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# S2 probability that 6 bulbs can run for t hours without failure is 0.95 watch

1. Really stuck on this one.

I am trying to work backwards, but the six bulbs is causing my brain problems.

Question is:

Light bulbs lifetime Normally distributed with mean 1100 hours and sd=80 hrs.

Second part of question is where a light fitting of six bulbs in now the subject.

What is the probability that the light fitting can run for t hours without any of the bulbs failing is 0.95. Find t.

Now if the question DIDN'T involve six bulbs I would simply find phi(z) = 0.95, z = 1.645, put that in the z equation to find x and you are done. But the six bulb part is confusing me. How do I incorporate the six bulbs issue into this???
2. (Original post by acomber)
Really stuck on this one.

I am trying to work backwards, but the six bulbs is causing my brain problems.

Question is:

Light bulbs lifetime Normally distributed with mean 1100 hours and sd=80 hrs.

Second part of question is where a light fitting of six bulbs in now the subject.

What is the probability that the light fitting can run for t hours without any of the bulbs failing is 0.95. Find t.

Now if the question DIDN'T involve six bulbs I would simply find phi(z) = 0.95, z = 1.645, put that in the z equation to find x and you are done. But the six bulb part is confusing me. How do I incorporate the six bulbs issue into this???
P(X>x0)= b
b6 = 0.95
the work from there...
3. (Original post by TeeEm)
P(X>x0)= b
b6 = 0.95
the work from there...
This is what I have so far. But it is wrong.

X~N(1100, 6400)

P(X > t) = a

a^6 = 0.95

a = 0.9915

phi(z) = 0.9915, z = 2.409

2.409 = (x - 1100) / 80

x = 2.409 x 80 + 1100 = 1292.72

But I am thinking this is the wrong way round. If you want six together NOT to fail then you would think there are many ways for just one to fail so the x should be somewhere on left.

Where have I gone wrong?
4. (Original post by acomber)
This is what I have so far. But it is wrong.

X~N(1100, 6400)

P(X > t) = a

a^6 = 0.95

a = 0.9915

phi(z) = 0.9915, z = 2.409

2.409 = (x - 1100) / 80

x = 2.409 x 80 + 1100 = 1292.72

But I am thinking this is the wrong way round. If you want six together NOT to fail then you would think there are many ways for just one to fail so the x should be somewhere on left.

Where have I gone wrong?

this is wrong ....

see if you can find it.
5. (Original post by TeeEm)
this is wrong ....

see if you can find it.
phi(z) = 0.9915, z = 2.409

to have 0.9915 of population NOT to fail you have to invert sign. ie z = -2.409

-2.409 = (x - 1100) / 80

x = -2.409 x 80 + 1100 = 907.28

Is that correct?

How would you explain in words why you have to negate z???
6. (Original post by acomber)
phi(z) = 0.9915, z = 2.409

to have 0.9915 of population NOT to fail you have to invert sign. ie z = -2.409

-2.409 = (x - 1100) / 80

x = -2.409 x 80 + 1100 = 907.28

Is that correct?

How would you explain in words why you have to negate z???
This is correct now

I do not have to explain it because the way I set the normal (forward and backward calculations) is very formal and everything is explainable in the algebra.

this is by no means of criticism of your work because 99.99..% of the teachers teach it like you are attempting it (I am unfortunately the odd one out...)
7. (Original post by TeeEm)
This is correct now

I do not have to explain it because the way I set the normal (forward and backward calculations) is very formal and everything is explainable in the algebra.

this is by no means of criticism of your work because 99.99..% of the teachers teach it like you are attempting it (I am unfortunately the odd one out...)
Do you have a web link to the way you teach it? I would be interested in having a look.
8. (Original post by acomber)
Do you have a web link to the way you teach it? I would be interested in having a look.

inverse calculation are in the second part
9. (Original post by TeeEm)

inverse calculation are in the second part
I like this approach. you have to define it at the start a lot more carefully.

But a question on Question 3 in the booklet.

My working is like this.

Jog for at least 14 minutes.

T = time spent jogging
T~N(22, 36)
P(T >= 14) = 1 - P(T < 14) = 1 - (z < (14-22)/6 )

= 1 - phi(-1.333)

Then I get to problem of -ve z and not sure how to deal with using this notation.

Your working seemed to anticipate -ve z like this:

= P(z > (14-22) / 6)

= phi(-1.33)
***
= phi(1.33)

= 0.9082

The switchover wasn't really described by the syntax.

I was wondering if there is a more explicit way of handling?
10. (Original post by acomber)
I like this approach. you have to define it at the start a lot more carefully.

But a question on Question 3 in the booklet.

My working is like this.

Jog for at least 14 minutes.

T = time spent jogging
T~N(22, 36)
P(T >= 14) = 1 - P(T < 14) = 1 - (z < (14-22)/6 )

= 1 - phi(-1.333)

Then I get to problem of -ve z and not sure how to deal with using this notation.

Your working seemed to anticipate -ve z like this:

= P(z > (14-22) / 6)

= phi(-1.33)
***
= phi(1.33)

= 0.9082

The switchover wasn't really described by the syntax.

I was wondering if there is a more explicit way of handling?
φ(-1.96) = φ(1.96) = 0.9750
because φ is EVEN

but φ-1(0.9750)=+- 1.96

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