Statistic and Probability Q3

2013-10-16 8:31 pm
Question 1
In a production process, a total number of 50 items were produced by machines M1, M2 and M3. Of those items being produced, 27 were made by machine M1, 13 were made by M2, and 10 were made by M3.
The three machines work independently, however, they do not work perfectly.
From the past experience, 9% of the items produced by M1 are defective, 7% of the items produced by M2 are defective, and 4% of the items produced by M3 are defective.
(a) Given that a randomly selected item is non-defective, what is the probability that it is produced by machine M2?
(b) If two items which are not produced by machine M3 are selected at random without replacement, what is the probability that at least one of them is non-defective?

Question 2

A machine produces vitamin C tablets, the diameters of which are normally distributed with mean 10mm and standard deviation 0.1mm. A tablet is acceptable if its diameter lies between 9.81mm and 10.19mm. To pass a quality test, at least 90% of the tablets in a pack should be acceptable.

(a) Determine the probability that a tablet is acceptable.
(b) What is the probability that a pack of 20 tablets will pass the quality test?
(c) What is the probability that a pack of 200 tablets will pass the quality test?


Question 3

A factory produces a certain type of ball bearings whose outer diameters are normally distributed with mean 52 mm and standard deviation 0.05 mm. A ball bearing of this type is suitable for manufacturing certain electric motors if its outer diameter is between 51.92 mm and 52.06 mm. Using normal approximation to binomial, find the minimum number of ball bearings that have to be produced so that the probability of having at least 500 ball bearings suitable for manufacturing the electric motors is at least 0.95.

回答 (2)

2013-10-17 11:22 am
✔ 最佳答案
Good, I will answer you tonight if no one help you.

(=^o^=)

2013-10-16 16:58:13 補充:
You can use normal approximation to binomial probability.
That is common.

I denote p as your answer to part (a).

X ~ B(200, p)

Approximately, X ~ N(200p, 200p(1-p))
Find Pr(X>=180) = Pr(Z > (179.5 - 200p)/sqrt[200p(1-p)] ) with continuity correction.

2013-10-17 03:20:00 補充:
嗯,你問了一連串的問題~
不如你先看看我的作答,然後再想一次明白與否。

最緊要必須要明白題目的情況是什麼和它究竟問什麼,否則什麼也不能做。

2013-10-17 03:22:02 補充:
Please read:


圖片參考:http://imgcld.yimg.com/8/n/HA00430218/o/20131017032122.jpg
2013-10-16 11:04 pm
Thx so much >~<"
I think I figure out how to do Q2 part a and b
but Q2 c ..... I really have no idea how to calculate it /-\
Do I have to add from P(x=180) +.... + P(x=200) ? =~+
OMG

2013-10-16 21:12:45 補充:
I can do Question 2 la thx

But how to do Q3 ar

I don't even understand what the question ask ? ~-~"
Mean is 52mm.
SD is 0.05mm.
then the suitable scope of value is between 51.92=< 52.06

What is 500 ?
the least no. of suitable ball bearings found ?

2013-10-16 21:17:06 補充:
What is 0.95 mean?
it means the probability of suitable ones found in the "n" produced ball bearings is 0.95 ?

Then, we have to find the value of "n" ?

Do I need to use equations, like " mean=np , SD= sq.root ( npq )
?


收錄日期: 2021-04-16 16:13:31
原文連結 [永久失效]:
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