IBDP Maths SL 4.1 Concepts of population, sample AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
To estimate the probability that a visitor will ride The Dragon:
- Total visitors sampled = 40
- Visitors who rode at least once:
- 16 rode once
- 13 rode twice
- 2 rode three times
- 3 rode four times
- Total riders = 16 + 13 + 2 + 3 = 34
Probability calculation:
\[ P(\text{riding}) = \frac{34}{40} = \frac{17}{20} = 0.85 \]
Thus, there’s an 85% probability a visitor will ride The Dragon.
To find the expected number of rides per visitor:
\[ E(X) = \sum [x \times P(X=x)] \]
Calculating each term:
| Rides (x) | Frequency | P(X=x) | x × P(X=x) |
|---|---|---|---|
| 0 | 6 | 6/40 | 0 |
| 1 | 16 | 16/40 | 16/40 |
| 2 | 13 | 13/40 | 26/40 |
| 3 | 2 | 2/40 | 6/40 |
| 4 | 3 | 3/40 | 12/40 |
Summing the contributions:
\[ E(X) = 0 + \frac{16}{40} + \frac{26}{40} + \frac{6}{40} + \frac{12}{40} = \frac{60}{40} = 1.5 \]
Therefore, the expected number of rides per visitor is 1.5.
To estimate the minimum number of runs needed:
- Calculate total expected rides:
\[ 1.5 \text{ rides/visitor} \times 1000 \text{ visitors} = 1500 \text{ rides} \]
- Determine runs required:
The Dragon carries 10 people per run, so:
\[ \frac{1500 \text{ rides}}{10 \text{ rides/run}} = 150 \text{ runs} \]
However, we must consider that:
- This is the minimum estimate based on averages
- Actual demand may fluctuate due to random variations
- The park might want to add a safety margin for peak times
Thus, while 150 runs is the theoretical minimum, practical operation might require slightly more.
▶️ Answer/Explanation
For any probability distribution, the sum of all probabilities must equal \( 1 \):
\[ p + (0.5 – p) + 0.25 + 0.125 + p^3 = 1 \]
Simplify the equation:
\[ 0.5 + 0.25 + 0.125 + p^3 = 1 \]
\[ 0.875 + p^3 = 1 \]
\[ p^3 = 0.125 \]
Solve for \( p \):
\[ p = \sqrt[3]{0.125} = 0.5 \]
Verification: Substituting back:
\( P(X=0) = 0.5 \)
\( P(X=1) = 0.5 – 0.5 = 0 \)
\( P(X=4) = (0.5)^3 = 0.125 \)
Sum \( = 0.5 + 0 + 0.25 + 0.125 + 0.125 = 1 \) ✓
The expected value \( \mu \) is calculated as:
\[ \mu = \sum [x \times P(X=x)] \]
Compute each term:
\[ = (0 \times 0.5) + (1 \times 0) + (2 \times 0.25) + (3 \times 0.125) + (4 \times 0.125) \]
\[ = 0 + 0 + 0.5 + 0.375 + 0.5 \]
\[ = 1.375 \text{ or } \frac{11}{8} \]
Thus, the expected value is \( \boxed{1.375} \).
To find \( P(X > \mu) \) where \( \mu = 1.375 \):
We consider values of \( X \) greater than \( 1.375 \), which are \( X = 2, 3, \) and \( 4 \).
Calculate the probability:
\[ P(X > 1.375) = P(X=2) + P(X=3) + P(X=4) \]
\[ = 0.25 + 0.125 + 0.125 \]
\[ = 0.5 \]
Thus, the probability is \( \boxed{0.5} \).
(a) \([2]\)
- M1: Correct equation setup \( (\sum P(X=x) = 1) \)
- A1: Correct solution \( (p = 0.5) \)
(b)(i) \([2]\)
- M1: Correct expected value formula application
- A1: Correct final answer \( (1.375 \text{ or } \frac{11}{8}) \)
(b)(ii) \([2]\)
- M1: Correct identification of \( P(X > \mu) \) cases
- A1: Correct final answer \( (0.5) \)
Total: 6 marks
- Probability Distributions: \( \sum P(X=x) = 1 \)
- Expected Value: \( \mu = E[X] = \sum x \cdot P(X=x) \)
- Cumulative Probability: \( P(X > a) = \sum_{x > a} P(X=x) \)
- Algebraic Solutions: Solving \( p^3 = 0.125 \)
▶️ Answer/Explanation
New mean calculation:
\[ \overline{x}_{\text{new}} = \frac{\sum_{i=1}^{n}(x_i + k)}{n} = \frac{\sum_{i=1}^{n}x_i + \sum_{i=1}^{n}k}{n} = \frac{\sum_{i=1}^{n}x_i}{n} + \frac{nk}{n} = \overline{x} + k \]
Thus, the new mean is \( \boxed{\overline{x} + k} \).
New standard deviation calculation:
\[ s_{\text{new}} = \sqrt{\frac{\sum_{i=1}^{n}[(x_i + k) – (\overline{x} + k)]^2}{n}} = \sqrt{\frac{\sum_{i=1}^{n}(x_i – \overline{x})^2}{n}} = s \]
Thus, the standard deviation remains \( \boxed{s} \).
New mean calculation:
\[ \overline{x}_{\text{new}} = \frac{\sum_{i=1}^{n}(kx_i)}{n} = \frac{k\sum_{i=1}^{n}x_i}{n} = k\overline{x} \]
Thus, the new mean is \( \boxed{k\overline{x}} \).
New standard deviation calculation:
\[ s_{\text{new}} = \sqrt{\frac{\sum_{i=1}^{n}(kx_i – k\overline{x})^2}{n}} = \sqrt{\frac{k^2\sum_{i=1}^{n}(x_i – \overline{x})^2}{n}} = k\sqrt{\frac{\sum_{i=1}^{n}(x_i – \overline{x})^2}{n}} = ks \]
Thus, the new standard deviation is \( \boxed{ks} \).
Variance relationship:
Since \( s_{\text{new}} = ks \), then:
\[ s_{\text{new}}^2 = (ks)^2 = k^2s^2 \]
Thus, the new variance is \( \boxed{k^2} \) times the original variance.
(a)(i) [2 marks]
- M1: Correct expansion of sum
- A1: Correct final result \( (\overline{x} + k) \)
(a)(ii) [2 marks]
- M1: Correct simplification inside square root
- A1: Correct conclusion \( (s) \)
(b)(i) [2 marks]
- M1: Correct factorization of \( k \)
- A1: Correct final result \( (k\overline{x}) \)
(b)(ii) [2 marks]
- M1: Correct factorization of \( k^2 \)
- A1: Correct final result \( (ks) \)
(b)(iii) [1 mark]
- A1: Correct variance relationship \( (k^2) \)
Total: 9 marks
- Mean Transformation: Additive \( (+k) \) vs multiplicative \( (\times k) \) changes
- Standard Deviation: Invariant under addition, scales linearly under multiplication
- Variance: Scales quadratically \( (k^2) \) under multiplication
- Summation Properties: Linearity of summation operations
▶️ Answer/Explanation
Mean (\( \mu_x \)):
\[ \mu_x = \frac{1+2+3+4}{4} = \frac{10}{4} = 2.5 \]
Variance (\( \sigma_x^2 \)):
\[ \sigma_x^2 = \frac{(1-2.5)^2 + (2-2.5)^2 + (3-2.5)^2 + (4-2.5)^2}{4} \]
\[ = \frac{2.25 + 0.25 + 0.25 + 2.25}{4} = \frac{5}{4} = 1.25 \]
Standard deviation: \( \sigma_x = \sqrt{1.25} \approx 1.11803 \)
\(\boxed{\mu_x = 2.5,\ \sigma_x^2 = 1.25}\)
Mean (\( \mu_y \)):
\[ \mu_y = \frac{2+3+7+8}{4} = \frac{20}{4} = 5 \]
Variance (\( \sigma_y^2 \)):
\[ \sigma_y^2 = \frac{(2-5)^2 + (3-5)^2 + (7-5)^2 + (8-5)^2}{4} \]
\[ = \frac{9 + 4 + 4 + 9}{4} = \frac{26}{4} = 6.5 \]
Standard deviation: \( \sigma_y = \sqrt{6.5} \approx 2.54951 \)
\(\boxed{\mu_y = 5,\ \sigma_y^2 = 6.5}\)
Correlation coefficient (\( r \)):
First calculate covariance:
\[ \text{Cov}(x,y) = \frac{(1-2.5)(2-5) + (2-2.5)(3-5) + (3-2.5)(7-5) + (4-2.5)(8-5)}{4} \]
\[ = \frac{4.5 + 1 + 1 + 4.5}{4} = \frac{11}{4} = 2.75 \]
Then:
\[ r = \frac{\text{Cov}(x,y)}{\sigma_x \sigma_y} = \frac{2.75}{(1.11803)(2.54951)} \approx 0.965 \]
\(\boxed{r \approx 0.965}\)
The correlation coefficient \( r \approx 0.965 \) indicates a:
\(\boxed{\text{Strong positive linear relationship between } x \text{ and } y}\)
(Since \( r \) is close to 1 and positive)
Regression line \( y \) on \( x \):
Slope (\( a \)):
\[ a = r \cdot \frac{\sigma_y}{\sigma_x} = 0.965 \cdot \frac{2.54951}{1.11803} \approx 2.2 \]
Intercept (\( b \)):
\[ b = \mu_y – a\mu_x = 5 – (2.2)(2.5) = -0.5 \]
Equation:
\(\boxed{y = 2.2x – 0.5}\)
Regression line \( x \) on \( y \):
Slope (\( c \)):
\[ c = r \cdot \frac{\sigma_x}{\sigma_y} = 0.965 \cdot \frac{1.11803}{2.54951} \approx 0.423 \]
Intercept (\( d \)):
\[ d = \mu_x – c\mu_y = 2.5 – (0.423)(5) \approx 0.385 \]
Equation:
\(\boxed{x = 0.423y + 0.385}\)
Inverse of (e):
Starting with \( y = 2.2x – 0.5 \):
\[ y + 0.5 = 2.2x \]
\[ x = \frac{y + 0.5}{2.2} = \frac{y}{2.2} + \frac{0.5}{2.2} \]
\[ x \approx 0.455y + 0.227 \]
Comparison with (f):
The inverse \( x \approx 0.455y + 0.227 \) is different from the regression \( x = 0.423y + 0.385 \) found in (f).
\(\boxed{\text{No, they are different functions}}\)