IB Mathematics AI – Topic 4

Types of Random Variables:

• Discrete Random Variable: Takes countable values (e.g., 0, 1, 2, 3, ...)
• Continuous Random Variable: Takes any value in an interval (e.g., height, time, temperature)

Notation:

Random variables are typically denoted by capital letters: X, Y, Z

Specific values are denoted by lowercase: x, y, z

Example: \(P(X = 3)\) means "probability that X equals 3"

Properties:

• Each probability must satisfy: \(0 \leq P(X = x_i) \leq 1\)
• Sum of all probabilities equals 1: \(\sum P(X = x_i) = 1\)

Expected Value (Mean):

\[ E(X) = \mu = \sum x_i \cdot P(X = x_i) \]

The long-term average value if the experiment is repeated many times

Variance:

\[ \text{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2 = \sum x_i^2 \cdot P(X = x_i) - \mu^2 \]

Standard Deviation:

\[ \sigma = \sqrt{\text{Var}(X)} \]

⚠️ Common Pitfalls & Tips:

• Always verify that probabilities sum to 1 before proceeding
• Expected value is NOT always a possible outcome (e.g., expected 2.5 children)
• For variance, calculate \(E(X^2)\) first, then subtract \([E(X)]^2\) – don't confuse order
• Standard deviation has same units as X; variance has squared units

Solution:

(a) Finding k:

Since probabilities must sum to 1:

\[ 0.1 + 0.3 + k + 0.2 = 1 \]

\[ 0.6 + k = 1 \]

\[ k = 0.4 \]

(b) Expected Value:

\[ E(X) = \sum x_i \cdot P(X = x_i) \]

\[ E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) \]

\[ E(X) = 0 + 0.3 + 0.8 + 0.6 = 1.7 \]

\(E(X) = 1.7\)

(c) Variance and Standard Deviation:

First, calculate \(E(X^2)\):

\[ E(X^2) = \sum x_i^2 \cdot P(X = x_i) \]

\[ E(X^2) = 0^2(0.1) + 1^2(0.3) + 2^2(0.4) + 3^2(0.2) \]

\[ E(X^2) = 0 + 0.3 + 1.6 + 1.8 = 3.7 \]

Now calculate variance:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81 \]

Standard deviation:

\[ \sigma = \sqrt{0.81} = 0.9 \]

\(\text{Var}(X) = 0.81\), \(\sigma = 0.9\)

Conditions for Binomial Distribution (BINS):

1. Binary outcomes: Only two possible outcomes (success or failure)
2. Independent trials: Outcome of one trial doesn't affect others
3. Number of trials fixed: n is predetermined
4. Same probability: p remains constant for all trials

Notation:

If X follows a binomial distribution:

\[ X \sim B(n, p) \]

where n = number of trials, p = probability of success

Probability Formula:

\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} = \frac{n!}{r!(n-r)!} p^r q^{n-r} \]

where \(r\) is the number of successes, \(q = 1-p\)

Mean and Variance:

\[ E(X) = np \]

\[ \text{Var}(X) = np(1-p) = npq \]

\[ \text{Standard Deviation} = \sqrt{npq} \]

⚠️ Common Pitfalls & Tips:

• Always use GDC for binomial probabilities – manual calculation is error-prone
• Check BINS conditions before applying binomial distribution
• "At least" and "at most" questions require cumulative probabilities
• \(P(X \geq r) = 1 - P(X \leq r-1)\) (useful for complement)
• "More than r" means \(X \geq r+1\), NOT \(X \geq r\)
• Know your GDC functions: binompdf for \(P(X=r)\), binomcdf for \(P(X \leq r)\)

Solution:

(a) Why binomial?

This satisfies all BINS conditions:

• Binary: Each shot is either successful or not (two outcomes)
• Independent: Each free throw is independent of the others
• Number fixed: n = 12 trials is predetermined
• Same probability: p = 0.7 remains constant for all attempts

Therefore, let X = number of successful shots, then \(X \sim B(12, 0.7)\)

(b) P(X = 9):

Using GDC: binompdf(12, 0.7, 9)

Or using formula:

\[ P(X = 9) = \binom{12}{9} (0.7)^9 (0.3)^3 \]

\[ P(X = 9) = \frac{12!}{9! \cdot 3!} (0.7)^9 (0.3)^3 = 220 \times 0.04035 \times 0.027 \]

\[ P(X = 9) = 0.240 \text{ (3 s.f.)} \]

(c) P(X ≥ 10):

"At least 10" means 10, 11, or 12 successes

Method 1 – Direct addition:

\[ P(X \geq 10) = P(X=10) + P(X=11) + P(X=12) \]

Using GDC: binompdf(12, 0.7, 10) + binompdf(12, 0.7, 11) + binompdf(12, 0.7, 12)

Method 2 – Using complement (easier):

\[ P(X \geq 10) = 1 - P(X \leq 9) \]

Using GDC: 1 - binomcdf(12, 0.7, 9)

\[ P(X \geq 10) = 1 - 0.7237 = 0.276 \text{ (3 s.f.)} \]

(d) Expected value and standard deviation:

Expected number of successes:

\[ E(X) = np = 12 \times 0.7 = 8.4 \text{ shots} \]

Variance:

\[ \text{Var}(X) = np(1-p) = 12 \times 0.7 \times 0.3 = 2.52 \]

Standard deviation:

\[ \sigma = \sqrt{2.52} = 1.59 \text{ shots (3 s.f.)} \]

Notation:

\[ X \sim N(\mu, \sigma^2) \]

where \(\mu\) = mean, \(\sigma^2\) = variance, \(\sigma\) = standard deviation

Key Properties:

• Symmetric about the mean \(\mu\)
• Bell-shaped curve (Gaussian curve)
• Mean = Median = Mode (all at center)
• Total area under curve = 1
• Curve never touches x-axis (extends to ±∞)
• 68% of data within 1 standard deviation of mean
• 95% of data within 2 standard deviations of mean
• 99.7% of data within 3 standard deviations of mean

Standard Normal Distribution:

Special case with \(\mu = 0\) and \(\sigma = 1\):

\[ Z \sim N(0, 1) \]

Standardization (Z-score):

\[ Z = \frac{X - \mu}{\sigma} \]

Converts any normal distribution to standard normal

Calculating Probabilities:

For \(X \sim N(\mu, \sigma^2)\):

• \(P(X < a)\): Use normalcdf(-∞, a, μ, σ) on GDC
• \(P(X > a)\): Use normalcdf(a, ∞, μ, σ) or \(1 - P(X < a)\)
• \(P(a < X < b)\): Use normalcdf(a, b, μ, σ)

Inverse Normal (Finding Values):

To find x given \(P(X < x) = p\): Use invNorm(p, μ, σ)

⚠️ Common Pitfalls & Tips:

• Always use GDC – no need to use z-tables in IB exams
• Remember: variance is \(\sigma^2\), not \(\sigma\) in the notation
• For "at least," use \(P(X \geq a) = 1 - P(X < a)\)
• Sketch a diagram to visualize the area you're finding
• Check if your answer makes sense (probabilities between 0 and 1)
• For inverse normal, input the probability, not the z-score

Solution:

Given: Let X = height of adult males, \(X \sim N(175, 8^2)\)

\(\mu = 175\) cm, \(\sigma = 8\) cm

(a) P(X > 185):

We need the probability of height greater than 185 cm.

Using GDC: normalcdf(185, 1×10⁹⁹, 175, 8)

Or: \(1 - \text{normalcdf}(-1 \times 10^{99}, 185, 175, 8)\)

Alternative using z-score (for understanding):

\[ Z = \frac{185 - 175}{8} = \frac{10}{8} = 1.25 \]

Then \(P(X > 185) = P(Z > 1.25)\)

\[ P(X > 185) = 0.106 \text{ or } 10.6\% \]

(b) P(170 < X < 180):

Using GDC: normalcdf(170, 180, 175, 8)

\[ P(170 < X < 180) = 0.468 \text{ or } 46.8\% \]

Interpretation: About 46.8% of adult males have heights between 170 cm and 180 cm.

(c) Height exceeded by 10%:

We need to find h such that \(P(X > h) = 0.10\)

This is equivalent to \(P(X < h) = 0.90\)

Using GDC: invNorm(0.90, 175, 8)

\[ h = 185.3 \text{ cm (1 d.p.)} \]

Interpretation: 10% of males are taller than 185.3 cm (or 90% are shorter than 185.3 cm).

Solution:

Step 1: Set up binomial

Let X = number of heads, \(X \sim B(100, 0.5)\)

We could use binomial, but with large n, normal approximation is easier.

Step 2: Check if normal approximation is valid

Normal approximation works well when \(np > 5\) and \(n(1-p) > 5\)

\(np = 100(0.5) = 50 > 5\) ✓

\(n(1-p) = 100(0.5) = 50 > 5\) ✓

Step 3: Find parameters for normal distribution

Mean: \(\mu = np = 100(0.5) = 50\)

Variance: \(\sigma^2 = np(1-p) = 100(0.5)(0.5) = 25\)

Standard deviation: \(\sigma = \sqrt{25} = 5\)

So approximate with \(X \sim N(50, 25)\)

Step 4: Apply continuity correction

Since binomial is discrete and normal is continuous, apply continuity correction:

\(P(45 \leq X \leq 55)\) becomes \(P(44.5 < X < 55.5)\)

Step 5: Calculate probability

Using GDC: normalcdf(44.5, 55.5, 50, 5)

\[ P(45 \leq X \leq 55) \approx 0.730 \text{ or } 73.0\% \]

Interpretation: There's approximately a 73% chance of getting between 45 and 55 heads in 100 coin tosses.

Notation:

\[ X \sim \text{Po}(\lambda) \]

where \(\lambda\) (lambda) = mean number of occurrences per interval

Probability Formula:

\[ P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!} \quad \text{for } r = 0, 1, 2, 3, \ldots \]

Mean and Variance:

\[ E(X) = \lambda \]

\[ \text{Var}(X) = \lambda \]

Unique property: mean equals variance

Common Applications:

• Number of phone calls received per hour
• Number of emails received per day
• Number of arrivals at a service point
• Number of defects per unit of product
• Number of accidents per week

⚠️ Common Pitfalls & Tips:

• λ must match the time interval in the question – adjust if necessary
• If λ = 3 per hour, then for 30 minutes use λ = 1.5
• Use poissonpdf for \(P(X=r)\) and poissoncdf for \(P(X \leq r)\)
• Always use GDC – manual calculations with factorials are tedious
• Poisson approximates binomial when n is large and p is small

Solution:

Given: Average rate = 4.5 calls per minute

Let X = number of calls per minute, \(X \sim \text{Po}(4.5)\)

(a) P(X = 6):

Using GDC: poissonpdf(4.5, 6)

Or using formula:

\[ P(X = 6) = \frac{e^{-4.5} \times 4.5^6}{6!} = \frac{e^{-4.5} \times 8303.77}{720} \]

\[ P(X = 6) = 0.128 \text{ or } 12.8\% \]

(b) P(X < 3):

"Fewer than 3" means 0, 1, or 2 calls

\(P(X < 3) = P(X \leq 2)\)

Using GDC: poissoncdf(4.5, 2)

\[ P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.174 \text{ or } 17.4\% \]

(c) At least 2 calls in 30 seconds:

Important: Adjust λ for the time interval!

30 seconds = 0.5 minutes

New rate: \(\lambda = 4.5 \times 0.5 = 2.25\) calls per 30 seconds

Let Y = number of calls in 30 seconds, \(Y \sim \text{Po}(2.25)\)

We need \(P(Y \geq 2) = 1 - P(Y \leq 1)\)

Using GDC: 1 - poissoncdf(2.25, 1)

\[ P(Y \geq 2) = 1 - 0.431 = 0.569 \text{ or } 56.9\% \]

Rules for Expected Values:

For random variables X and Y, and constants a, b:

1. Constant multiple:

\[ E(aX) = a \cdot E(X) \]

2. Sum/Difference (always, independent or not):

\[ E(X \pm Y) = E(X) \pm E(Y) \]

3. Linear combination:

\[ E(aX + bY) = a \cdot E(X) + b \cdot E(Y) \]

4. Constant addition:

\[ E(X + c) = E(X) + c \]

Rules for Variances:

1. Constant multiple:

\[ \text{Var}(aX) = a^2 \cdot \text{Var}(X) \]

2. Constant addition (variance unchanged):

\[ \text{Var}(X + c) = \text{Var}(X) \]

3. Sum/Difference (only if X and Y are independent):

\[ \text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y) \]

Note: Variance adds even when subtracting random variables!

4. Linear combination (if independent):

\[ \text{Var}(aX + bY) = a^2 \cdot \text{Var}(X) + b^2 \cdot \text{Var}(Y) \]

Sum of Independent Normal Variables:

If \(X \sim N(\mu_1, \sigma_1^2)\) and \(Y \sim N(\mu_2, \sigma_2^2)\) are independent:

\[ X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2) \]

\[ X - Y \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2) \]

⚠️ Common Pitfalls & Tips:

• Critical: Variances ADD even when subtracting variables
• Adding constants doesn't change variance
• Multiplying by constant squares the variance: \(\text{Var}(2X) = 4\text{Var}(X)\)
• For variances to add, variables must be independent
• Expected values always add/subtract regardless of independence

Solution:

Given:

Let A = mass from Farm A: \(A \sim N(150, 20^2)\)

Let B = mass from Farm B: \(B \sim N(180, 25^2)\)

(a) Total mass T = A + B:

Mean:

\[ E(T) = E(A + B) = E(A) + E(B) = 150 + 180 = 330 \text{g} \]

Variance:

\[ \text{Var}(T) = \text{Var}(A) + \text{Var}(B) = 20^2 + 25^2 = 400 + 625 = 1025 \]

Standard deviation:

\[ \sigma_T = \sqrt{1025} = 32.0 \text{g (3 s.f.)} \]

Therefore: \(T \sim N(330, 1025)\)

(b) P(T > 350):

Using GDC: normalcdf(350, 1×10⁹⁹, 330, 32.0)

\[ P(T > 350) = 0.266 \text{ or } 26.6\% \]

(c) Difference D = B - A:

Mean:

\[ E(D) = E(B - A) = E(B) - E(A) = 180 - 150 = 30 \text{g} \]

Variance (note: still ADD):

\[ \text{Var}(D) = \text{Var}(B) + \text{Var}(A) = 625 + 400 = 1025 \]

Standard deviation:

\[ \sigma_D = \sqrt{1025} = 32.0 \text{g} \]

Therefore: \(D \sim N(30, 1025)\)

Note: Same standard deviation as the sum, but different mean!

Discrete Distributions

• Expected Value: \(\sum x_i P(x_i)\)
• Variance: \(E(X^2) - [E(X)]^2\)

Binomial: \(X \sim B(n,p)\)

• Mean: \(np\)
• Variance: \(np(1-p)\)
• Use GDC for probabilities

Normal: \(X \sim N(\mu, \sigma^2)\)

• 68-95-99.7 rule
• Symmetric, bell-shaped
• Use normalcdf/invNorm

Poisson: \(X \sim Po(\lambda)\)

• Mean = Variance = \(\lambda\)
• Adjust \(\lambda\) for time interval
• Use poissonpdf/cdf