I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.
Week 11
We looked at the normal distribution in more detail and talked about sampling.
Week 12
We will talk about sampling in more details and also introduce control charts.
Independent Learning: Exercices
You are supposed to be working outside of class and I am supposed to help you with this. Working outside of class means doing the exercises in the notes. Any work that is handed up will be corrected by me. Also you can ask me a question here on this site and I will answer it ASAP.
Questions that you can do at this point include:
- P. 125 Q. 1-21, particularly Q. 5-10
I wouldn’t necessarily advise looking at ALL of these questions: I would focus on the exam questions over the other questions.
Academic Learning Centre
Those in danger of failing need to use the Academic Learning Centre. As you can see from the timetable there is evening support. You will get best results if you come to the helpers there with specific questions.
Math.Stack Exchange
If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly.
Additional Notes: E-Books
If you look in the module descriptor, you will see there is some suggested reading. Of course I think my notes are perfect but if you can look here, search for ‘Bird Higher Engineering’ you will see that the library have an E-Book resource.
J.P. McCarthy: Math Page 
6 comments
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April 15, 2014 at 10:48 am
Student
Hi J.P,
I was doing a normal distribution question: P.125, Q.6, it also on the exam paper at the back of the book as Q. 1(b).
It’s looking for out of 20 holes drilled what is the nearest whole number that have the diameter between 4.048 and 4.056.The answer is 15 on the book
I started off by getting the
values for each which… 4.048 was -0.7143 which i rounded off to 0.71 and i got .7611 and the other 4.056 was 2.143 which i rounded off to 2.1 and got .9821 from the tables………Is this right so far?
Then i calculated .9821-(1- .7611) = .7432 now this is where i don’t know what to do next but i got an answer of 15 and i think it might be pure luck so if you could tell me if it’s right or the right way of doing it
What i did is I divided the mean 4.05 by my answer .7432 = 5.4494
and took if away from the 20 holes drilled which was 20 – 5.4494 = 14.55
which when you round if off gives you 15 holes
Is this right? Thank you again for you time
Kind regards.
April 15, 2014 at 10:52 am
J.P. McCarthy
Looks pretty good down to 0.7432 but after that you might have gotten the correct answer but it doesn’t make a whole pile of sense to me!
What you should have is that where
is the diameter, ![\mathbb{P}[4.048\leq X\leq 4.056]=0.7432](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BP%7D%5B4.048%5Cleq+X%5Cleq+4.056%5D%3D0.7432&bg=ffffff&fg=545454&s=0&c=20201002)
. This means that the probability that the diameter is between those limits is 0.7432: in other words 74.32% of the objects have a diameter between those two.
If you have 20 then you expect 0.7432(20)=14.864 of the objects to have a diameter in that range which rounds up to 15.
Regards,
J.P.
April 15, 2014 at 11:50 am
Student
Thanks J.P,
I knew there was an easy way about it.
I’m just stuck with one more and I’m finished with this type of question. On the same page, Q. 5 at the end it asks: “Determine
if the consumption is in the interval 
, on 90% of the days”.
How is this done?
April 15, 2014 at 11:57 am
J.P. McCarthy
Translate the question into one about
values. For what value of 
does 90% of the 
values fall inside 
?
Can you see that
? So we have for normal data 90% of the data falls within 1.65 standard deviations of the mean.
Now translate this back into the
picture where 
will be 1.65 standard deviations.
Regards,
J.P.
April 16, 2014 at 12:30 pm
Student
Hi J.P,
Could you give me solutions for P.117 Q. 10, P.125 Q. 5 (d) and P.126 Q. 8 (d) if you get a chance thanks!
April 16, 2014 at 12:48 pm
J.P. McCarthy
P.117 Q.10.
Picture yourself in the printing room. Four students didn’t print anything, seven students printed one file (seven files per 11 students), ten students printed two files so that is 20 files (27 files per 21 students), three students printed three files so that is nine files (36 files per 24 students), one student printed four files (40 files per 25 students).
So altogether we have 40 files per 25 students that is an average rate of
files per student per hour.
Now let![X\sim \text{Poi}[1.6]](https://s0.wp.com/latex.php?latex=X%5Csim+%5Ctext%7BPoi%7D%5B1.6%5D&bg=ffffff&fg=545454&s=0&c=20201002)
be the number of files sent to the printer by a student in an hour. We are interested in ![\mathbb{P}[X=2\cup X=3]=\mathbb{P}[X=2]+\mathbb{P}[X=3]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BP%7D%5BX%3D2%5Ccup+X%3D3%5D%3D%5Cmathbb%7BP%7D%5BX%3D2%5D%2B%5Cmathbb%7BP%7D%5BX%3D3%5D&bg=ffffff&fg=545454&s=0&c=20201002)
, which is given by
P. 125 Q. 5 (d)
Translate the question into one about
values. For what value of 
does 90% of the 
values fall inside 
?
Can you see that
? So we have for normal data 90% of the data falls within 1.65 standard deviations of the mean.
Now translate this back into the![X\sim N[8000,1200]](https://s0.wp.com/latex.php?latex=X%5Csim+N%5B8000%2C1200%5D&bg=ffffff&fg=545454&s=0&c=20201002)
picture where 
will be 1.65 standard deviations.
P.126 Q. 8 (d)
Translate the question into one about
values. For what value of 
is the probability that 
is less than 
is 0.05/
Can you see that
? So we have for normal data 5% of the data is smaller than the mean less 1.65 standard deviations.
Now translate this back into the![X\sim N[150,15]](https://s0.wp.com/latex.php?latex=X%5Csim+N%5B150%2C15%5D&bg=ffffff&fg=545454&s=0&c=20201002)
picture where 
will be 
.
Regards,