MATH6038: Spring 2018, Week 11

Student Feedback

If you would like to submit anonymous feedback on this module/lecturer, you may do so here. This link will be open until Friday May 11 2018.

Week 11

We looked at the normal distribution.

In Maple looked at Binomial and Poisson random variables.

Week 12 – Maple Test

The Maple Test should take no more than one hour but I am giving ye extra time. For various reasons, I have decided to schedule next week’s class as:

19:05 – 20:25: Sampling and Control Charts
20:25 – 20:45: Break
20:45 – 22:00: Maple Test

The Maple Test will be open book. You have a sample Maple Test (this is also in the notes) with solutions (*the first with(Statistics) should be with(LinearAlgebra)). The Maple Test will not include anything from Chapter 2 (Lab 4).

We will speak about sampling in more detail and also introduce control charts.

Week 13 – Review

We will hold a review class on Wednesday 9 May in the usual room. First off, the layout of your exam is the same as Autumn 2016 (in the back of your notes): do question one worth 50/100 and two out of questions two, three, four; each worth 25/100.

I will field any questions ye might have at this time and if there are no questions we will do this exam paper. The best possible thing for your study is to do this exam paper and then on Wednesday see how you got on.

Maple Catch Up

If you have missed a lab you have two options: either download Maple onto your own machine (instructions may be found here) or come into CIT at another time to use Maple.

Go through the missed lab on your own, doing all the exercises in Maple. Save the worksheet and email it to me.

The deadline for Maple Catch up is Friday May 11 2018.

Independent Learning

Questions you can do include:

After Week 11: P. 124, Q. 1-10 (this is loads: more is Q. 11-21)
After Week 10: P. 102, Q. 1-4; P. 107, Q. 1-7; P. 111, Q. 1-12 (this is loads: more is Q. 13-16); P. 115, Q. 1-15
After Week 9: P. 92, Q. 1-10 (not too important); P. 96, Q. 1-6 (this is loads: more is Q. 7-13)
After Week 8*: P. 89, Q. 1-3
After Week 7*: P. 89, Q. 1 (a), (b); Q. 2 (b); Q. 3 (b)
After Week 6*: P. 74, Q. 1-4; P. 77, Q. 1-3
After Week 5*: P.44, Q. 1-3, Q. 4-5 more abstract. P.47, Q. 1-3, Q. 4 more abstract. P.56, Q. 1-3, Q. 4 more abstract. P.69, Q. 9 is an important question. A $2\times 2$ version might be

Use only determinants to determine if the following homogeneous system of linear equations has non-zero solutions:

$2x+y=0$

$-6x-3y=0$

After Week 4: P. 41, Q. 1-4
After Week 3: P. 28, Q. 1-5, 6-9 have answers with Q. 7 a harder question. P. 34 exercises.
After Week 2: P. 18 Q. 2
After Week 1: P. 18 Q. 1, 3 – 6. Harder questions are 7 and 8. For those who do not yet have the manual, see here.

I am not suggesting you should do all of these. It is recommended by the module descriptor that you do two hours of independent and directed learning every week but of course this isn’t feasible for everyone.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

2 comments

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May 16, 2018 at 7:50 am

Student

Hi J.P.,

Just going through some past paper questions. I attached a Q2 and Q4 example I worked out, can you have a look at them?

Also in the past papers the formula for Confidence Interval does not seem to there…we will definitely have this?

I’m also having a little trouble with Reliability Block Diagrams where there is series and parallel together…you might do a calculation for this so I can see were I’m going wrong; it’s just this type I’m having issue with…much appreciated

Thanks.

May 16, 2018 at 8:24 am

J.P. McCarthy

I’m afraid I have no idea what year the Q. 2 and Q. 4 you sent on are from and so can’t give very useful comments.

For Q. 2, the table looks fine except that 13 is not the midpoint of any bin and so can’t be an Assumed Mean. You want to midpoint of 10 and 14 which is 12. Note that you won’t have ‘gaps’ in the data like 0-4 | 5-9 | etc. but such data would instead be presented as -0.5-4.5 | 4.5-9.5 | 9.5-14.5 | etc. or probably redone with 0-5 | 5-10 | 10-15 | etc.

For Q. 4, not seeing the question makes it impossible to comment accurately but the work looks good except the UICL is not 101.32 but 101.87. This means that all the sample means should be within the limits but in your graph they are not. Assuming the UICL WAS 101.32, as you graphed it, you should have said that the process is under control with two warnings at samples 5 & 7.

The formula for the Confidence Interval will be there.

Regarding Summer 2012, Q. 1 (f). The following is a very complete answer: I would be happy with the correct values of $\mathbb{P}[\mathcal{C}_1]$ and $\mathbb{P}[\mathcal{C}_2]$ .

Looking first at $\mathcal{S}_1$ . Separate this block diagram into component 1, $A$ , and component 2, $\mathcal{C}_2$ , consisting of the components $B,C,D$ .

$A$ and $\mathcal{C}_2$ are in series and so:

$\mathbb{P}[\mathcal{S}_1]=\mathbb{P}[A\cap \mathcal{C}_2]=\mathbb{P}[A]\times \mathbb{P}[\mathcal{C}_2]$ .

Now, component two consists of two sub components: $\mathcal{X}$ consisting of $B$ and $C$ in series, and component $D$ .

Component $\mathcal{X}$ can be replaced by a single component of reliability:

$\mathbb{P}[\mathcal{X}]=\mathbb{P}[B\cap C]=\mathbb{P}[B]\times\mathbb{P}[C]=0.8\times 0.8=0.64$ .

We find the reliability of $\mathcal{C}_2$ by finding the probability that it fails. Component 2 fails if component $\mathcal{X}$ fails or component $D$ fails.

$\mathbb{P}[\text{not-}\mathcal{C}_2]=\mathbb{P}[(\text{not-}\mathcal{X})\cap(\text{not-})D]=$

$\mathbb{P}[\text{not-}\mathcal{X}]\times \mathbb{P}[\text{not-}D]=(1-0.64)\times(1-0.9)=0.036$ .

Therefore $\mathbb{P}[\mathcal{C}_2]=1-0.036=0.964$ and so

$\mathbb{P}[\mathcal{S}_1]=0.9\times 0.964=0.8676$ .

Consider now $\mathcal{S}_2$ . This consists of $\mathcal{C}_1$ comprised of $B,C,D$ in series. Thus

$\mathbb{P}[\mathcal{C}_1]=\mathbb{P}[B\cap C\cap D]=\mathbb{P}[B]\times\mathbb{P}[C]\times \mathbb{P}[D]$
$=0.8\times0.8\times0.9=0.576$ .

Now for $\mathcal{S}_2$ to fail both $\mathcal{C}_1$ and $A$ must fail:

$\mathbb{P}[\text{not-}\mathcal{S}_2]=\mathbb{P}[(\text{not-}\mathcal{C}_1)\cap(\text{not-}A)]=0.424\times 0.1=0.0424$ .

Therefore $\mathbb{P}[\mathcal{S}_2]=1-0.0424=0.9576$ .

Therefore $\mathcal{S}_2$ is more reliable than $\mathcal{S}_1$ .

Regards,
J.P.

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MATH6038: Spring 2018, Week 11

Student Feedback

Week 11

Week 12 – Maple Test

Week 13 – Review

Maple Catch Up

Independent Learning

Student Resources

Recent Comments

Categories

J.P. McCarthy Maths

2 comments

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MATH6038: Spring 2018, Week 11

Student Feedback

Week 11

Week 12 – Maple Test

Week 13 – Review

Maple Catch Up

Independent Learning

Student Resources

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J.P. McCarthy Maths

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