jack_chay's methods question thread

[b^3]

Firstly writing down the key information we know.

That is our variable follows a normal distribution.

Converting to a standard normal distribution using z scores gives

Drawing out a quick normal distribution graph for the last piece of information, since it's symmetrical about the mean (for the standard normal our mean is zero) we can deduce that

Now that will give us the area that is between the mean and the mean plus . When we use the inverse normal function on the calculator, it starts at and finds the value corresponding to the area given under the normal distribution from to that value. So we need to add the region that is below the mean, which we know is . So we have

Putting it into the calculator, on the Ti-nspire it's [Menu] [5] [5] [3], gives .
That is



NOTE: You could do it without converting to the standard normal distribution, as there is that option for it on the calculator, it's just that, sometimes marks are given for doing that. (The reason people used to convert to the standard norm was back when we didn't have calculators to do that for us, and they'd convert it across and look up the probability table for it). I've done it that way anyways, it just makes the question a little bit longer, but the key thing here is probably sketching out the curve to see how you can manipulate it. Hope that helps

EDIT: Fixed in the definition of the normal distribution, thanks silverpixeli

[Phy124]

So it's saying that the probability of the the diameter not being withing d cm of the mean, 3, is 0.25. This is given by the blue area below.



Using this we can calculate the value of 3-d. We know that the area below 3-d is 0.125 (half of 0.25). so now we can use the inversenorm function on the calculator with mean 3, standard deviation of 0.002 and area below of 0.125, voiding 2.9977, hence d = 3-2.9977 = 0.0023.

Oops it appears as though I refreshed the page more than once and missed the message that someone posted before awks.

[silverpixeli]

Firstly writing down the key information we know.

That is our variable follows a normal distribution.

I think the second parameter in the notation for describing a Normal Distribution is supposed to be the variance, rather than the standard deviation, ie


Just to be technically correct

[b^3]

I think the second parameter in the notation for describing a Normal Distribution is supposed to be the variance, rather than the standard deviation, ie


Just to be technically correct

Yeah it is too. Some books use , others use , for VCE it's meant to be . I was actually thinking about the binomial setup when I typed that originally... but anyways, I'll fix it up above. Thanks silverpixeli

EDIT: I forgot to put the square on the for vce part above. My bad.

[jack_chay]

thank you guys it was a big help... I get it now

[jack_chay]

This isn't a probability question, but:

The number of tonnes of coal, P, produced by x miners per shift is given by the rule:
P = x^2/90 * (56 − x) where 1 ≤ x ≤ 40

Write down an expression in terms of x for the average production per man in the
shift. Denote the average production per man by A (tonnes).

[silverpixeli]

The number of tonnes of coal, P, produced by x miners per shift is given by the rule:
P = x^2/90 * (56 − x) where 1 ≤ x ≤ 40

Write down an expression in terms of x for the average production per man in the
shift. Denote the average production per man by A (tonnes).

Cool question

Okay, so what do we need? The production A (tonnes) of ONE man.

What do we know? we know that the number of men is x, and that the production of x men is P

If we have x men producing P tonnes and we want the amount (A) that ONE man produces, we'll divide by x because that's effectively splitting the total amount produced between all the men.

As a result, A = P/x = [x^2/90 * (56 − x)]/x = x/90 * (56 − x)



(not sure if the (56-x) is under the fraction or not, but I'm going to assume it isnt, the following example works to conceptualise what's going on here either way)

We could try generating a value of x, say 6, to see what's actually happening here, if the above isn't clear

P(6) is the production of 6 men, and it's equal to 6^2/90 * (56 − 6)=36/90 * 50 = 20 tonnes
now we know that 6 men will produce 20 tonnes, so the average per man is going to be A=(amount)/(number of men)=20/6=about 3.33 tonnes per man
(each of the 6 men produce an average of 3.33 tones to 2 decimal places)

So what we did was A=P(6)/6 and that's why it makes sense to just go A=P(x)/x

Uhh, not to sure about this last bit, but I think it's worth saying that we can only divide by x because x can't be 0, and that's because of the domain 1 ≤ x ≤ 40

Hopefully I'm right, and hopefully that helps

[jack_chay]

oh my gosh! that is so true

thank you thank you thank you so much silverpixeli

[jack_chay]

this question seems a little weird, don't I just make f'(x) = 0, the answer would be 2/3, but B.O.B says it's k <-2

Consider the family of quadratic functions with rule
f (x) = (k + 2)x2 + (6k − 4)x + 2
where k is an arbitrary constant.

For what values of k is the turning point a local maximum?

thanks in advance

[Phy124]

For the turning point to be a local maximum the parabola must be negative, i.e. the coefficient of the x² term must be negative.

So

[jack_chay]

oh yeah...

thank you so much Snow Red

[jack_chay]

Show that
dy
dx = 0 ⇔ 2 loge x = −x, x > 0

I was just wondering what that symbol between the 0 and 2 is (it sort of looks like an arrow pointing to both sides)

and how would understanding that, help to explain why this means that the local minimum of y =(1/x) + ex lies in the interval (0, 1).

Thank you in advance

[jack_chay]

I found out what the double arrow is, but i'm not sure how it would answer the explain why bit: explain why this means that the local minimum of y =(1/x) + ex lies in the interval (0, 1)

I found that dy/dx of the equation y =(1/x) + ex is when  2 loge x = −x,

not sure what to do from there

thanks in advance

[jack_chay]

um... don't worry about the previous question i posted, I found the answer. Instead I'm having difficulty with this question. It seems so simple, but i don't get it.
A particle is moving along a path with equation y = (x^2 + 24)^1/2

When the particle is at the point with coordinates (5, 7), y is increasing at a rate of
10 units per second. At what rate is x increasing?

[darklight]

um... don't worry about the previous question i posted, I found the answer. Instead I'm having difficulty with this question. It seems so simple, but i don't get it.
A particle is moving along a path with equation y = (x^2 + 24)^1/2

When the particle is at the point with coordinates (5, 7), y is increasing at a rate of
10 units per second. At what rate is x increasing?

We want to find dx/dt. We are given dy/dt. Therefore, we apply the chain rule:
dx/dt = dy/dt *dx/dy

We are given dy/dt = 10.
y= (x^2 + 24)^1/2
dy/dx = 1/2 * (x^2 + 24)^-1/2 * 2x
          = x * (x^2 +24)^-1/2
dy/dx (5) = 5 * (49)^-1/2
               = 5/7

We don't want dy/dx(5) though! We want dx/dy(5) - therefore, we flip it to get "7/5."
Therefore, dx/dt = 10 * 7/5
= 14

Therefore, x is increasing at a rate of 14 units per second.