Probability/Statistics Question

het SG

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!

FirePig SG

anyone know how to do tis har?

Qn:

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) =0.95

P( 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.

maurizio13 SG

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

eagle SG

so solved already right?

Me can take break le

FirePig SG

so solved already right?

Me can take break le

No. I am not sure of my answer. Would need another point of view.

eagle SG

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)

het SG

should be P( Z < [(100a-2000)]/[30.sqrt(a)]) > 0.95

P( 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..

eagle SG

anyone know how to do tis har?

Qn:

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.

eagle SG

btw, the value of your z, when you substitute back your a, you should get a probability greater than 0.95

FirePig SG

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.

FirePig SG

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

het SG

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?

eagle SG

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...

eagle SG

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.

het SG

should be P( Z < [(100a-2000)]/[30sqrt(a)]) > 0.95

P( 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?

eagle SG

You can try to sub back and try

18 will not give you the required probability

het SG

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~!

FirePig SG

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.

FirePig SG

Sorry for the extremely slow replies as my computer is lagging. I am reading something and is checking the forum from time to time.

maurizio13 SG

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