VCE Methods Question Thread!

[Rishi97]

If Z is a standard normal variable, and Pr (-z < Z < z) = 0.48, then the value of z, correct to 4 decimal places is:
A. 0.6433
B. 0.0502
C. 0.3156
D. 0.3874
E. 0.7063

Thanks

[Yacoubb]

If Z is a standard normal variable, and Pr (-z < Z < z) = 0.48, then the value of z, correct to 4 decimal places is:
A. 0.6433
B. 0.0502
C. 0.3156
D. 0.3874
E. 0.7063

Thanks

1-0.48 = 0.52
0.52/2= 0.26
InvNorm indicates that z= -0.643345 for Probability of 0.26. Given symmetry, the answer is z=0.643345 (A).

[RKTR]

If Z is a standard normal variable, and Pr (-z < Z < z) = 0.48, then the value of z, correct to 4 decimal places is:
A. 0.6433
B. 0.0502
C. 0.3156
D. 0.3874
E. 0.7063

Thanks

Pr(0<Z<z)=0.24
Pr(Z<z)=0.50+0.24 =0.74 and solve for z

edit: got beaten by 10s lol

[jw12]

Could someone explain how these two are equal? Thanks.

When you take a sin graph translated by pi/2 it is the same as the cos graph i.e. sin(x+pi/2) = cos(x)

[IndefatigableLover]

If Z is a standard normal variable, and Pr (-z < Z < z) = 0.48, then the value of z, correct to 4 decimal places is:
A. 0.6433
B. 0.0502
C. 0.3156
D. 0.3874
E. 0.7063

Thanks

The answer is A?
So essentially the area between -z < Z < z = 0.48 however if you split the area outside this domain (that is the area outside is 0.52), then you will have an area of 0.26.

So chuck that value into Inverse Normal under Area with mean = 0 and SD = 1 and you'll get -0.6433 (which is answer A).

To check, if you put it in Normal CDF with the lower and upper bound with your -0.6433 and 0.6433, you'll get 0.48

EDIT: Well that's awkward :S

[Camo15]

 What's the most methodical way to go about differentiating and graphing functions with a modulus in them?

Especially for when the modulus is only around a small portion of the equation, like this:

[Yacoubb]

The answer is A?
So essentially the area between -z < Z < z = 0.48 however if you split the area outside this domain (that is the area outside is 0.52), then you will have an area of 0.26.

So chuck that value into Inverse Normal under Area with mean = 0 and SD = 1 and you'll get -0.6433 (which is answer A).

To check, if you put it in Normal CDF with the lower and upper bound with your -0.6433 and 0.6433, you'll get 0.48

Yeah I got A too same methods #nopunintended

[swagsxcboi]

What's the most methodical way to go about differentiating and graphing functions with a modulus in them?

Especially for when the modulus is only around a small portion of the equation, like this:

I find the easiest way is to split the mod into two equations by making it into a hybrid, then diff-ing it.

[Kaleidoscope]

When you take a sin graph translated by pi/2 it is the same as the cos graph i.e. sin(x+pi/2) = cos(x)

Ohh of course, thanks

[Yacoubb]

What's the most methodical way to go about differentiating and graphing functions with a modulus in them?

Especially for when the modulus is only around a small portion of the equation, like this:

Splitting up the function into

x^2 - 6x +5 for x>0
x^2 -6(-x) + 5 = x^2 + 6x + 5 for x<0

Then know that you cannot differentiate the function at any cusps/endpoints/points of discontinuity, and then derive them like you normally would. So, f'(x) becomes:

2x -6 for x>0
2x + 6 for x<0

Then graph that

[myanacondadont]

Hellllloooo. Can someone explain Q10 d in VCAA 2011 exam 1?

I transposed and got But I can't solve that and had no idea - Not sure what they did exactly, they did some crazy ass transposing?

[Yacoubb]

Hellllloooo. Can someone explain Q10 d in VCAA 2011 exam 1?

I transposed and got But I can't solve that and had no idea - Not sure what they did exactly, they did some crazy ass transposing?

Yeah something like that needs you to employ pythagoras' theorem.

So construct a triangle, and find sinthetaband costheta. Once you've done that, refer back to 10a (acostheta and asintheta). Multiply the given value by the specified a value, and then use them to get your final answer.

Is that clear?

[speedy]

For this, would you guys work out f(f^-1(x))? Or can we just say it's x?

[IndefatigableLover]

For this, would you guys work out f(f^-1(x))? Or can we just say it's x?
(Image removed from quote.)

I'd go RHS = f(f^-1(x)) = x
And then go LHS = [And solve until you get LHS = RHS]

Though to be honest for this question, the inverse is the same as the actual function anyway so it wouldn't hurt in actually solving it imo.

[Yacoubb]

For this, would you guys work out f(f^-1(x))? Or can we just say it's x?
(Image removed from quote.)

It is a show question.

Call f(f(x)) = LHS
Solve it.

Call f(f^-1(x)) = RHS
Solve it

Say LHS = RHS. Therefore f(f(x)) = f(f^-1(x))