1092 Calculus B 01-03 Final Exam
June 24, 2021
There are SIX questions in this examination.
1. Consider the differential equation ydy dt =√
1− y2.
(a) (6 pts) If the constant function y(t) = a is a solution of the equation, find the value of a.
(b) (10 pts) Given that y(0) = 1
2, solve y(t) for t near 0.
考慮微分方程式 ydy dt =√
1− y2。
(a) (6 pts) 如果常數函數 y(t) = a 是方程式的解，求a 的值。
(b) (10 pts) 給定 y(0) = 1
2，求 y(t)， 當t在0附近。
2. (16 pts) Solve
⎧⎪⎪⎪⎨⎪⎪⎪
⎩
y′(t) = t(t2− y(t)), y(0) = 3.
(16 pts) 解
⎧⎪⎪⎪⎨⎪⎪⎪
⎩
y′(t) = t(t2− y(t)), y(0) = 3.
3. (8 pts) Suppose that X and Y are independent and E(X) = 1, E(Y ) = 2, Var(X) = 5, Var(Y ) = 7.
Compute E((2X − Y )2).
(8 pts) 已知 X，Y 為兩個獨立的隨機變數，而且 E(X) = 1，E(Y ) = 2， Var(X) = 5， Var(Y ) = 7。
求 E((2X − Y )2)。
4. Suppose that X is a random variable with probability density function fX(t) = ce−t2+8t20 for some constant c> 0.
(a) (8 pts) Find the constant c such that ∫
∞
−∞
fX(t)dt = 1. (Hint: ∫−∞∞e−t2dt=√ π) (b) (8 pts) Find E(X) and Var(X).
(c) (6 pts) Suppose that X1,⋯, Xn are independent random variables and Xi ∼ X for 1 ≤ i ≤ n.
Use Chebyshev’s inequality to estimate the size of n such that we can derive P( ∣X1+ ⋯ + Xn
n − E(X)∣ < 0.05 ) > 0.9.
已知隨機變數 X 的機率密度函數為 fX(t) = ce−t2+8t20 ，其中 c> 0 是一個常數。
(a) (8 pts) 求常數 c 使得 ∫−∞∞fX(t)dt = 1. (提示: ∫−∞∞e−t2dt=√ π) (b) (8 pts) 求 E(X) 和 Var(X).
(c) (6 pts) 假設 X1,⋯, Xn 為互相獨立的隨機變數而且 Xi ∼ X， 1 ≤ i ≤ n。 利用 Chebyshev 不等 式，估計 n 需要多大，我們才能保證
P( ∣X1+ ⋯ + Xn
n − E(X)∣ < 0.05 ) > 0.9.
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5. Let f(t) =⎧⎪⎪
⎨⎪⎪⎩
c⋅ te−t, t> 0 0, t≤ 0.
(a) (8 pts) Find constant c such that f(t) is a probability density function.
(b) (4 pts) Suppose that X and Y are independent with probability density functions fX = fY = f.
Let Z = X + Y . Find the plane region D in the xy-plane such that the distribution function of Z, FZ(t) = P(X + Y ≤ t), is ∬DfX(x)fY(y)dxdy.
(c) (10 pts) Find the probability density function of Z in part (b).
令 f(t) =⎧⎪⎪
⎨⎪⎪⎩
c⋅ te−t, t> 0 0, t≤ 0.
(a) (8 pts) 求常數 c 使得 f(t) 是一個機率密度函數。
(b) (4 pts) 已知隨機變數 X 和 Y 是獨立的而且它們的機率密度函數為 fX = fY = f。 令 Z = X+Y 。求 xy 平面的區域D，使得 Z的分配函數， FZ(t) = P(X+Y ≤ t)， 是重積分 ∬DfX(x)fY(y)dxdy。
(c) (10 pts) 求(b)小題中 Z的機率密度函數。
6. The number of flaws in a carpet appears to be Poisson distributed at a rate of one every 6m2. (a) (8 pts) Find the probability that a 3m by 4m carpet contains more than 2 flaws.
(b) (8 pts) There are two carpets with sizes 2m by 4m and 2m by 5m. Find the probability that these two carpets together contain less than 2 flaws.
假設地毯上的瑕疵個數呈 Poisson 分配，而且平均每 6平方公尺有一個瑕疵。
(a) (8 pts) 求一條 3公尺寬 4公尺長的地毯有超過 2個瑕疵的機率。
(b) (8 pts) 有兩條地毯，各是 2公尺寬 4公尺長和 2公尺寬 5公尺長。求這兩條地毯上總共的瑕疵數少 於 2的機率。
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