Weekly Practice Questions for Exam C

[sam_broverman]

Frankie,

The level of difficulty is variable, but I think,
on average, these questions are about the same
level as actual exam questions.

Sam Broverman

[sam_broverman]

You can find the question at:
http://www.sambroverman.com/may1c.pdf

Sam Broverman
2brove@rogers.com

[alphaace]

Hi,

I agree with your solution up to the point ln (n) < 4.605. However if the max integer value of n is as the solution says 100, then ln(n) = 4.605170186 which is greater than 4.605. So shouldn't n really be 99?

Thanks,
Greg

[sam_broverman]

Greg,

You are right. Based on the Exam C Chi-Square table
value of 9.210 , we get n < 99.99 , so the max n should have
been 99. I chose numbers that cut the inequality
too close. I'll post a correction. Thanks for mentioning that.

SB

[sam_broverman]

You can find the question at:
http://www.sambroverman.com/may8c.pdf


Sam Broverman
2brove@rogers.com

[alphaace]

I am a bit confused why for the final answer you put f '(0) = -.8. I thought we were asked to find the natural cubic spline so wouldn't we in fact need three functions there in terms of x?

Thanks,
Greg R.

[sam_broverman]

What I meant was to ask for was f'(0) .
I have posted a revised version of the question.

SB

[Sherri]

the question has not pointed out that we must use natural cubic spline!
hence i can use interpolation polynomials, and i work it out that
f(x) = x^3 - x (it fits the sample well )
so f'(0)=-1.
doesn't it make sense?
thanks

[sam_broverman]

Sherri,

Your answer makes sense. My revision to the
question doesn't make sense. When I revised
the question, somehow I deleted the part of
the question indicating that it is a natural
cubic spline that is to be found. I'll revise
the revision shortly. Thanks for pointing that out.

Sam Broverman
2brove@rogers.com

[sam_broverman]

You can find the question at:
http://www.sambroverman.com/may15c.pdf

Good luck to all taking the exam.
I will resume posting questions of the week in
a month or so.


Sam Broverman
2brove@rogers.com