:

f(x) = C·x¹·e^-x/2

Compare to: x^α-1·e^-λx

α - 1 = 1 → α = 2
λ = 1/2

Gamma PDF Normalizing Constant:

C = λ^α / Γ(α)

📊 Visual: The PDF Shape

f(x) = (1/4)x·e^-x/2 peaks at x = 2

🧮 Integration by Parts: Detailed

Integration by Parts Formula:

∫u dv = uv - ∫v du

Setup (LIATE Rule):

u = x	dv = e^-x/2dx
du = dx	v = -2e^-x/2

💡 Tip: Why v = -2e^-x/2?

When integrating e^-x/2, the chain rule gives us -1/2.

So we divide by -1/2, which is multiply by -2!

Step-by-Step Flow

1 ∫x·e^-x/2dx

↓

2 = x·(-2e^-x/2) - ∫(-2e^-x/2)dx

↓

3 = -2x·e^-x/2 + 2·(-2e^-x/2)

↓

4 = -2e^-x/2(x + 2)

📈 Part (b): Find CDF F(x)

CDF Definition:

F(x) = ∫₀^x f(t) dt

Substitute our PDF:

F(x) = ∫₀^x (1/4)t·e^-t/2 dt

Using our integral result:

= (1/4)[-2e^-t/2(t+2)]₀^x

Evaluate bounds:

At t=x: -(1/2)e^-x/2(x+2)

At t=0: -(1/2)e⁰(0+2) = -1

F(x) = -(1/2)e^-x/2(x+2) - (-1)

✅ Final Answer (b):

F(x) = 1 - (1 + x/2)e^-x/2

for x > 0

✓ Quick Verification

F(0) = 1 - 1·1 = 0 ✓

F(∞) = 1 - 0 = 1 ✓

F'(x) = f(x) ✓

🔬 Interactive PDF & CDF Explorer

Toggle between PDF and CDF views:

Mean (E[X])

α/λ = 2/(1/2)

Variance

α/λ² = 2/(1/4)

Mode (peak)

(α-1)/λ = 1/(1/2)

📊 Probability Calculator

Find P(X ≤ x) where x =

P(X ≤ x) =

0.594

P(X > x) =

0.406

Formula

1-(1+x/2)e^-x/2

📏 Question 5: Uniform Distribution & Ratios

The Problem:

A point is chosen at random on a line of length L.

What is the probability that the ratio of the shorter segment to the longer segment is less than 1/4?

🎯 Interactive Line Cutter

Drag the point to cut the line and see the ratio change!

X = 0.25L

(shorter)

L-X = 0.75L

(longer)

📊 Ratio Analysis

Shorter / Longer

0.333

Target: < 1/4

0.25

💡 Key Insight

The ratio is < 1/4 when the cut is very close to either end of the line!

🧮 Algebraic Manipulation

We want:

m/M < 1/4

Step 1: Note that M = L - m

m/M < 1/4

Step 2: Substitute M = L - m

m/(L - m) < 1/4

Step 3: Cross multiply (since L-m > 0)

4m < L - m

Step 4: Solve for m

5m < L → m < L/5

🎯 Key Result:

Shorter segment < L/5

💡 What does this mean?

The cut must be in one of two regions:

Near left end: X < L/5
Near right end: X > 4L/5

Both make the shorter segment < L/5!

📊 Calculate the Probability

Favorable regions:

Left: X ∈ (0, L/5) → length L/5

Right: X ∈ (4L/5, L) → length L/5

Total favorable length:

L/5 + L/5 = 2L/5

Total possible length:

✅ Final Answer:

P = 2/5

= 0.4 = 40%

🎨 Visual: Favorable Regions

Favorable

2L/5 = 40%

Unfavorable

3L/5 = 60%

🔬 Interactive Ratio Explorer

See how different ratio thresholds affect the probability:

Ratio threshold (r): m/M < r

r = 0.25 = 1/4

Shorter must be < this fraction of L:

1/5

Probability:

2/5 = 40%

Green regions show where the cut satisfies the ratio condition

🔔 Question 6: Normal Distribution

The Problem:

X ~ N(500, 20²) — Mean μ = 500, Standard Deviation σ = 20

(a) Find P(X > 530)

(b) Find P(480 < X < 520)

🔑 The Standardization Process

The Z-Score Formula:

Z = (X - μ) / σ

Why Standardize?

📊 Converts any Normal to Standard Normal (μ=0, σ=1)
📋 Allows us to use the Z-table (Φ function)
🔢 Z tells us "how many standard deviations from mean"

Our Parameters:

Mean (μ)

500

Std Dev (σ)

📊 Normal Distribution Visual

X ~ N(500, 400) — The bell curve centered at 500

📐 Part (a): P(X > 530)

Step 1: Standardize

Z = (530 - 500) / 20 = 30/20 = 1.5

Step 2: Rewrite probability

P(X > 530) = P(Z > 1.5)

Step 3: Use complement rule

P(Z > 1.5) = 1 - P(Z ≤ 1.5) = 1 - Φ(1.5)

Step 4: Look up Φ(1.5) in Z-table

Φ(1.5) ≈ 0.9332

✅ Answer (a):

P(X > 530) = 0.0668

≈ 6.68%

🎨 Visual: Shaded Area

Orange area = P(X > 530) ≈ 6.68%

📐 Part (b): P(480 < X < 520)

Step 1: Standardize both bounds

Z₁ = (480 - 500)/20 = -1

Z₂ = (520 - 500)/20 = +1

Step 2: Rewrite probability

P(480 < X < 520) = P(-1 < Z < 1)

Step 3: Calculate using Φ

P(-1 < Z < 1) = Φ(1) - Φ(-1)

💡 Symmetry Trick

Φ(-z) = 1 - Φ(z)

So: Φ(-1) = 1 - Φ(1) = 1 - 0.8413 = 0.1587

Step 4: Calculate

= Φ(1) - (1 - Φ(1)) = 2Φ(1) - 1
= 2(0.8413) - 1 = 0.6826

✅ Answer (b):

P(480 < X < 520) = 0.6826

≈ 68.26%

🎨 Visual: Shaded Area

Purple area = P(480 < X < 520) ≈ 68.26%

📚 The 68-95-99.7 Rule!

This is the famous "empirical rule":

• ≈68% within 1σ of mean ✓
• ≈95% within 2σ of mean
• ≈99.7% within 3σ of mean

📋 Z-Table Reference & Tips

Common Z-Values

Z	Φ(z)	1-Φ(z)
0.0	0.5000	0.5000
0.5	0.6915	0.3085
1.0	0.8413	0.1587
1.5	0.9332	0.0668
1.645	0.9500	0.0500
1.96	0.9750	0.0250
2.0	0.9772	0.0228

🔄 Key Formulas

P(Z ≤ z) = Φ(z)

P(Z > z) = 1 - Φ(z)

Φ(-z) = 1 - Φ(z) (symmetry)

P(a < Z < b) = Φ(b) - Φ(a)

✨ Pro Tips

• Always draw the curve and shade the region!
• Z = 1.645 gives 5% in the tail (one-sided)
• Z = 1.96 gives 2.5% in each tail (95% CI)
• Memorize: Φ(1) ≈ 0.84, Φ(2) ≈ 0.98

🧮 Interactive Normal Calculator

Adjust parameters and see probabilities update in real-time!

Mean (μ):

Std Dev (σ):

Lower bound:

Upper bound:

Z (lower)

-1.00

Z (upper)

+1.00

P(lower < X < upper)

68.27%

⏳ Question 7: Exponential Waiting Time

The Problem:

Bus arrival time X ~ Exp(λ = 1/10). Average wait = 10 minutes.

(a) Find P(X > 15)

(b) Find P(X > 15 | X > 5) — Given you've waited 5 min, prob of waiting 15+ min total?

🎯 Exponential Distribution Basics

Key Facts: X ~ Exp(λ)

• PDF: f(x) = λe^-λx for x ≥ 0

• CDF: F(x) = 1 - e^-λx

• Mean: E[X] = 1/λ

• Variance: Var(X) = 1/λ²

Our Parameters:

1/10

Mean

10 min

💡 Survival Function

For exponential:

P(X > x) = e^-λx

📊 Exponential PDF (λ = 1/10)

The exponential decays quickly — most waits are short!

📐 Part (b): P(X > 15 | X > 5)

Apply memoryless property:

P(X > 15 | X > 5)

= P(X > 5 + 10 | X > 5)

= P(X > 10)

Calculate P(X > 10):

P(X > 10) = e^-(1/10)(10)

= e^-1

✅ Answer (b):

P(X > 15 | X > 5) = e^-1

≈ 0.3679 ≈ 36.79%

📋 Summary

P(X > 15)

e^-1.5 ≈ 22.3%

P(X > 15 | X > 5)

e^-1 ≈ 36.8%

📊 Compare the Probabilities

🤔 Why is (b) > (a)?

Part (b) asks about waiting 10 more minutes (from minute 5).

Part (a) asks about waiting 15 total minutes (from minute 0).

The conditional probability "resets" the clock!

🔬 Interactive Exponential Explorer

Explore the memoryless property with different parameters:

λ (rate parameter):

λ = 0.10

Already waited (s):

s = 5 min

Additional wait (t):

t = 10 min

Mean wait time

10 min

P(X > s+t)

22.31%

e^-1.5

P(X > s+t | X > s)

36.79%

= P(X > t) by memoryless

P(X > t)

36.79%

e^-1

✓ Memoryless Verified: P(X > s+t | X > s) = P(X > t)

🔄 Question 8: Transformation of Variable

The Problem:

Let X ~ Exp(1). Find the PDF of Y = ln(X).

🎯 The CDF Method

General Strategy

When Y = g(X), find PDF of Y by:

Find CDF: F_Y(y) = P(Y ≤ y)
Express in terms of X
Differentiate to get PDF: f_Y(y) = F'_Y(y)

Our Setup:

X ~ Exp(1)

f_X(x) = e^-x for x > 0

F_X(x) = 1 - e^-x for x > 0

Y = ln(X)

💡 Range Check

Since X ∈ (0, ∞):

Y = ln(X) ∈ (-∞, ∞)

The log transforms positive reals to all reals!

📊 Visualizing the Transform

X ~ Exp(1): PDF peaks at 0, decays exponentially

📐 Step 2: Differentiate to Find PDF

PDF = derivative of CDF:

f_Y(y) = d/dy F_Y(y)

Differentiate F_Y(y) = 1 - e^{-e^y}:

f_Y(y) = d/dy (1 - e^{-e^y})

🔧 Chain Rule Setup

Let u = e^y, so we have -e^-u

d/dy(-e^-u) = e^-u · du/dy

= e^{-e^y} · e^y

Simplify:

f_Y(y) = e^y · e^{-e^y}

📝 Chain Rule Visualization

Outer function:

-e^-u → derivative: e^-u

Inner function:

u = e^y → derivative: e^y

e^{-e^y} · e^y

e^{y - e^y}

✅ Final Answer

PDF of Y = ln(X):

f_Y(y) = e^{y - e^y}

for y ∈ (-∞, ∞)

🎓 Fun Fact: This is the Gumbel Distribution!

The Gumbel distribution is used in:

Extreme value theory
Modeling maximum/minimum values
Flood analysis, earthquake magnitudes

📋 Summary of Method:

Start with CDF: F_Y(y) = P(Y ≤ y)
Rewrite in terms of X: P(ln X ≤ y) = P(X ≤ e^y)
Use known CDF: F_X(e^y) = 1 - e^{-e^y}
Differentiate using chain rule

📊 The Gumbel PDF

Notice: Y can be negative (when X < 1)!

Mode

y = 0

Mean

y ≈ -0.577

(Euler's constant γ)

🔬 Interactive Transformation Explorer

See how the exponential distribution transforms under Y = ln(X):

Input: X ~ Exp(1)

Select a value of X:

X = 1.00

Output: Y = ln(X)

Y = ln(1.00) = 0.00

X value

1.00

Y = ln(X)

0.00

f_X(x) = e^-x

0.368

f_Y(y) = e^{y-e^y}

0.368

🔑 Key Relationship:

Notice how probability "mass" is conserved during transformation. The relationship is:

f_Y(y) = f_X(e^y) · |d(e^y)/dy| = e^{-e^y} · e^y

🧠 Question 1: True/False Intuition Check

📊 Visual Demo: Linearity of Expectation

📊 Visual Demo: PDF Can Be > 1

📈 Question 2: Finding PDF Parameters

🎯 Strategy

💡 Visual Intuition

📐 Integral 1: Area = 1

📐 Integral 2: E[X] = 5/4

🔧 Solve the System

📊 Part (b): Compute Var(X)

🔬 Interactive PDF Explorer

📞 Question 3: Conditional Probability with Poisson

🎯 What We're Being Asked

📊 Visual: Poisson(3) Distribution

📐 Conditional Probability Formula

🎨 Venn Diagram View

📊 Poisson PMF Refresher

📋 PMF Values Table

🔢 Final Calculation

📊 Probability Breakdown

🧮 Interactive Poisson Calculator

📉 Question 4: Gamma/Exponential PDF

🔍 Pattern Recognition

📊 Visual: The PDF Shape

📐 Part (a): Find C

🧮 Integration by Parts: Detailed

Step-by-Step Flow

📈 Part (b): Find CDF F(x)

✓ Quick Verification

🔬 Interactive PDF & CDF Explorer

📊 Probability Calculator

📏 Question 5: Uniform Distribution & Ratios

🎯 Interactive Line Cutter

📊 Ratio Analysis

📐 Mathematical Setup

🎨 Visual: Which is shorter?

🧮 Algebraic Manipulation

📊 Calculate the Probability

🎨 Visual: Favorable Regions

🔬 Interactive Ratio Explorer

🔔 Question 6: Normal Distribution

🔑 The Standardization Process

📊 Normal Distribution Visual

📐 Part (a): P(X > 530)

🎨 Visual: Shaded Area

📐 Part (b): P(480 < X < 520)

🎨 Visual: Shaded Area

📋 Z-Table Reference & Tips

🧮 Interactive Normal Calculator

⏳ Question 7: Exponential Waiting Time

🎯 Exponential Distribution Basics

📊 Exponential PDF (λ = 1/10)

📐 Part (a): P(X > 15)

📊 Visual: P(X > 15)

🧠 The Memoryless Property

🎨 Visual Metaphor

📐 Part (b): P(X > 15 | X > 5)

📊 Compare the Probabilities

🔬 Interactive Exponential Explorer

🔄 Question 8: Transformation of Variable

🎯 The CDF Method

📊 Visualizing the Transform

📐 Step 1: Find CDF of Y

🎨 Visual: ln(X) Transformation

📐 Step 2: Differentiate to Find PDF

📝 Chain Rule Visualization

✅ Final Answer

📊 The Gumbel PDF

🔬 Interactive Transformation Explorer

Input: X ~ Exp(1)

Output: Y = ln(X)