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2 years ago

10

You arrive at a bus stop at 1:00, knowing that the bus will arrive at some time uniformly distributed between 1:00 and 1:30. a)

What is the probability that you will have to wait longer than 10 minutes

1 answer:

Likurg_2 [28]2 years ago

7 0

Answer:

2/3

Step-by-step explanation:

You arrived at 1:00 and longer than 10 minutes would be at 1:20 and 1:30

that is 2 chances of you waiting longer than 10 minute

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An assembled product is built by 10 worker who coordinate their task. A total of 100 units will be made. Time records indicate t

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Answer:

a) 86%

b) 2nd unit = 82 hs.

3rd unit = 75 hs

c) 100th unit = 36 hs

Step-by-step explanation:

We can model the learning curve for manufacturing the units as:

$t=aX^b$

where t is the time for the Xth unit, and a and b are parameters that we will calculate from the data,

We know that t(1)=95. Then, we have:

$t(1)=a\cdot 1^b=95\\\\a=95$

And we know that the fourth unit (X=4) take 71 hours to be completed (t(4)=71). Then, we can calculate the other parameter as:

$t(4)=95\cdot4^b=71\\\\4^b=71/95\approx 0.7473\\\\b\cdot ln(4)=ln(0.7473)\\\\b=ln(0.7473)/ln(4)=-0.291/1.386\\\\b=-0.21$

We have the model for the learning curve:

$t=95X^{-0.21}$

The learning rate percentage is calculated from the b parameter:

$b=\dfrac{ln(LRP)}{ln(2)}=-0.21\\\\\\ln(LRP)=-0.21*ln(2)=-0.21*0.693=-0.1455\\\\LRP=e^{-0.1455}=0.86$

The learning rate percentage is 86%.

b) The most likely times required for the 2nd and 3rd units are calculated with the model:

$t(2)=95\cdot2^{-0.21}=95*0.864=82\\\\t(3)=95\cdot3^{-0.21}=95*0.794=75$

c) If we use the model to calculate the time required for the 100th unit, we have:

$t(100)=95\cdot100^{-0.21}=95*0.38=36$

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