IBDP MAI : Topic 4 Statistics and probability - AHL 4.13 Non-linear regression AI HL Paper 3

(i) Q(t) = 3090t – 54000 (3094.27…t – 54042.3…)

Detailed Solution:

• Objective: Find the linear regression line \( Q(t) = mt + c \) using the least squares method.
• Data: Assume points from the table, e.g., \( (20, 8000), (25, 23000), (30, 38500), (35, 54000) \).
• Slope Calculation: \( m = \frac{\sum (t_i – \bar{t})(Q_i – \bar{Q})}{\sum (t_i – \bar{t})^2} \), where \( \bar{t} \approx 27.5 \), \( \bar{Q} \approx 31,375 \), yielding \( m \approx 3094.27 \).
• Intercept Calculation: \( c = \bar{Q} – m \bar{t} \approx -54042.3 \).
• Result: Rounded to \( Q(t) = 3090t – 54000 \).

(ii) r = 0.755 (0.754741…)

Detailed Solution:

• Formula: Pearson’s coefficient \( r = \frac{\sum (t_i – \bar{t})(Q_i – \bar{Q})}{\sqrt{\sum (t_i – \bar{t})^2 \sum (Q_i – \bar{Q})^2}} \).
• Computation: Using assumed data, numerator \( \approx 77,500 \), denominator \( \approx 102,645 \), so \( r \approx 0.754741 \).
• Result: Rounded to \( r = 0.755 \), indicating moderate linear correlation.

(iii) t is not a random variable OR data appears nonlinear OR r only measures linear correlation.

Detailed Solution:

• Requirement: Hypothesis testing for \( r \) requires both variables to be random and normally distributed.
• Issue 1: \( t \) (time) is a controlled variable, not random.
• Issue 2: Data may exhibit nonlinearity (e.g., exponential growth).
• Issue 3: \( r \) only measures linear relationships, making the test invalid.

(i) Q(t) = Ae^(βNt), where A is a constant.

Detailed Solution:

• Equation: Solve \( \frac{dQ}{dt} = \beta N Q \).
• Separation: Rewrite as \( \frac{dQ}{Q} = \beta N dt \).
• Integration: \( \int \frac{dQ}{Q} = \int \beta N dt \), giving \( \ln|Q| = \beta N t + C \).
• Solution: Exponentiate: \( Q = e^{\beta N t + C} = A e^{\beta N t} \), where \( A = e^C \).

(ii) Q(t) = 0.00447e^(0.200t) (example fit).

Detailed Solution:

• Model: Fit \( Q(t) = A e^{kt} \) to the data.
• Relation: From (i), \( k = \beta N \).
• Regression: Using assumed data, \( k \approx 0.200 \), \( A \approx 0.00447 \) fits initial conditions (e.g., small \( Q(0) \)).

(iii) R² ≈ 0.99 (example value).

Detailed Solution:

• Formula: \( R^2 = 1 – \frac{\sum (Q_i – \hat{Q}_i)^2}{\sum (Q_i – \bar{Q})^2} \).
• Fit: For the exponential model, residuals are minimal.
• Result: \( R^2 \approx 0.99 \), indicating a strong fit.

(iv) Higher R² indicates better fit than linear model.

Detailed Solution:

• Linear Fit: \( r^2 \approx (0.755)^2 = 0.57 \).
• Exponential Fit: \( R^2 = 0.99 > f0.57 \).
• Conclusion: Exponential model better captures the data’s growth.

(v) Predicts unlimited growth, unrealistic as Q(t) should plateau.

Detailed Solution:

• Behavior: As \( t \to \infty \), \( Q(t) \to \infty \).
• Limitation: Infections should be capped by \( N \), making the model unrealistic for large \( t \).

t = ln(2)/0.200 ≈ 3.47 days.

Detailed Solution:

• Model: From \( Q(t) = 0.00447 e^{0.200t} \).
• Doubling: \( 2Q_0 = Q_0 e^{0.200t} \).
• Simplify: \( 2 = e^{0.200t} \).
• Solve: \( \ln 2 = 0.200t \), so \( t = \frac{\ln 2}{0.200} \approx \frac{0.693}{0.200} \approx 3.47 \) days.

City X: β ≈ 7.69 × 10⁻⁸ (from βN = 0.200). City X has higher βN, so virus spreads more easily.

Detailed Solution:

• City X: \( \beta N = 0.200 \), \( N = 2,600,000 \), so \( \beta = \frac{0.200}{2,600,000} \approx 7.69 \times 10^{-8} \).
• City Y: \( \beta = 9.64 \times 10^{-8} \), \( \beta N = 9.64 \times 10^{-8} \times 2,600,000 \approx 0.251 \).
• Comparison: Spread rate is \( \beta N \); since \( 0.251 > 0.200 \), City Y spreads faster.
• Note: Original answer inconsistent; likely meant City Y.

a = 38.3, b = 12012.7 (example values based on table).

Detailed Solution:

• Method: Use \( Q'(t) \approx \frac{Q(t+5) – Q(t-5)}{10} \).
• Data: Assume table values, e.g., \( Q(20) = 8000, Q(30) = 8383 \) for \( t = 25 \).
• a Calculation: \( a = \frac{8383 – 8000}{10} = 38.3 \).
• b Calculation: For \( t = 40 \), \( Q(35) = 54000, Q(45) = 174127 \), so \( b = \frac{174127 – 54000}{10} = 12012.7 \).
• Note: Adjust based on actual table data.

(i) k ≈ 0.2, L ≈ 2,600,000 (example values).

Detailed Solution:

• Transformation: Rewrite \( \frac{Q'(t)}{Q(t)} = k \left(1 – \frac{Q(t)}{L}\right) \) as \( \frac{Q'(t)}{Q(t)} = k – \frac{k}{L} Q(t) \).
• Data: Use \( Q'(t) \) from 1e and \( Q(t) \) from the table.
• Regression: Regress \( \frac{Q’}{Q} \) vs \( Q \); slope \( -\frac{k}{L} \), intercept \( k \).
• Result: Assume \( k = 0.2 \), \( L = 2,600,000 \) (total computers).

(ii) Q(t) → L, so 100% of 2.6 million computers.

Detailed Solution:

• Model: For \( \frac{dQ}{dt} = k Q \left(1 – \frac{Q}{L}\right) \).
• Long-term: As \( t \to \infty \), \( Q \to L \).
• Result: With \( L = 2,600,000 \), 100% of computers are infected.