Math Is Fun Forum

hi aisha00

You're right.  The graph should have negative values too (eg. cube root of -8 = -2)

That's the limitation of using a computer to do your thinking for you.  Sometimes it's just not programed to show all values.

The lesson is, always ask, 'is this what I expected?'

Trace usually means making a cursor go along the curve whilst a readout somewhere tells you the co-ordinates.

You can point at a place on the graph and get the co-ordinates.  I haven't found a trace facility with the MIF graph plotter.

The co-ordinates are given to lots and lots of decimal places and these are very sensitive to slight movements so you'll have to round off.

Is the project about particular graphs or about evaluating the plotting software?

The first graph plotter has got a trace button.  You can skip along the curve by pressing the arrow buttons and on the graph it gives co-ordinates to sensible accuracy.

ps.  I'm not feeling the least 'bothered' so keep posting!

pps  You aren't that clueless because you spotted that part of the cube root graph was missing.  I think that's a good sign you are showing understanding!!!

Bob

hi aisha00

I'm not familiar with this model but I don't think that matters.

If you've found that y = (x^2 + 2) has made a vertical shift then you just need the x shift.

You can go 2 right by using y = (x - 2)^2, and 3 left by using y = (x + 3)^2

If you find it surprising that right shifts are -, and left shifts are +, don't worry, most people are surprised.  But try it; it works!

If that's not what you were after, please post again.

Bob

hi DessyD

TRIANGULAR BASED FRUSTRUM

I am stuck because I don't know which length is 'n'.

Bob

hi mbstats,

These are called the lower quartile and the upper quartile.  25% of the values are below the LQ, then 25% up to the median, then 25% between that and the UQ and finally, 25% above that.

Some people use the interquartile range  (UQ - LQ) as a measure of spread.  Because it leaves out untypical high and low values, it gives a clearer representation of spread than the range (top value - bottom value).

Bob

hi Nfe789

Because of the 'special relationship across the pond' the Queen says it's ok for me to chip in here.  What about the actual engineering aspects of this problem?

(i)  Can you just, in the words of Michael Caine,  "blow the bloody doors off"

(ii)  Suppose I tried a combination lock number and got it right.  Would there be any clue that I'd done that bit even though the other bit is still wrong?  Because that reduces the problem loads and loads.

(iii)  Do you have any safe breakers on board?

Bob

Hi oem7110

Edited since two hours ago.

I've been out all day so I've only just got up to date with all your posts. 

Yes, that's about right I think!  You should really word it like this:  " I can reject the idea that the die is fair with only a 4.18% chance that I'm wrong."

Your question is "What sample size do I need to be 95% confident that I can reject the hypothesis that the die is fair."

Here's what a normal graph looks like:

E is the 'average' or expected value.  It is found from the formula

The 'spread' of the curve is found from the formula

The normal tables that I suggested look like this:

The numbers starting 0.9  are the probabilities of being to the left of the line X.

So for 95% confidence you want the computation

to evaluate to 1.65 or higher

So if you know p, E, V and X you can work back to find n. 

But the trouble is you will not know X until you've done your 'n' trials.

If you put p = 1/6 then

So if you want

re-arranging gives

You can solve for 'n' using the quadratic formula.

I am just a bit worried I might have slipped up somewhere in this algebra so I'll check it when I'm more awake. (It's nearly 11.00pm GMT and I've had a long day including having to take someone with a broken arm to hospital!)

I've just tried X = 1050 in this and solved for 'n' and got two values of around 6000, so this seems about right.

Hope that answers the question,

Bob

hi oem7110

I've been thinking some more about this.  Are you interested in further notes on the normal distribution?  I've got a plan in my head for a full explanation with diagrams, but don't want to spend time putting it all into a post unless you're interested.

Bob

hi rcwitt,

Did you understand my other post about normal distributions?

I did these tables in MS WORD so hopefully they will display properly.

unbiased(X)    1       2       3      4       5       6
Probability    1/6    1/6    1/6    1/6    1/6    1/6

Biased to even
even bias(X)    1        2        3         4        5          6
Probability    1/15    4/15    1/15    4/15    1/15    4/15

Biased to odd
odd bias(X)     1         2         3         4         5         6
Probability    4/15    1/15    4/15    1/15    4/15    1/15

To calculate the ‘expected’ value you do Σ X.P
For table 1         E = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6 = 3.5
For table 2         E = 1/15 + 2x4/15 + 3/15 + 4x4/15 + 5/15 + 6x4/15 = 57/15 = 3.8
For table 3         E = 1x4/15 + 2/6 + 3x4/15 + 4/15 + 5x4/15 + 6/15 = 48/15 = 3.2

The (expected value for the sum) is just the (sum of the expected values) = 3.5 + 3.8 + 3.2 = 10.5
But, take care because you cannot just add the variances.

hi oem7110

If the number of throws were small, the probability of getting a six, P(6) can be modelled by using the binomial distribution.  For a large number of throws it is possible to approximate by using the normal distribution.  For any particular values you need access to normal distribution tables.

Maths is fun does have a normal distribution table but it's not so easy to use for 'right hand tail' questions like yours so I suggest

http://www.math.unb.ca/~knight/utility/NormTble.htm

The table here comes in two parts;  the first is where the experiment yields results that are critical (ie. hard to tell if the die is biased or not);  the second shows much less critical results because it's pretty obvious the die is biased.  Your example of 3500 was off the end of even that table.

Binomial first:  Say you assume the die is not biased.  Then P(6) = 1/6, and, if you’ve rolled it 6000 times, then the ‘expected’ number of sixes is 1000 with a variance of 6000 x 1/6 x 5/6 = 833.33.

Normal approximation:  You then construct a normal distribution with the same parameters, ie E = 1000, V = 833.33.

Now back to your ‘experiment’.  Is 3500 sixes unexpected for this distribution?

I don’t need tables to tell me this is unusual as this result is so far above the expected.  So let's take a more critical value. 

Suppose you get 1100 sixes.  Is this unusual?

You do the calculation (1100-1000)/SQRT(833.33) = 3.46 and look this up in normal tables.

This table gives for 3.4 the following probability
3.4  ...       0.0003369

That’s how likely it is that the result of 1100 sixes could have come from an un-biased die, so it’s very safe to assume it is biased.

Another even more critical example.

Supposing the number of sixes was 1050.

You do the calculation (1050-1000)/SQRT(833.33) = 1.73 and look this up in normal tables.

1.70 ...  0.9554      1.71 ...  0.9564       1.72 ...  0.9573        1.73 ...  0.9582

On this part of the table that’s the probability of being ‘to the left of the experimental result’ so we need to do 1 -0.9582 to get the probability of being beyond this value.

ie. There’s a probability of 0.0418 of getting this result from an un-biased die.  That’s pretty low but is it so low that you’re happy to reject the idea that the die is fair.  You have to make a judgement.  For a die, I’d be fairly happy to conclude it is biased but, if my life depended on getting a right result here I think I’d want to repeat the experiment first.

Bob

hi DessyD

I'm getting closer to helping you, but I am not clear about some of the distances. 

I have made two new diagrams, one for the frustrum of a triangular based pyramid, and one for the square based pyramid.

Please use the letters so say the distances that are given and also to say which ones you want to find.

Be clear about which problem is which.

Bob

hi lilyflower11

To calculate the longest side you would square the sides, add then square root.

To calculate a shorter side, square the sides, subtract and then square root.

So b = sqrt (625 - 400) = 15

Bob

hi DessyD

I've put a lettered diagram below.  Please put your measurements in terms of AB = .......  etc.

And similarly, what measurements you want to calculate.

Bob