IB Mathematics SL 4.4 Linear correlation of bivariate data AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
To estimate \( A \) (arm span) from \( F \) (foot length), we must use the regression line of \( x \) on \( y \) (where \( x \) is the variable to be predicted)[cite: 1395]:
\( A = 2.89F + 99.3 \)
Substitute \( F = 19.8 \):
\( A = 2.89(19.8) + 99.3 = 57.222 + 99.3 = 156.522 \text{ cm} \)
To three significant figures: \( A \approx 157 \text{ cm} \).
Answer: \( \boxed{157 \text{ cm}} \)

(b)
The mean point \( (\bar{A}, \bar{F}) \) is the intersection of the two regression lines[cite: 1331]. Solve the system of equations:
1) \( \bar{F} = 0.335\bar{A} – 32.6 \)
2) \( \bar{A} = 2.89\bar{F} + 99.3 \)
Substitute (1) into (2):
\( \bar{A} = 2.89(0.335\bar{A} – 32.6) + 99.3 \)
\( \bar{A} = 0.96815\bar{A} – 94.214 + 99.3 \)
\( \bar{A} – 0.96815\bar{A} = 5.086 \)
\( 0.03185\bar{A} = 5.086 \)
\( \bar{A} = \frac{5.086}{0.03185} \approx 159.686 \text{ cm} \)
Now find \( \bar{F} \):
\( \bar{F} = 0.335(159.686) – 32.6 \approx 20.8948 \text{ cm} \)
To three significant figures: \( \bar{A} \approx 160 \text{ cm}, \bar{F} \approx 20.9 \text{ cm} \).
Answer: \( \boxed{\bar{A} = 160 \text{ cm}, \ \bar{F} = 20.9 \text{ cm}} \)

(c)
Given \( H \sim N(163, \sigma^2) \) and \( P(153 < H < 173) = 0.88 \)[cite: 1383].
Due to symmetry, the area in each tail is \( \frac{1 – 0.88}{2} = 0.06 \).
Thus, \( P(H < 153) = 0.06 \).
Find the \( z \)-score for a cumulative probability of 0.06[cite: 1406]:
\( z \approx -1.55477 \)
Using the formula \( z = \frac{x – \mu}{\sigma} \):
\( -1.55477 = \frac{153 – 163}{\sigma} \)
\( -1.55477 = \frac{-10}{\sigma} \)
\( \sigma = \frac{10}{1.55477} \approx 6.4318 \text{ cm} \)
To three significant figures: \( \sigma \approx 6.43 \text{ cm} \).
Answer: \( \boxed{6.43 \text{ cm}} \)