Chocoholic's Questions

[Mao]

working now~

if thats any comfort?

[Chocoholic]

Thanks Mao! I don't think I would of been able to think of that...it actually looked gibberish. So yeah, thanks for translating the gibberish to english.

[Chocoholic]

working now~

if thats any comfort?

Yeah, now i'll be able to sleep at night. lol

[Chocoholic]

ok, time to revive my old thread...it's not that I haven't had problems since May, I've just been to lazy to post.

Monique is practising goaling for netball. She knows from past experience that her chances of making any one shot is about 70%. Her coach has asked her to keep practising until she scores 50 goals. How many shots would she need to attempt to ensure that the probability of making at least 50 shots is more than .99

I'm usually ok with questions where 'n' is unkown. But how on earth do i do this....life would of been much easier if she just listened to her coach and shot and stopped at exactly 50.

[Chocoholic]

Monique is practising goaling for netball. She knows from past experience that her chances of making any one shot is about 70%. Her coach has asked her to keep practising until she scores 50 goals. How many shots would she need to attempt to ensure that the probability of making at least 50 shots is more than .99

Thanks for your help Mao, but I'm a bit confused now, because it says to 'find the probability of making at least 50 shots' so wouldn't that mean Pr(X(equal to or greater than) 50). So how come you made the binomial as if it was Pr(X=50). Sorry for being a bit very annoying, but i don't really get why the working out is like that.

P.S. The first time I did this question I did it like you Mao, but then I noticed the 'find the probability of making at least 50 shots' so that's the part that confuses me. Does she shoot exactly 50 times, or at leas(implying equal or greater than).

Thanks!!!

[Mao]

yeah, i was confused myself, which is why i deleted it.

[Chocoholic]

yeah, i was confused myself, which is why i deleted it.

Thanks for your help Mao. I think they mad a typo, if not, that's a lot of working out which I don't think we'll ever have to do for methods. 

[Mao]

seeing that the inequality will be quite complex:



i don't see how anyone can solve that using methods knowledge [if possible at all], hence I have not attempted trying to solve this [but open invitations to anyone willing to try]

I assume that you are doing MM-CAS?
anyhow, the command "binomcdf" can be used. If you have TI-Stat/list editor app installed on your TI89:

Code: [Select]

tistat.binomcdf(n,0.7,50,n)|n=..
by trial and error [placing values into ".."], the number 86 is your friend.

[no idea as for how TI84 would go... probably a similar method via trial and error]

and for the fun of it:

is as simple as I got it to go.. hehe

[Chocoholic]

no, I'm not doing CAS but I do have 'binomcdf" installed. But seeing as I don't do MM-CAS should I just disregard this question sice I do normal MM.

Once again thanks a lot for your patience and helpfullness!!

[excal]

Try using the table set:

http://vcenotes.com/forum/index.php/topic,2146.msg57498.html#msg57498

[/0]

Alternatively, Stats-List Editor -> F5: Binomial Cdf
Then just test different numbers of trials and home in on the answer lol...

[Mao]

using table works

Code: [Select]

y1=binomcdf(x,0.7,50,x)

then in table view, scroll down to 86. that's the first value that reaches >0.99

[/0]

using table works

Code: [Select]

y1=binomcdf(x,0.7,50,x)

then in table view, scroll down to 86. that's the first value that reaches >0.99

What is the last x in y1=binomcdf(x,0.7,50,x)?

[Chocoholic]

using table works

Code: [Select]

y1=binomcdf(x,0.7,50,x)


hmm I tried doing this and it came up with ERR:ARGUMENT.

Anyhow, I know officially know how to go about answering this....Mission Accomplished!!!
Thanks heaps Mao, Excalibur and DivideBy0

[Mao]

out of x trials, succeed 50 to x trials (i.e. 50 out of x to x out of x)