问一个MLC的问题，急求解答

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Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20%
of them make only a deposit, 30% make only a withdrawal and the remaining 50% are there
only to complain. Deposit amounts are distributed with mean 8000 and standard deviation
1000. Withdrawal amounts have mean 5000 and standard deviation 2000.
The number of customers and their activities are mutually independent.
Using the normal approximation, calculate the probability that for an 8-hour day the total
withdrawals of the bank will exceed the total deposits.
(A) 0.27
(B) 0.30
(C) 0.33
(D) 0.36
(E) 0.39
这是MLC Sample中的182题，我觉得官方的解答有问题，不知是不是自己的概念错
以下是官方部分解答，有问题的地方我做了注释
Split into three independent processes:
Deposits, with λ* = (0.2)(100)(8) =160 per day
Withdrawals, with λ* = (0.3)(100)(8) = 240 per day
Complaints. Ignore, no cash impact.
For aggregate deposits,
E(D) = (160)(8000) =1,280,000
Var (D) = (160) (1000)^2 + (160) (8000)^2 有问题，方差不是这么算的
=1.04×10^10
For aggregate withdrawals
E(W) = (240)(5000) =1,200,000
Var (W) = (240)( 2000)^2 + (240)( 5000)^2 同样，这里也有问题
= 0.696×10^10
Var (W − D) = 0.696×10^10 +1.04×10^10 =1.736×10^10 这里就更加不对了
求高手指点啊

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关键词：急求解答 MLC 求解答 Independent distributed 解答 MLC

哪里有问题？？
泊松过程的方差计算。。Var=泊松参数*（索赔额方差+索赔额期望平方）（也就是E（X^2））
还有var（aX+bY）=a^2var（X）+b^2var(Y)

2# 红军波波
Var(aX+bY)=a^2Var(X)+b^2Var(Y)这里没有问题
问题是为什么Var=E(X^2)?
Var不是应该为E(X^2)-E(X)^2吗？而且这里E(X)不为零啊
求指点

3# a936300153
不是等于E（x^2）是等于E（x^2）乘以泊松参数。。
用条件方差算一下就出来了

4# 红军波波
哦，我想明白了，非常非常感谢啊～