A technician assembles electronic devices. The mean time required to assemble a

Question

A technician assembles electronic devices. The mean time required to assemble a device is 15.0 minutes, with a standard deviation of 3.0 minutes. The devices are all the same, and the time required to assemble one is independent of the time to assemble any other. (a) What is the expected value of the time needed to produce 25 devices? What is the standard deviation of this quantity? (b) The technician can go home at the end of his 6-hour shift or when he’s completed 25 devices, whichever comes first. Use the central limit theorem to estimate the probability that he goes home early. A technician assembles electronic devices. The mean time required to assemble a device is 15.0 minutes, with a standard deviation of 3.0 minutes. The devices are all the same, and the time required to assemble one is independent of the time to assemble any other. (a) What is the expected value of the time needed to produce 25 devices? What is the standard deviation of this quantity? (b) The technician can go home at the end of his 6-hour shift or when he’s completed 25 devices, whichever comes first. Use the central limit theorem to estimate the probability that he goes home early. A technician assembles electronic devices. The mean time required to assemble a device is 15.0 minutes, with a standard deviation of 3.0 minutes. The devices are all the same, and the time required to assemble one is independent of the time to assemble any other. (a) What is the expected value of the time needed to produce 25 devices? What is the standard deviation of this quantity? (b) The technician can go home at the end of his 6-hour shift or when he’s completed 25 devices, whichever comes first. Use the central limit theorem to estimate the probability that he goes home early.

Explanation / Answer

mean = 15
std. dev. = 3

(a) mean time required to produce 25 devices = 15*25 = 375 mins
Standard deviation = 3*sqrt(25) = 15 mins

(b)
The technician will go home if he completes assembly of 25 devices in less than 6 hours i.e. 6*60 = 360 mins.
here we need to find the probability of assembling devices in less than 360 mins using Central Limit theorem. Mean = 375 and std. dev = 15

P(X<360) = P(z<(360-375)/15) = P(z<-1) = 0.1587

Hence 0.1587 is the probability that the technichian will go home early

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