IB Mathematics AHL 4.14 Linear transformation AI HL Paper 2- Exam Style Questions- New Syllabus
Protective barriers to shield against radiation are built using two parallel concrete panels with a lead sheet between them, as shown in the diagram.
The thickness of a concrete panel is modelled by a normal distribution with mean \( 350 \, \text{mm} \) and standard deviation \( 10 \, \text{mm} \).
(a) Determine the probability that a randomly selected concrete panel is less than \( 340 \, \text{mm} \) in thickness. [2]
(b) Identify the endpoints of the interval, symmetric about the mean, such that 95% of the panels have a thickness within this interval. [3]
Emma assumes the lead sheet is also modelled by a normal distribution, but with mean \( 100 \, \text{mm} \) and standard deviation \( 5 \, \text{mm} \), and is independent of the thickness of the panels.
Let \( W \) be the random variable representing the total thickness of the barrier, measured in mm.
(c) (i) Given that the thicknesses of any two concrete panels are independent, calculate Emma’s value for the mean and standard deviation of \( W \). [2]
(ii) Hence find \( P(780 < W < 810) \). [7]
There are concerns that the mean and standard deviation for Emma’s model of the lead sheet are inaccurate. However, her assumption that the model is normal and the thickness of the lead is independent of the thickness of the concrete panels still holds.
Upon review, it is found that the total thickness of the barrier is normally distributed with mean \( 810 \, \text{mm} \) and standard deviation \( 16 \, \text{mm} \). The model for the thickness of a concrete panel remains unchanged.
(d) Use the results for the sum of independent random variables to find a revised value for:
(i) the mean of the thickness of the lead sheet. [2]
(ii) the standard deviation of the thickness of the lead sheet. [2]
Under this revised model, 80% of the lead sheets have a thickness less than \( k \, \text{mm} \).
(e) Calculate the value of \( k \). [1]
▶️ Answer/Explanation
(a) Let \( X \) represent the thickness of a concrete panel, \( X \sim N(350, 10^2) \). To find \( P(X < 340) \), standardize:
\( Z = \frac{X – \mu}{\sigma} = \frac{340 – 350}{10} = -1 \)
Using standard normal tables, \( P(Z < -1) = 1 – P(Z < 1) = 1 – 0.8413 = 0.1587 \).
\( P(X < 340) = 0.159 \) (to 3 decimal places). (M1 A1)[2 marks]
(b) For a 95% symmetric interval, the area in each tail is \( \frac{1 – 0.95}{2} = 0.025 \). Thus, \( P(X < k) = 0.975 \), corresponding to \( z = 1.96 \) (from standard normal tables).
Lower endpoint: \( 350 – 1.96 \times 10 = 350 – 19.6 = 330.4 \, \text{mm} \).
Upper endpoint: \( 350 + 1.96 \times 10 = 350 + 19.6 = 369.6 \, \text{mm} \).
Interval: \( 330 < X < 370 \) (or \( 330.4 < X < 369.6 \)). (M1 A1A1)[3 marks]
(c) (i) Let \( C_1, C_2 \sim N(350, 10^2) \) be the thicknesses of the two concrete panels, and \( L \sim N(100, 5^2) \) be the lead sheet thickness. Since they are independent, \( W = C_1 + C_2 + L \).
Mean: \( E(W) = E(C_1) + E(C_2) + E(L) = 350 + 350 + 100 = 800 \). (M1 A1)
Variance: \( \text{Var}(W) = \text{Var}(C_1) + \text{Var}(C_2) + \text{Var}(L) = 10^2 + 10^2 + 5^2 = 100 + 100 + 25 = 225 \).
Standard deviation: \( \text{SD}(W) = \sqrt{225} = 15 \). (M1 A1)
(ii) Since \( W \sim N(800, 15^2) \), find \( P(780 < W < 810) \).
Standardize: \( Z_1 = \frac{780 – 800}{15} = -\frac{20}{15} = -1.333 \), \( Z_2 = \frac{810 – 800}{15} = \frac{10}{15} = 0.667 \).
\( P(780 < W < 810) = P(-1.333 < Z < 0.667) = P(Z < 0.667) – P(Z < -1.333) \).
From standard normal tables: \( P(Z < 0.667) \approx 0.7475 \), \( P(Z < -1.333) = 1 – P(Z < 1.333) \approx 1 – 0.9088 = 0.0912 \).
\( P(-1.333 < Z < 0.667) = 0.7475 – 0.0912 = 0.6563 \).
\( P(780 < W < 810) = 0.656 \) (to 3 decimal places). (M1 A1)[7 marks]
(d) Given \( W \sim N(810, 16^2) \), and \( C_1, C_2 \sim N(350, 10^2) \), with \( W = C_1 + C_2 + L \).
(i) Mean: \( E(W) = E(C_1) + E(C_2) + E(L) \).
\( 810 = 350 + 350 + E(L) \implies E(L) = 810 – 700 = 110 \). (A1)
(ii) Variance: \( \text{Var}(W) = \text{Var}(C_1) + \text{Var}(C_2) + \text{Var}(L) \).
\( 16^2 = 10^2 + 10^2 + \text{Var}(L) \implies 256 = 100 + 100 + \text{Var}(L) \implies \text{Var}(L) = 256 – 200 = 56 \).
Standard deviation: \( \text{SD}(L) = \sqrt{56} \approx 7.4833 \approx 7.48 \). (A1)[4 marks]
(e) Let \( L \sim N(110, \sqrt{56}^2) \). Find \( k \) such that \( P(L < k) = 0.8 \).
From standard normal tables, \( P(Z < 0.8416) \approx 0.8 \), so \( z = 0.8416 \).
\( k = \mu + z \cdot \sigma = 110 + 0.8416 \cdot \sqrt{56} \approx 110 + 0.8416 \cdot 7.4833 \approx 110 + 6.298 \approx 116.298 \).
\( k = 116 \) (to nearest mm). (A1)[1 mark]