好深既statictics 2 --- probability

[匿名]

2008-05-07 8:44 am

4. Grade point averages of students on a large campus follow a normal distribution with mean 2.6 and standard deviation 0.5.

a. One student is chosen at random from this campus. What is the probability that this student has a grade point average higher than 3.0?
b. What is the minimum grade point average needed for a student’s grade point average to be among the highest 10% of the campus?
c. Two students are chosen at random from this campus. What is the probability that at least one of them has a grade point average higher than 3.0?

5. Suppose that a new computerized claims processing system has been installed by a major health insurance company. Only 40 % of the claims require work by a human claims processor when this system is used. On a particular day 100 claims arrived for processing. What is the probability that:

(1) there are between 37 and 43 (inclusive) claims that require work by a human?
(2) there are at most 38 claims that need the attention of a human?
(3) there are more than 42 claims that require work by a human?

6. A particular industrial product is shipped in lots of 20. Testing to determine whether an item is defective is costly; hence, the manufacturer samples production rather than using a 100% inspection plan. A sampling plan constructed to minimize the number of defectives shipped to customers’ calls for sampling five items from each lot and rejecting the lot if more than one defective is observed. (If rejected, each item in the lot is then tested.) if a lot contains four defectives, what is the probability that it will be accepted?

回答 (1)

魏王將張遼

2008-05-07 6:00 pm

✔ 最佳答案

(4a) A grade point of 3.0 has a standard score of:
(3.0 - 2.6)/0.5 = 0.8
So the probability that the chosen student has a grade point higher than 3.0 is:
P(σ > 0.8) = 0.2119
(4b) According to the normal distribution table, the required standard score for satisfying the criteria is 1.2817
So the required grade point is 2.6 + 1.2817 x 0.5 = 3.3 (rounded up)
(4c) Alternatively, we find out the probability that BOTH of them don't have a grade point average higher than 3.0 which is:
(1 - 0.2119)2 = 0.6211
So the probability that at least one of them has a grade point average higher than 3.0 is 1- 0.6211 = 0.3789
(5) We may use the Poisson distribution with parameter λ = 40 which is the expected no. of occurence.
So for (1), the probability is:
(4037/37! + 4038/38! + 4039/39! + 4040/40! + 4041/41! + 4042/42! + 4043/43!) x e-40
= 0.4199
For (2), the probability is:
Σ (k = 0 to 38) [(40k/k!) x e-40] = 0.4160
For (3), the probability is:
Σ (k = 43 to 100) [(40k/k!) x e-40] = 0.2770
(6) The lot will be accepted if no or one defective item is found.
We can use the Binomial distribution for it:
No. of possible combinations in choosing 5 out of 20 = 20C5
For no defective item found, no. of possible combinations is 16C4 which is equivalent to choosing 4 components out of 16 non-defective.
For one defective item found, no. of possible combinations is 16C3 x 4C1 which is equivalent to choosing 3 components out of 16 non-defective and 1 out of 4 defective.
Hence the required probability is:
(16C4 + 16C3 x 4C1)/20C5 = 0.2619

2008-05-07 10:03:18 補充：
Sorry for the mistake in Q6:

For no defective item found, no. of possible combinations is 16C5 which is equivalent to choosing 5 components out of 16 non-defective.

2008-05-07 10:03:22 補充：
For one defective item found, no. of possible combinations is 16C4 x 4C1 which is equivalent to choosing 4 components out of 16 non-defective and 1 out of 4 defective.

Hence the required probability is:
(16C5 + 16C4 x 4C1)/20C5 = 0.7513

參考: My Maths knowledge

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