F.5 Maths M1 distribution(3)

For Machine A, the number of defective article is
X ~ B(100, 0.2).

E(X) = 100(0.2) = 20
Var(X) = 100(0.2)(0.8) = 16

Also note that
Var(X) = E{[X - E(X)]²} = E[(X - 20)²]

The expected daily maintenance cost of Machine A is
$ E[40(X - 20)²]
= $40 E[(X - 20)²]
= $40 Var(X)
= $40 (16)
= $640

For Machine B, the number of defective article is
X ~ B(80, 0.25).

E(X) = 80(0.25) = 20
Var(X) = 80(0.25)(0.75) = 15

Also note that
Var(X) = E{[X - E(X)]²} = E[(X - 20)²]

The expected daily maintenance cost of Machine B is
$ E[40(X - 20)²]
= $40 E[(X - 20)²]
= $40 Var(X)
= $40 (15)
= $600

2015-05-02 17:30:17 補充：
Remark 1
You would wonder this question can only be done as above because 20 is just coincidentally the mean value of the distribution.

How about if the mean is not 20?

You can consider
Var(X) = E(X²) - [E(X)]²
Then, E(X²) can be found as Var(X) + [E(X)]².

2015-05-02 17:31:12 補充：
Next, E[40(X - 20)²]
= 40 E(X² - 40X + 400)
= 40 [E(X²) - 40E(X) + 400]
= ...

2015-05-02 17:34:20 補充：
Remark 2
You asked what f(x) is

Machine A
X ~ B(100, 0.2), so
f(x) = (100Cx) (0.2)^x (0.8)^(100 - x) for x = 0, 1, 2, ..., 100.

Machine B
X ~ B(80, 0.25), so
f(x) = (80Cx) (0.25)^x (0.75)^(80 - x) for x = 0, 1, 2, ..., 80.

but it is inefficient to use f(x) to get expected value.

2015-05-02 17:34:52 補充：
Remark 2
You asked what f(x) is

Machine A
X ~ B(100, 0.2), so
f(x) = (100Cx) (0.2)^x (0.8)^(100 - x) for x = 0, 1, 2, ..., 100.

Machine B
X ~ B(80, 0.25), so
f(x) = (80Cx) (0.25)^x (0.75)^(80 - x) for x = 0, 1, 2, ..., 80.

but it is inefficient to use f(x) to get expected value.

2015-05-03 14:36:59 補充：
～～～　可能會字數滿額，請看意見欄　～～～

2015-05-03 14:37:55 補充：
Therefore, for discrete random variable,
we should use E(X²) and [E(X)]² to find Var(X).

> No, it depends no situation.
> There is no must to use this for discrete or continuous r.v.

2015-05-03 14:38:55 補充：
For Binomial and Geometric distribution,
we should use Var(X) and [E(X)]² to find E(X²).

> Not necessarily.
> We do this because we know the mean and variance of certain famous distribution.
> In general, you need to be flexible in using the relationship:
> Var(X) = E(X²) - [E(X)]²

2015-05-03 14:40:19 補充：
> There is not a must to use what method.
> It is important for you to know the whole picture, rather than study like an exam machine, that is, see A and do B, see C and do D.
> This harms you.

2015-05-03 14:41:13 補充：
(I tried to substitute 100 and 80 into 'x²' to find E(X²) for machine A and B respectively 0_0)

> This is not useful this time.
> If you use this method, you need to add up 100 and 80 times for those terms.