statistic question

[匿名]

2012-11-30 4:23 pm

1.A computer company finds that the time taken to complete a backup job has a normal distribution with mean 50 minutes and standard deviation 10 minutes.

(a)What percentage of jobs takes less than 35 minutes to finish?

(b)This morning, Sammy has to backup 16 computers. What is the probability the average backup time of 16 computers is more than 54.5 minutes?

The company charges $100 if it takes less than 35 minutes, $200 if it takes between 35 minutes and 60 minutes, and $300 if it takes more than 60 minutes to finish the back up job.

(c) Find the expected charge for a backup job.

A new software package can reduce the back up time such that only 10% of jobs take more than 60 minutes to finish. Suppose the standard deviation remains to be 10 minutes.

(d) Find the new mean back up time.

回答 (1)

用戶25975

2012-11-30 4:58 pm

✔ 最佳答案

a) 35 minutes have a z-score of (35 - 50)/10 = -1.5

So the percentage is P(z < -1.5) x 100% = 6.68%

b) The mean time follows N(50, 100/16)

So the standard deviation of the mean time = √(100/16) = 2.5 minutes

54.5 minutes have a z-score of (54.5 - 50)/2.5 = 1.8

So the required probability is P(z > 1.8) = 0.0359

c) 60 minutes have a z-score of (60 - 50)/10 = 1

So percentage of jobs having mean time > 60 mins is P(z > 1) x 100% = 15.87%

Hence the expected charge is:

0.0668 x 100 + (1 - 0.0668 - 0.1587) x 200 + 0.1587 x 300 = $209.19

d) With P(z > 1.282) = 0.1, 60 minutes should have a z-score of 1.282 and hence:

(60 - μ)/10 = 1.282 where μ is the new mean time.

μ = 47.18 minutes

參考: 原創答案

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