Homework 12 for MATH/STAT 414, Spring 2024 Section 002
1. Let A be the rectangle in the (x,y)-plane given by a ≤ x ≤ b and 0 ≤ y ≤ c. Let B be the region under the curve y = f(x) (and above the x-axis) for a ≤ x ≤ b, so the integral 1a(b) f(x)dx is the area of region B . In practice, it is not unusual that we cannot evaluate this integral analytically because the function f(x) is too complicated. In this case, we seek for a numerical (approximate) solution. One possible strategy is to take random samples from A, calculate the proportion of the samples that also fall into the area B, and multiply the proportion by the area of A. This is depicted in the following figure; points in B are in black, and points not in B are in white.
To see why this works, suppose we pick i.i.d. points (X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ) uniformly in the rectangle A. Define indicator r.v.s I1 ,..., In by letting Ij = 1 if (Xj , Yj ) is in B and Ij = 0 otherwise. Then the Ij are Bernoulli r.v.s whose success probability θ is precisely the ratio of the area of B to the area of A. That is,
Then, the proportion of the random points that fall into the area B is Σj(n)=1 Ij . Multiplying
this proportion by the entire area of A is
(a) (3 pts) Explain why this quantity converges to the desired area B (=1a(b) f(x)dx).
Hint: If Xn → k as n → ∞, then cXn → ck as n → ∞ .
(b) (3 pts) Find an approximate distribution of the numerical solution,
as n → ∞ .
2. From past experience, we know that the test score of a student taking their final examination is a random variable with mean 80 (out of 100) and variance 16.
(a) (3 pts) Find an upper bound for the probability that a student’s test score will exceed 90.
(b) (3 pts) Find an lower bound for the probability that a student will score between 70 and 90.
(c) (3 pts) How many students would have to take the examination to ensure, with probability at least 0.95, that the class average would be within 5 of 80?
(d) (3 pts) Using the CLT, find the approximate probability that the class average would be within 5 of 80. The number of students is determined in (c). You can leave your answer like P(a < Z < b) with clearly specified constants a and b. Here Z is a standard Normal r.v.
3. (3 pts) The times to process orders at the service counter of a pharmacy are exponentially distributed with mean 20 minutes and they are independent to each other. If 100 customers visit the counter in a 2-day period, what is the probability that at least half of them need to wait more than 20 minutes? Feel free to use a Normal approximation.
Hint: Let Xj be the waiting time of the j-th customer, and let us define an indicator Ij that is 1 if the j-th customer waits more than 20 minutes and 0 otherwise. Then the probability of interest is P (Σj(1) Ij ≥ 50). Since the sum of 100 indicators is discrete, do not forget to apply the continuity correction.
4. Each of the batteries in a collection of 50 batteries is equally likely to be either a type A or a type B battery. Type A batteries last for an amount of time that has mean 40 and standard deviation 10; type B batteries last for an amount of time that has mean 30 and standard devi- ation 5. Let Xi be the lifetime of the i-th battery (either type A or B), and the lifetimes of 40 batteries are independent to each other.
(a) (3 pts) Find the mean and variance of Xi using the Adam’s and Eve’s laws.
Hint: Let IA be an indicator of whether the chosen battery is type A or not. By the Adam’s and Eve’s laws, we can say
E(X1 ) = E(E(X1 | IA )) = E(X1 | IA = 1)P(IA = 1) + E(X1 | IA = 0)P(IA = 0).
Var(X1 ) = E(Var(X1 | IA )) + Var(E(X1 | IA ))
= Var(X1 | IA = 1)P(IA = 1) + Var(X1 | IA = 0)P(IA = 0)
+ {E(X1 | IA = 1)2 P(IA = 1) + E(X1 | IA = 0)2 P(IA = 0) − E(X1 )2 }.
(b) (3 pts) Approximate the probability that the total life of all 50 batteries exceeds 1700. You can leave your answer in the following format, without calculating the exact number.
where Z ∼ N(0, 1) with a clearly specified constant c.
(c) (3 pts) Suppose it is known that 20 of the batteries are type A and 30 are type B . Now approximate the probability that the total life of all 50 batteries exceeds 1700. You can also leave your answer, saying that the probability is approximately the same as P(Z > c′ ). Without loss of generality, let us assume that the first 20 batteries are type A batteries and the last 30 are type B .
Hint: If X ∼ N(a,b) and independently Y ∼ N(c,d), then (X + Y) ∼ N(a + c, b + d). This property holds for approximate distributions.