Probability/Statistics Question
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anyone know how to do tis har?
Qn:
A certain component is critical to operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and standard deviation 30 hours, how many of these components must be in stock so that probability that system is in continual operation for the next 2000 hours is at least 0.95?
i tried findin P(X>2000) = P( Z > [(2000-100a)]/[30.sqrt(a)]) = 0.95, where a is the no. of components required in stock.
then, P( Z < [(2000-100a)]/[30.sqrt(a)]) = 0.05 gif [(2000-100a)]/[30.sqrt(a)]=1.645, but ah i cant solve from here liaoz! anyone know where i got stuck?
rather, e value i obtain for a is not correct when I work back to verify ans, meaning when i sub a into P( Z > [(2000-100a)]/[30.sqrt(a)]), i dun get 0.95 as ans!
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Originally posted by het:
anyone know how to do tis har?
Qn:
A certain component is critical to operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and standard deviation 30 hours, how many of these components must be in stock so that probability that system is in continual operation for the next 2000 hours is at least 0.95?
i tried findin P(X>2000) = P( Z > [(100a-2000)]/[30.sqrt(a)]) = 0.95, where a is the no. of components required in stock.
then, P( Z < [(100a-2000)]/[30.sqrt(a)]) = 0.05 gif [(100a-2000)]/[30.sqrt(a)]=1.645, but ah i cant solve from here liaoz! anyone know where i got stuck?
rather, e value i obtain for a is not correct when I work back to verify ans, meaning when i sub a into P( Z > [(2000-10a)]/[30.sqrt(a)]), i dun get 0.95 as ans!
should be P( Z < [(100a-2000)]/[30sqrt(a)]) > 0.95 because P(Z< 1.645) =0.95P( Z > [(100a-2000)]/[30sqrt(a)]) < 0.05 because P(Z > 1.645) = 0.05
Once you get a from [100a-2000]/30sqrt(a) > 1.645, you will get your solution.
You will get a > 22.3 round up a = 23. The other solution is rejected.
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lifetime of component = 100 ± 1.96 x 30/[42.68]^(0.5)
lifetime of component = 100 ± 9
shortest lifespan = 91 hours
for 2,000 hours operation with C.I. of 95%, you need:
2,000 hours /91 hours = 21.98 components
approximately 22 components.
n = [(1.96 x 100)/(30)]^2
n = 42.68
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so solved already right?
Me can take break le
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Originally posted by eagle:
so solved already right?
Me can take break le
No. I am not sure of my answer. Would need another point of view. -
Originally posted by FirePig:
No. I am not sure of my answer. Would need another point of view.But I always always do badly in statistics de wor... Last time C maths, I chose all mechanics for my options... F maths also 5 mechanics (the max) and 2 stats (the min)
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Originally posted by FirePig:
should be P( Z < [(100a-2000)]/[30.sqrt(a)]) > 0.95P( Z > [(100a-2000)]/[30.sqrt(a)]) < 0.05
Once you get a from [100a-2000]/30sqrt(a) > 1.645, you will get your solution.
You will get a > 22.3 round up a = 23. The other solution is rejected.
i dun understand why you use "<" instead of ">"..btw, wat is e solution u rejected? your 1st ans was 18 - y u wanna reject 18 and not 22.got reason?
i tink maurizio13 make more sense cos he oni consider the shortest lifespan case..
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Originally posted by het:
anyone know how to do tis har?
Qn:
A certain component is critical to operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and standard deviation 30 hours, how many of these components must be in stock so that probability that system is in continual operation for the next 2000 hours is at least 0.95?
i tried findin P(X>2000) = P( Z > [(2000-100a)]/[30.sqrt(a)]) = 0.95, where a is the no. of components required in stock.
then, P( Z < [(2000-100a)]/[30.sqrt(a)]) = 0.05 gif [(2000-100a)]/[30.sqrt(a)]=1.645, but ah i cant solve from here liaoz! anyone know where i got stuck?
rather, e value i obtain for a is not correct when I work back to verify ans, meaning when i sub a into P( Z > [(2000-100a)]/[30.sqrt(a)]), i dun get 0.95 as ans!
I guess you are trying to do normal distribution
Why is your standard deviation [30.sqrt(a)]? Shouldn't it be just 30a?
Even if you take variance, the property is
var (cX) = c^2 var(x)
standard deviation is just the square root of the above. -
btw, the value of your z, when you substitute back your a, you should get a probability greater than 0.95
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Originally posted by het:
i dun understand why you use "<" instead of ">"..btw, wat is e solution u rejected? your 1st ans was 18 - y u wanna reject 18 and not 22.got reason?
i tink maurizio13 make more sense cos he oni consider the shortest lifespan case..
this question will yield a quadratic curve 10000a^2 - 402435a + 4000000 > 0,
When you solve it, you will get a > 22.3 and a < 17.9.
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Originally posted by eagle:
I guess you are trying to do normal distribution
Why is your standard deviation [30.sqrt(a)]? Shouldn't it be just 30a?
Even if you take variance, the property is
var (cX) = c^2 var(x)
standard deviation is just the square root of the above.
I also get stuck by this. But i think this is a binomial distribution.If n is large enough, the skew of the distribution is not too great, and a suitable continuity correction is used, then an excellent approximation to B(n, p) is given by the normal distribution
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i tink maurizio13 is right...juz dun understand how he calculate n..as in, got any web links tat can explain why n is calculated tat way? i mean, how u obtain tat as e sample size leh?
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Originally posted by FirePig:
I also get stuck by this. But i think this is a binomial distribution.If n is large enough, the skew of the distribution is not too great, and a suitable continuity correction is used, then an excellent approximation to B(n, p) is given by the normal distribution
just now mistake... 30 sqrt (a) is correct
We are not taking var (cX) here, but c Var(x) instead. This is because the X here are all independent variables... My mistake...
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Originally posted by het:
i tink maurizio13 is right...juz dun understand how he calculate n..as in, got any web links tat can explain why n is calculated tat way? i mean, how u obtain tat as e sample size leh?
does it make sense if the number required is more than 40?
I would say FirePig is correct.
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Originally posted by FirePig:
should be P( Z < [(100a-2000)]/[30sqrt(a)]) > 0.95P( Z > [(100a-2000)]/[30sqrt(a)]) < 0.05
Once you get a from [100a-2000]/30sqrt(a) > 1.645, you will get your solution.
You will get a > 22.3 round up a = 23. The other solution is rejected.
P( Z > [(2000-100a)]/[30.sqrt(a)]) --> P( Z < [(100a-2000)]/[30sqrt(a)]) . nth wrong there..BOTH r correct cos u merely reversed the 2000-100a to 100a-2000. prob solved liaoz, but still wonderin..how come take a=23 and not a=18? i mean, how u know which value to take?
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Originally posted by het:
P( Z > [(2000-100a)]/[30.sqrt(a)]) --> P( Z < [(100a-2000)]/[30sqrt(a)]) . nth wrong there..BOTH r correct cos u merely reversed the 2000-100a to 100a-2000. prob solved liaoz, but still wonderin..how come take a=23 and not a=18? i mean, how u know which value to take?
You can try to sub back and try
18 will not give you the required probability
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so u r sayin to verify,we muz sub in and check both values to see which is right?
thx everyone for helping to solve tis problem~!
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Originally posted by het:
P( Z > [(2000-100a)]/[30.sqrt(a)]) --> P( Z < [(100a-2000)]/[30sqrt(a)]) . nth wrong there..BOTH r correct cos u merely reversed the 2000-100a to 100a-2000. prob solved liaoz, but still wonderin..how come take a=23 and not a=18? i mean, how u know which value to take?
this question will yield a quadratic curve (2000-100a)>1.645*30*sqrt(a)
(2000-100a)^2>[1.645*30*sqrt(a)]^2
10000a^2 - 402435a + 4000000 > 0
This is actually a U-shaped curve where if you want to have your answer > 0, you have to get inequalities using the two end points,
When you solve it, you will get a > 22.3 and a < 17.9
Why take a =23 and not a=17? By intuition...you need a number that is at least greater than some number.
When you square a number, you tend to get unwanted solution. we want to show a number 100a is always greater than 2000 and calculate its probability so it is more accurate to write 100a first. This way, you will know 100a=2300 > 2000.
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Sorry for the extremely slow replies as my computer is lagging. I am reading something and is checking the forum from time to time.
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Originally posted by het:
i tink maurizio13 is right...juz dun understand how he calculate n..as in, got any web links tat can explain why n is calculated tat way? i mean, how u obtain tat as e sample size leh?
slide 55 to 60
http://www70.homepage.villanova.edu/hae-kyong.bang/17-Determining%20Sample%20Size.PPT
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Originally posted by maurizio13:
Formula is [zs/E]^2
you should sub in s=30 and not 100 in your first post and E will be unknown. Anyway, these are two different questions. We are not looking for a sample size based on error bound in this question. We are looking for sample size based on the fact that it must exceed 2000 with 95% confidence.
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Originally posted by FirePig:
Formula is [zs/E]^2
you should sub in s=30 and not 100 in your first post and E will be unknown. Anyway, these are two different questions. We are not looking for a sample size based on error bound in this question. We are looking for sample size based on the fact that it must exceed 2000 with 95% confidence.
think you are right.
