IBDP Maths SL 4.10 Equation of the regression line AA HL Paper 2- Exam Style Questions- New Syllabus
Question
The data analysis produced the following linear regression models:
- The regression line of \( F \) on \( A \): \( F = 0.335A – 32.6 \)
- The regression line of \( A \) on \( F \): \( A = 2.89F + 99.3 \)
It is known that both lines intersect at the mean point \( (\bar{A}, \bar{F}) \).
Syllabus Topic Codes (IB Mathematics AA HL):
• SL 4.10: Equation of the regression line of \( x \) on \( y \); use for prediction purposes — part (a)
• SL 4.3: Measures of central tendency (mean) — part (b)
• SL 4.9: The normal distribution and curve; diagrammatic representation; normal probability calculations — part (c)
• SL 4.12: Standardization of normal variables (z-values); inverse normal calculations – part (c)
▶️ Answer/Explanation
(a)
We are given foot length \( F \) and want to estimate arm span \( A \).
The appropriate line is \( A \text{ on } F \): \( A = 2.89F + 99.3 \).
Substitute \( F = 19.8 \):
\( A = 2.89(19.8) + 99.3 = 57.222 + 99.3 = 156.522 \) cm
Rounded to three significant figures: \( \boxed{157 \, \text{cm}} \)
(b)
The mean point \( (\bar{A}, \bar{F}) \) lies on both regression lines.
Equate the two lines at the mean:
From \( F = 0.335A – 32.6 \) and \( A = 2.89F + 99.3 \), substitute the first into the second:
\( A = 2.89(0.335A – 32.6) + 99.3 \)
\( A = 0.96815A – 94.214 + 99.3 \)
\( A – 0.96815A = 5.086 \)
\( 0.03185A = 5.086 \)
\( A \approx 159.686 \) cm
Then \( F = 0.335(159.686) – 32.6 \approx 53.495 – 32.6 = 20.895 \) cm
Rounded to three significant figures: \( \boxed{\bar{A} = 160 \, \text{cm}}, \quad \boxed{\bar{F} = 20.9 \, \text{cm}} \)
(c)
METHOD 1 (Using symmetry)
88% between 153 and 173 ⇒ 12% outside ⇒ 6% below 153 and 6% above 173.
Let \( Z \) be standard normal. Then \( P(Z < z) = 0.06 \).
Using inverse normal: \( z \approx -1.55477 \).
For \( H \sim N(163, \sigma^2) \), \( \frac{153 – 163}{\sigma} = -1.55477 \)
\( -10/\sigma = -1.55477 \)
\( \sigma = 10/1.55477 \approx 6.4318 \) cm
Rounded to three significant figures: \( \boxed{6.43 \, \text{cm}} \)
METHOD 2 (Using normal CDF directly)
\( P(153 < H < 173) = 0.88 \)
Using calculator: normcdf(153, 173, 163, \( \sigma \)) = 0.88
Solve: \( \sigma \approx 6.4318 \) cm
Rounded: \( \boxed{6.43 \, \text{cm}} \)