advertisement

Physics 2BL Homework Set 03 Taylor Problems: 5.2, 5.6, 5.20, 5.36 5.2: fk 0.5 0.4 0.3 Interval for t (in minutes) 0-10 Number of counts 9 10-20 6 20-30 3 30-40 1 40-50 1 0.2 0.1 HL tk t (minutes) minutes 10 20 30 40 50 60 5.6: (A) f(t) = 1 τ e -t / τ t (minutes) ∞ ∫ f (t )dt = 1 (B) Eqn. 5.13 states: −∞ We are told f(t) = 0 for t < 0 in our case, so ∞ ∞ ∞ −∞ 0 0 ∫ f (t )dt = ∫ f (t )dt = ∫ (1 / τ )Exp(−t / τ )dt ∞ ∫ −∞ −∞ −∞ 0 0 Let u = -t/ τ, du = -(1/ τ)dt → dt = - τ du f (t )dt = ∫ (−τ / τ )e u du = − ∫ e u du =e 0 − e −∞ = 1 − 0 = 1 , as expected ∞ (C) Eqn. 5.15 states: t = ∫ tf (t )dt −∞ Since f(t) = 0 for t < 0, we have ∞ ∞ −∞ 0 ∫ tf (t )dt = ∫ (t / τ ) Exp(−t / τ )dt −∞ ∞ ∞ −∞ 0 ∫ tf (t )dt = ∫ tf (t )dt Let u = -t/ τ, du = -(1/ τ)dt → dt = - τ du −∞ 0 0 −∞ = ∫ ue u (−τ )du = −τ ∫ ue u du = τ ∫ ue u du 0 ⎡ = τ ⎢ue u ⎣ 0 −∞ ⎤ − ∫ e u du ⎥ = τ ⎡0 − 0 − e u ⎢⎣ −∞ ⎦ 0 0 −∞ ⎤ = τ [1 − 0] = τ ⎥⎦ 5.20: Mean height h = 69”, σ = 2” (A) How many men between 67” and 71” = h ± 1σ 68% should be within the range of h ± 1σ . → N = 1000 · 68% = 680 men (B) How many men taller than 71” = h + 1σ If 68% should be within the range of h ± σ , 32% should be outside that range. Assuming it is symmetric, 16% should be lower than h − σ and 16% should be above h +σ . → N = 1000 · 16% = 160 men (C) How many men taller than 75” = h + 3σ For same reasons as above, half of the men outside the range of h ± 3σ → N = (1/2) (1-0.997) · 1000 = 1.5 ≈ 2 men (D) How many men between 65” and 67” = (h − 2σ ) and (h − σ ) How many inside h + 2σ = (0.9545) 1000 = 954.5 How many inside h + σ = (0.68) 1000 = 680 (1/2) (955 – 680) = 137.5 ≈ 138 men 5.36: x A = 13 ± 1 , x B = 15 ± 1 (A) x A − xB = 2 ± 12 + 12 = 2 ± 1.4 x A − xB 2 = 1.4 σ 1.4 Prob(outside 1.4 σ) = 1 – Prob(within 1.4 σ) = 1 – 0.8385 ≈ 0.16 Prob(outside 1.4 σ) = 16% Assuming 5% tolerance, this discrepancy is not significant. (B) t = =