Probability

I have managed to do part (i).
If L1 and L2 denote the lifetime of D1 and D2 respectively then:
Pr(L1 > 148) = 0.081, Pr(L2 > 148) = 0.067. So D1 should be chosen.

For parts (ii) and (iii) I'm not sure how to formulate what is asked.
In (ii), it seems to me it is a conditional probability of the form Pr(L2 >286.5 given...)

Any ideas?
Thanks

Original post by John taylor

I have managed to do part (i).
If L1 and L2 denote the lifetime of D1 and D2 respectively then:
Pr(L1 > 148) = 0.081, Pr(L2 > 148) = 0.067. So D1 should be chosen.

For parts (ii) and (iii) I'm not sure how to formulate what is asked.
In (ii), it seems to me it is a conditional probability of the form Pr(L2 >286.5 given...)

Any ideas?
Thanks

I think for part (ii) you need to find P(D1 + D2 > 286.5).

Then for part (iii) you need to find P(|D1 - D2|) >5.4.

Reply 2

John taylor

OP

1

Original post by SherlockHolmes

I think for part (ii) you need to find P(D1 + D2 > 286.5).

Then for part (iii) you need to find P(|D1 - D2|) >5.4.

Thanks. I'll try to make sense of that.

Reply 3

Kevin De Bruyne

21

Original post by John taylor

I have managed to do part (i).
If L1 and L2 denote the lifetime of D1 and D2 respectively then:
Pr(L1 > 148) = 0.081, Pr(L2 > 148) = 0.067. So D1 should be chosen.

For parts (ii) and (iii) I'm not sure how to formulate what is asked.
In (ii), it seems to me it is a conditional probability of the form Pr(L2 >286.5 given...)

Any ideas?
Thanks

Think of how you could combine the lengths of times.

Reply 4

John taylor

OP

1

Also, I know that the D1 - D2 is also normally distributed with the mean and variance easily determined.
But how do u deal with the absolute value?

Reply 5

SherlockHolmes

13

Original post by John taylor

Also, I know that the D1 - D2 is also normally distributed with the mean and variance easily determined.
But how do u deal with the absolute value?

If P(|A-B|) < C, then this is equivalent to:

-C < P(A-B) < C

Reply 6

John taylor

OP

1

I see but shouldn't the probability be Pr(abs(D1 - D2) > 5.4).
I.e. the inequality should be within the probability itself.

Reply 7

John taylor

OP

1

The the distribution of D1- D2 can be determined. But how do u handle the absolute value of it...

Reply 8

atsruser

11

Original post by SherlockHolmes

I think for part (ii) you need to find P(D1 + D2 > 286.5).

Don't you need P(D1 + D2 > 286.5 | D1 < 286.5) ?

Reply 9

SherlockHolmes

13

Original post by John taylor

I see but shouldn't the probability be Pr(abs(D1 - D2) > 5.4).
I.e. the inequality should be within the probability itself.

I am not too sure what you mean but does this help?

P(|A-B|) > C is equivalent to:

P(A-B) > C or P(A-B) < -C

Reply 10

John taylor

OP

1

Original post by atsruser

Don't you need P(D1 + D2 > 286.5 | D1 < 286.5) ?

I thought it was a given probability too. I not sure...

Reply 11

SherlockHolmes

13

Original post by atsruser

Don't you need P(D1 + D2 > 286.5 | D1 < 286.5) ?

Original post by John taylor

I thought it was a given probability too. I not sure...

I am not too sure now but I think what atsruser has suggested is correct though. Sorry for the confusion above.

Reply 12

John taylor

OP

1

Thinking about it for a couple minutes Pr(D1 + D2 > 286.5) makes sense.

Reply 13

John taylor

OP

1

I have another probability question, and again i'm not sure what the probabilities there after and how to determine the distribution that is required to solve find the probabilities themselves. From my notes, "normal approximation" could either mean a normal approximation to a poisson or a normal approximation to a binomial. Here i think if i'm not wrong we are dealing with a sample size and probability so the normal approximation to the binomial seems fit. But the question is what are the parameters of this binomial distribution and what are the probabilities we need to calculate in (i) and (ii).

Original post by John taylor

Thinking about it for a couple minutes Pr(D1 + D2 > 286.5) makes sense.

That calculation doesn't seem to tie in with the wording of the question: "What is the probability that the second component is still running after 286.5 hours?".

To my mind, that implies that the first component must have failed by the time we reach 286.5 hours, else the second component won't yet have started running, let alone be "still" running.

But then this is probability, which is more about deciphering subtleties in carefully worded statements than doing maths. So who knows?

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