Questions and Answers
Part a) Analysis of Weights
A sample of eleven weights (in kg) is given: 60, 72, 65, 68, 70, 100, 62, 75, 78, 80, 83.
- Calculate Q1, Q2, Q3, and IQR.
- Determine potential outliers using the 1.5 × IQR rule.
Goal: To find the quartiles (spread) and check for any unusual values (outliers).
i) Calculate Q1, Q2, Q3, and IQR
Step 1: Sort the data.
(There are 11 numbers total)
Step 2: Find the Median (Q2).
The exact middle number is the 6th number.
Q2 (Median) = 72
Step 3: Find the First Quartile (Q1).
Q1 is the middle of the lower half (numbers left of 72): 60, 62, 65, 68, 70.
Q1 = 65
Step 4: Find the Third Quartile (Q3).
Q3 is the middle of the upper half (numbers right of 72): 75, 78, 80, 83, 100.
Q3 = 80
Step 5: Calculate Interquartile Range (IQR).
This is the spread of the middle 50%.
IQR = Q3 - Q1 = 80 - 65
IQR = 15
ii) Check for Outliers
The Rule: Values are outliers if they are more than 1.5 × IQR away from Q1 or Q3.
- Fence Distance: 1.5 × 15 = 22.5
- Lower Limit: Q1 - 22.5 = 65 - 22.5 = 42.5
- Upper Limit: Q3 + 22.5 = 80 + 22.5 = 102.5
Conclusion: Looking at our data (60 to 100):
- Is anything below 42.5? No.
- Is anything above 102.5? No. (100 is close, but safe!)
There are no outliers.
Part b) Astronaut Training Times
The following data shows the time (in minutes) it took for a group of astronauts to complete various training simulations: 18, 25, 19, 32, 27, 22, 29, 30, 24, 26.
Calculate the mean and standard deviation of these times.
Goal: Calculate the Mean (average) and Standard Deviation (spread).
(n = 10 numbers)
1. Calculate the Mean
Sum of all numbers = 252
Mean = Sum ÷ Count = 252 ÷ 10
Mean = 25.2 minutes
2. Calculate Standard Deviation
We look at how far each number is from the mean (25.2) and square it.
| Data Point (x) | Distance (x - 25.2) | Squared Distance |
|---|---|---|
| 18 | -7.2 | 51.84 |
| 25 | -0.2 | 0.04 |
| 19 | -6.2 | 38.44 |
| 32 | 6.8 | 46.24 |
| 27 | 1.8 | 3.24 |
| 22 | -3.2 | 10.24 |
| 29 | 3.8 | 14.44 |
| 30 | 4.8 | 23.04 |
| 24 | -1.2 | 1.44 |
| 26 | 0.8 | 0.64 |
| SUM | 189.6 |
Formula Step 1 (Variance): Sum ÷ (n - 1)
189.6 ÷ 9 = 21.067
Formula Step 2 (Standard Deviation): Square root of Variance
√21.067 ≈ 4.59
Standard Deviation = 4.59 minutes
📚 Recommended Reference Material
For a complete beginner, I highly recommend OpenStax Introductory Statistics. It is free, high-quality, and widely used.
- Download PDF for Free: OpenStax Introductory Statistics
- Read: Chapter 2.3 (Mean/Median) and Chapter 2.7 (Standard Deviation/Outliers)
Probability Distributions
Part a) Normal Distribution
Life of an instrument: Mean (μ) = 12 months, Standard Deviation (σ) = 2 months.
- Find probability it lasts less than 7 months.
- Find probability it lasts between 7 and 12 months.
Goal: Use the Z-score formula to standardize values and find probabilities from the Normal Table.
[Image of standard normal distribution curve]i) Probability (Life < 7 months)
Step 1: Calculate the Z-score.
The Z-score tells us how many standard deviations "7" is away from the mean "12".
Z = (7 - 12) / 2
Z = -5 / 2 = -2.5
Step 2: Look up Z = -2.5 in the Standard Normal Table.
The area to the left of Z = -2.5 corresponds to the probability.
ii) Probability (7 < Life < 12 months)
Logic: We know the total area to the left of the mean (12) is 0.5 (50%). We just calculated the small tail area to the left of 7.
To find the area between 7 and 12, we subtract the small tail from the half.
- Probability (X < 12): 0.5000 (The Mean)
- Probability (X < 7): 0.0062 (From part i)
Answer: 49.38% probability.
Part b) Binomial vs. Poisson
Biased coin with p = 0.15 flipped n = 40 times.
Find the probability of getting exactly 5 heads using i) Binomial and ii) Poisson.
i) Binomial Distribution
Formula: P(X=k) = nCk × pk × (1-p)n-k
- n = 40, k = 5, p = 0.15
- (1-p) = 0.85
Calculation:
P(X=5) = 40C5 × (0.15)5 × (0.85)35
P(X=5) = 658,008 × 0.0000759 × 0.0034
ii) Poisson Approximation
Condition: Poisson estimates Binomial when 'n' is large and 'p' is small. We use the mean (λ) as the parameter.
λ = n × p
λ = 40 × 0.15 = 6
Formula: P(X=k) = (e-λ × λk) / k!
P(X=5) = (e-6 × 65) / 120
P(X=5) = (0.002479 × 7776) / 120
Note: The values are close (16.9% vs 16.1%), showing that Poisson is a reasonable approximation here.
📚 Recommended Reference
For probability distributions, OpenStax Statistics is excellent.
- Chapter 6: The Normal Distribution (Z-scores)
- Chapter 4: Discrete Random Variables (Binomial/Poisson)
- View Online Textbook
Naive Bayes Classification
Problem Statement
A company wants to classify incoming emails as "Spam" or "Not Spam" (Ham) using a Naïve Bayes classifier. Given the training dataset below, predict the label for the new email:
"Congratulations! You have won money quickly"
| Email ID | Text (Bag of Words) | Label |
|---|---|---|
| 1 | Congratulations, you have won a free prize | Spam |
| 2 | Monthly meeting scheduled for Monday | Not Spam |
| 3 | Earn money quickly with this simple trick | Spam |
| 4 | Project report attached. Review by Friday | Not Spam |
Goal: Calculate the posterior probability for both classes (Spam and Not Spam) and assign the class with the higher probability.
Step 1: Pre-processing & Priors
First, we calculate the Prior Probabilities based on the document counts.
- Total Documents: 4
- Spam Documents: 2 (Email 1, Email 3)
- Not Spam Documents: 2 (Email 2, Email 4)
P(Not Spam) = 2/4 = 0.5
Step 2: Vocabulary & Word Counts
We tokenize the training data to create a Vocabulary (V) and count words per class.
| Spam Bag of Words | Congratulations, you, have, won, a, free, prize, Earn, money, quickly, with, this, simple, trick |
| Not Spam Bag of Words | Monthly, meeting, scheduled, for, Monday, Project, report, attached, Review, by, Friday |
- Total Words in Spam (NSpam): 14
- Total Words in Not Spam (NNotSpam): 11
- Vocabulary Size (|V|): 25 (14 unique spam words + 11 unique non-spam words)
Step 3: Likelihoods with Laplace Smoothing
Test Sentence Tokens: "Congratulations", "You", "have", "won", "money", "quickly".
Note: We perform case-insensitive matching (e.g., "You" = "you").
Formula: $P(w|Class) = \frac{count(w, Class) + 1}{N_{Class} + |V|}$
Class: Spam
Denominator: $14 + 25 = 39$
- "Congratulations": (1+1)/39 = 2/39
- "You": (1+1)/39 = 2/39
- "have": (1+1)/39 = 2/39
- "won": (1+1)/39 = 2/39
- "money": (1+1)/39 = 2/39
- "quickly": (1+1)/39 = 2/39
Class: Not Spam
Denominator: $11 + 25 = 36$
- "Congratulations": (0+1)/36 = 1/36
- "You": (0+1)/36 = 1/36
- "have": (0+1)/36 = 1/36
- "won": (0+1)/36 = 1/36
- "money": (0+1)/36 = 1/36
- "quickly": (0+1)/36 = 1/36
Step 4: Final Probability Scores
Score(Spam): $0.5 \times (\frac{2}{39})^6$
Calculation: $0.5 \times (0.0513)^6 \approx \mathbf{9.1 \times 10^{-9}}$
Score(Not Spam): $0.5 \times (\frac{1}{36})^6$
Calculation: $0.5 \times (0.0278)^6 \approx \mathbf{2.3 \times 10^{-10}}$
Conclusion
Since the score for Spam is significantly higher than Not Spam, the email is classified as:
[cite_start]SPAM [cite: 1]
Bayes' Theorem: Medical Diagnosis
Problem Statement
A new test is developed to detect a rare genetic disorder. The disorder affects 0.1% of the population. The test has:
- Sensitivity (true positive rate): 98%
- Specificity (true negative rate): 97%
Apply Bayes' theorem to identify the probability that a person has the disorder if their test result is positive.
Goal: Calculate the Posterior Probability, which is P(Disease | Positive Test).
Step 1: Define Events & Probabilities
Let's convert the percentages into probabilities (decimals) and define our terms.
| Term | Symbol | Value |
|---|---|---|
| Prevalence (Prior) | P(Disease) | 0.1% = 0.001 |
| No Disease | P(No Disease) | 1 - 0.001 = 0.999 |
| Sensitivity | P(Pos | Disease) | 98% = 0.98 |
| Specificity | P(Neg | No Disease) | 97% = 0.97 |
P(Pos | No Disease) = 1 - Specificity = 1 - 0.97 = 0.03
Step 2: Apply Bayes' Theorem
The formula for the probability of having the disease given a positive test is:
P(Pos)
Expand the denominator P(Pos):
P(Pos) = [True Positives] + [False Positives]
P(Pos) = [P(Pos|Disease) × P(Disease)] + [P(Pos|No Disease) × P(No Disease)]
Step 3: Calculation
1. Numerator (True Positives)
0.98 × 0.001
= 0.00098
2. False Positives
0.03 × 0.999
= 0.02997
3. Final Division
Total Probability of Positive Test = 0.00098 + 0.02997 = 0.03095
P(Disease | Pos) = 0.00098 ÷ 0.03095
≈ 0.03166
Final Answer
The probability that a person actually has the disorder given a positive test result is approximately:
3.17%
(Note: This low probability, despite high sensitivity/specificity, is due to the "Base Rate Fallacy" — the disease is extremely rare.)
Discrete Random Variables
Q5) Probability Mass Function
Given: A machine learning model predicts the number of errors (X) in a document. X takes values {0, 1, 2, 3} with the following PMF:
- P(X=0) = k
- P(X=1) = 2k
- P(X=2) = 3k
- P(X=3) = 4k
- P(X=other) = 0
a) Find the value of k
Rule: The sum of all probabilities in a valid probability mass function must equal 1.
$\sum P(X=x) = 1$
$P(0) + P(1) + P(2) + P(3) = 1$
$k + 2k + 3k + 4k = 1$
$10k = 1$
b) Compute Expectation E(X)
Formula: $E(X) = \sum [x \cdot P(X=x)]$
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.1 | 0 |
| 1 | 0.2 | 0.2 |
| 2 | 0.3 | 0.6 |
| 3 | 0.4 | 1.2 |
Total Sum = $0 + 0.2 + 0.6 + 1.2$
c) Meaning of E(X) in context
The expected value represents the long-term average.
d) Compute Variance(2X + 3)
Step 1: Calculate $E(X^2)$
$E(X^2) = \sum [x^2 \cdot P(x)]$
- $0^2(0.1) = 0$
- $1^2(0.2) = 0.2$
- $2^2(0.3) = 4(0.3) = 1.2$
- $3^2(0.4) = 9(0.4) = 3.6$
$E(X^2) = 0 + 0.2 + 1.2 + 3.6 = \mathbf{5.0}$
Step 2: Calculate Variance of X, $Var(X)$
Formula: $Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = 5.0 - (2.0)^2$
$Var(X) = 5.0 - 4.0 = \mathbf{1.0}$
Step 3: Apply Variance Property
Property: $Var(aX + b) = a^2 Var(X)$
Here, $a = 2$ and $b = 3$.
$Var(2X + 3) = 2^2 \times Var(X)$
$= 4 \times 1.0$
Joint Probability & Expectation
Q6) Joint Probability Distribution
Experiment: A fair coin is tossed 3 times.
- X: Random variable for the number of tails.
- Y: Random variable for winnings based on the position of the 1st tail.
Rules for Y:
- 1st Tail on Toss 1: Win $3 (Y=3)
- 1st Tail on Toss 2: Win $2 (Y=2)
- 1st Tail on Toss 3: Win $1 (Y=1)
- No Tail: Lose $2 (Y=-2)
Step 1: Analyze Sample Space
There are $2^3 = 8$ possible outcomes. Let's list them and determine X and Y for each.
| Outcome | X (Count Tails) | 1st Tail Pos | Y (Winnings) |
|---|---|---|---|
| H H H | 0 | None | -2 |
| H H T | 1 | 3rd | 1 |
| H T H | 1 | 2nd | 2 |
| H T T | 2 | 2nd | 2 |
| T H H | 1 | 1st | 3 |
| T H T | 2 | 1st | 3 |
| T T H | 2 | 1st | 3 |
| T T T | 3 | 1st | 3 |
i) Joint Probability Function P(X, Y)
We group the outcomes by their (X, Y) pairs. Since each individual outcome has a probability of 1/8:
| X \ Y | -2 | 1 | 2 | 3 | Total P(X) |
|---|---|---|---|---|---|
| 0 | 1/8 (HHH) | 0 | 0 | 0 | 1/8 |
| 1 | 0 | 1/8 (HHT) | 1/8 (HTH) | 1/8 (THH) | 3/8 |
| 2 | 0 | 0 | 1/8 (HTT) | 2/8 (THT, TTH) | 3/8 |
| 3 | 0 | 0 | 0 | 1/8 (TTT) | 1/8 |
Interpretation: The table shows the probability of obtaining a specific number of tails (X) and a specific winning amount (Y) simultaneously. For example, P(X=2, Y=3) = 2/8 means there is a 25% chance of getting exactly 2 tails where the first tail occurs on the first toss.
ii) Marginal Probability Function for X
This is obtained by summing the probabilities across the rows (as shown in the "Total P(X)" column above).
- P(X=0) = 1/8 = 0.125
- P(X=1) = 3/8 = 0.375
- P(X=2) = 3/8 = 0.375
- P(X=3) = 1/8 = 0.125
iii) Expectation E(X)
Formula: $E(X) = \sum [x \cdot P(X=x)]$
$E(X) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8})$
$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$
$E(X) = \frac{12}{8} = 1.5$
Probability & Conditional Events
Question 2
A study on ethics in the workplace by the Ethics Resource Center and Kronos, Inc., revealed that 35% of employees admit to keeping quiet when they see co-worker misconduct. Suppose 75% of employees who admit to keeping quiet when they see co-worker misconduct call in sick when they are well. In addition, suppose that 40% of the employees who call in sick when they are well admit to keeping quiet when they see co-worker misconduct. If an employee is randomly selected, determine the following probabilities:
- The employee calls in sick when well and admits to keeping quiet when seeing co-worker misconduct.
- The employee admits to keeping quiet when seeing co-worker misconduct or calls in sick when well.
- Given that the employee calls in sick when well, he or she does not keep quiet when seeing co-worker misconduct.
- The employee neither keeps quiet when seeing co-worker misconduct nor calls in sick when well.
- The employee admits to keeping quiet when seeing co-worker misconduct and does not call in sick when well.
- Event Q: Employee keeps quiet about misconduct. ($P(Q) = 0.35$)
- Event S: Employee calls in sick when well.
- Conditional 1: "75% of employees who keep quiet... call in sick" $\rightarrow P(S|Q) = 0.75$
- Conditional 2: "40% of employees who call in sick... keep quiet" $\rightarrow P(Q|S) = 0.40$
a) P(S and Q)
This asks for the intersection: The employee calls in sick AND keeps quiet.
Formula: $P(S \cap Q) = P(S|Q) \times P(Q)$
Calculation: $0.75 \times 0.35 = \mathbf{0.2625}$
We need the total probability of calling in sick $P(S)$ for the next steps.
Since $P(Q|S) = \frac{P(S \cap Q)}{P(S)}$, we can rearrange to find $P(S)$:
$P(S) = \frac{P(S \cap Q)}{P(Q|S)} = \frac{0.2625}{0.40} = \mathbf{0.65625}$
b) P(Q or S)
This asks for the union: Keeps quiet OR calls in sick.
Formula: $P(Q \cup S) = P(Q) + P(S) - P(Q \cap S)$
Calculation: $0.35 + 0.65625 - 0.2625 = \mathbf{0.74375}$
c) P(not Q | S)
Given they called in sick ($S$), what is the probability they do NOT keep quiet ($Q'$)?
Formula: $P(Q'|S) = 1 - P(Q|S)$
Calculation: $1 - 0.40 = \mathbf{0.60}$
d) P(not Q and not S)
Neither keeps quiet nor calls in sick. This is the complement of the union ($Q \cup S$).
Formula: $P(Q' \cap S') = 1 - P(Q \cup S)$
Calculation: $1 - 0.74375 = \mathbf{0.25625}$
e) P(Q and not S)
Admits to keeping quiet ($Q$) but does NOT call in sick ($S'$). This is "Q only".
Formula: $P(Q \cap S') = P(Q) - P(Q \cap S)$
Calculation: $0.35 - 0.2625 = \mathbf{0.0875}$
📚 Recommended Reference
For conditional probability and Bayes' theorem, OpenStax Statistics is a great resource.
- Chapter 3: Probability Topics (Conditional Probability, Contingency Tables)
- View Online Textbook
Probability & Statistics Analysis
Part a) Poisson Distribution
Based on past experience, it is assumed that the number of flaws per foot in rolls of grade 2 paper follows a Poisson distribution with a mean of 0.2 flaw per foot. What is the probability that in a:
- 1-foot roll, there will be at least 2 flaws?
- 50-foot roll, there will be greater than or equal to 5 flaws and less than or equal to 8 flaws
i) 1-foot roll, at least 2 flaws
Parameter: Since the roll is 1 foot long, the mean ($\lambda$) remains 0.2.
Goal: Find $P(X \ge 2)$. Using the complement rule: $1 - P(X < 2) = 1 - [P(0) + P(1)]$
Formula: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$
- $P(0) = \frac{e^{-0.2} (0.2)^0}{0!} = 0.8187$
- $P(1) = \frac{e^{-0.2} (0.2)^1}{1!} = 0.1637$
Probability: 0.0176 (or 1.76%)
ii) 50-foot roll, 5 to 8 flaws
Parameter: For 50 feet, the new mean is $\lambda = 0.2 \times 50 = 10$.
Goal: Find $P(5 \le X \le 8)$. This is the sum of probabilities for X = 5, 6, 7, and 8.
Using cumulative probability logic: $P(X \le 8) - P(X \le 4)$
Probability: 0.3036 (or 30.36%)
Part b) Law of Total Probability
An individual has 3 different email accounts. Most of her messages, in fact 70%, come into account #1, whereas 20% come into account #2 and the remaining 10% into account #3. Of the messages into account #1, only 1% are spam, whereas the corresponding percentages for accounts #2 and #3 are 2% and 5%, respectively. What is the probability that a randomly selected message is spam?
Calculation
We use the Law of Total Probability to find the overall probability of Spam ($S$).
Let $A_i$ be the event the message is from Account $i$.
- Account 1: $P(S|A_1)P(A_1) = 0.01 \times 0.70 = 0.007$
- Account 2: $P(S|A_2)P(A_2) = 0.02 \times 0.20 = 0.004$
- Account 3: $P(S|A_3)P(A_3) = 0.05 \times 0.10 = 0.005$
$0.007 + 0.004 + 0.005 = 0.016$
Probability: 0.016 (or 1.6%)
📚 Recommended Reference Material
For Poisson distributions and Total Probability, OpenStax Introductory Statistics is an excellent free resource.
- Chapter 4.6: Poisson Distribution - Explains the formula for events occurring over time/space.
- Chapter 3.2: Independent and Mutually Exclusive Events (Total Probability rules).
- View Online Textbook (Poisson)
Joint & Hypergeometric Distributions
4. a) Joint Probability Mass Function
Given: Joint PMF $P(X=x, Y=y) = k(x^2 + y^2)$.
Note: The specific range for x and y was missing from your prompt. I will solve this assuming the standard small range $x \in \{1, 2\}$ and $y \in \{1, 2\}$ to demonstrate the method.
i) Find the value of constant k
Rule: The sum of all probabilities in a joint distribution must equal 1.
$\sum \sum P(x, y) = 1$
Using range $x \in \{1, 2\}, y \in \{1, 2\}$:
1 = k(1²+1²) + k(1²+2²) + k(2²+1²) + k(2²+2²)
1 = k(2) + k(5) + k(5) + k(8)
1 = 20k
Answer: $k = \frac{1}{20}$ (or 0.05)
ii) Find Marginal Probability Functions
For X (Sum over Y):
- $P_X(1) = P(1,1) + P(1,2) = 2k + 5k = 7k = \mathbf{7/20}$
- $P_X(2) = P(2,1) + P(2,2) = 5k + 8k = 13k = \mathbf{13/20}$
For Y (Sum over X):
- $P_Y(1) = P(1,1) + P(2,1) = 2k + 5k = 7k = \mathbf{7/20}$
- $P_Y(2) = P(1,2) + P(2,2) = 5k + 8k = 13k = \mathbf{13/20}$
4. b) Hypergeometric Distribution
Given: Lot of 10 items, 3 defective. Sample of 4 drawn without replacement.
Find: The probability distribution of X (number of defectives).
iii) Probability Calculation
Parameters: Total ($N=10$), Defective ($K=3$), Good ($7$), Sample ($n=4$).
Formula: $P(X=x) = \frac{\binom{K}{x} \binom{N-K}{n-x}}{\binom{N}{n}}$
Denominator (Total Ways): $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
| X (Defectives) | Calculation | Probability |
|---|---|---|
| 0 | $\frac{\binom{3}{0}\binom{7}{4}}{210} = \frac{1 \times 35}{210}$ | 0.1667 (1/6) |
| 1 | $\frac{\binom{3}{1}\binom{7}{3}}{210} = \frac{3 \times 35}{210}$ | 0.5000 (1/2) |
| 2 | $\frac{\binom{3}{2}\binom{7}{2}}{210} = \frac{3 \times 21}{210}$ | 0.3000 (3/10) |
| 3 | $\frac{\binom{3}{3}\binom{7}{1}}{210} = \frac{1 \times 7}{210}$ | 0.0333 (1/30) |
📚 Recommended Reference Material
For Joint distributions and Sampling Without Replacement, refer to OpenStax Introductory Statistics.
- Chapter 4.4: Geometric and Hypergeometric Distributions.
- View Online Textbook
Weather Prediction Analysis
Q5) If the weather is Snowy...
Question: If the weather is Snowy, then the player will play or not?
Step 1: The Full Dataset
Here are all the parameters provided in your image. We combine the left and right columns into a single list to see the total history.
| # | Weather | Player Play? |
|---|---|---|
| 1 | Sunny | Yes |
| 2 | Rainy | No |
| 3 | Cloudy | Yes |
| 4 | Sunny | No |
| 5 | Sunny | Yes |
| 6 | Snowy | No |
| 7 | Rainy | No |
| 8 | Cloudy | Yes |
| 9 | Sunny | Yes |
| 10 | Snowy | No |
| 11 | Cloudy | Yes |
| 12 | Rainy | No |
| 13 | Snowy | No |
| 14 | Snowy | Yes |
Step 2: Filter for "Snowy" Condition
The question asks specifically about the condition "If the weather is Snowy". Therefore, we filter the dataset to look only at the highlighted rows where Weather = Snowy. The Sunny, Rainy, and Cloudy days are ignored for this specific calculation because they are not the current weather condition.
| Filtered Instances | Weather | Player Play? |
|---|---|---|
| 6 | Snowy | No |
| 10 | Snowy | No |
| 13 | Snowy | No |
| 14 | Snowy | Yes |
Step 3: Calculate Probabilities
From the filtered table above:
- Total "Snowy" days = 4
- Days where Play = "No" = 3
- Days where Play = "Yes" = 1
$P(\text{Play=No} | \text{Snowy}) = \frac{3}{4} = 0.75$ (75%)
$P(\text{Play=Yes} | \text{Snowy}) = \frac{1}{4} = 0.25$ (25%)
Conclusion
Since the probability of No (0.75) is significantly higher than Yes (0.25), the prediction is:
The player will NOT play.
📚 Recommended Reference Material
This problem demonstrates Conditional Probability, often used in Naïve Bayes classifiers.
- OpenStax Introductory Statistics:
- Chapter 3.1: Terminology (Conditionals).
- Chapter 3.2: Independent and Mutually Exclusive Events.
- View Probability Chapter
Bayes' Theorem & Normal Distribution
Part a) Incidence of a Rare Disease
Given:
- Prevalence $P(D) = 1/1000 = 0.001$
- Sensitivity (Positive if Disease) $P(Pos|D) = 0.99$
- False Positive Rate (Positive if Healthy) $P(Pos|D') = 0.02$
Step 1: Calculate Probability of Positive Test
Using the Law of Total Probability:
$P(Pos) = P(Pos|D)P(D) + P(Pos|D')P(D')$
Note: $P(D') = 1 - 0.001 = 0.999$
$P(Pos) = 0.00099 + 0.01998$
$P(Pos) = 0.02097$
Step 2: Apply Bayes' Theorem
We need to find $P(D|Pos)$.
$$P(D|Pos) = \frac{P(Pos|D) \times P(D)}{P(Pos)}$$
$$P(D|Pos) = \frac{0.00099}{0.02097}$$
Part b) Electrical Resistors
Given: Normal Distribution with Mean ($\mu$) = 40 ohms, Standard Deviation ($\sigma$) = 2 ohms.
i) Percentage Exceeding 43 Ohms
Find $P(X > 43)$.
Z-Score Formula: $Z = \frac{X - \mu}{\sigma}$
$Z = \frac{43 - 40}{2} = \frac{3}{2} = 1.5$
From Z-table, area to the left of Z=1.5 is approx 0.9332.
Area to the right = $1 - 0.9332 = 0.0668$
ii) Percentage Exceeding 43 Ohms (Nearest Ohm)
If measured to the nearest ohm, a value "exceeds 43" if the rounded integer is 44 or higher.
The boundary for rounding to 44 starts at 43.5.
So, we calculate $P(X \ge 43.5)$.
Z-Score: $Z = \frac{43.5 - 40}{2} = \frac{3.5}{2} = 1.75$
From Z-table, area to the left of Z=1.75 is approx 0.9599.
Area to the right = $1 - 0.9599 = 0.0401$
📚 Recommended Reference Material
OpenStax Introductory Statistics:
- Chapter 3.2: Conditional Probability and Bayes' Theorem.
- Chapter 6.2: Using the Normal Distribution (Z-scores).
- View Normal Distribution Chapter