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 =
=