I believe I have part a, I watched a video for the first part though I used the equation I saw in a video I got an answer 5 times bigger than the actual answer. I need help with B
#Simple Statistics Question
raven cape
fervent trellis
so i think
main umbra
I believe I have part a, I watched a video for the first part though I used the ...
What did you get for (a)? Just to make sure you understood it correctly.
raven cape
What did you get for (a)? Just to make sure you understood it correctly.
for a my working was ((5!)/(1!*4!)) x 0.0668 x (1-0.0668)^4 = 0.253 then I checked online and saw that someone had worked it out without the number of combinations and got 0.0507 (5 times smaller)
so I'm not really too sure
I checked this video and it told me to do the first method https://www.youtube.com/watch?v=8Vqa3irjRgs
I haven't done this topic in months
so I completely forgot it
main umbra
for a my working was ((5!)/(1!*4!)) x 0.0668 x (1-0.0668)^4 = 0.253 then I check...
Combinations won't work for (a), since not only it needs to be just once, but on the FIRST day. So, the probability would be p(1 - p)^4, where p is the probability of water being hot on a selected day.
raven cape
what about b
main umbra
We do the following:
P(more than on 1 day) = 1 - P(on 0 or 1 day) = 1 - P(on 0 days) - P(on 1 day)
And here you use the formulas with combinations, since the positions of the days don't matter, only the amount.
raven cape
this is confusing
main umbra
Well, it's just the application of two formulas:
- P(A) = 1 - P(¬A)
- P(A∪B) = P(A) + P(B), as long as A and B are disjoint.
raven cape
We do the following:
P(more than on 1 day) = 1 - P(on 0 or 1 day) = 1 - P(on 0 d...
whats confusing me is the day numbers
main umbra
The day numbers don't matter in (b), only the amount of days does.
raven cape
so $1-(1-0.0668)-(0.0507)$?
quaint peakBOT
Yooda
main umbra
No.
P(on 0 days) = (1 - p)^5
P(on 1 day) = 5p(1 - p)^4
So:
P(more than on 1 day) = 1 - (1 - p)^5 - 5p(1 - p)^4 = 1 - (1 + 4p)(1 - p)^4
Now just substitute p = 0.0668.
main umbra
You just need some practice.
raven cape
I've been slacking off revision for a week
I have exams in 3 now
I need to start
for this question can I use X~B(20, 0.45)?
and I'm also lost again
main umbra
First, use Bin(20, 0.45) to find the probability that a store has 12 or more wins. Take that as p.
Then, use Bin(8, p) to find the probabilities that at least 2 stores are "winners".
main umbra
CDF.
Binomial is discrete, so its PDF isn't too useful. PMF is, but here we are looking at cumulatove probabilities.
raven cape
ok I got an extremely small answer
x10^-7 small
What I did was (X = 12 N = 20 P = 0.45) on binomial CD gave me 0.9419 then I did it again with (X = 2 N = 8 P = 0.9419) which gave me 9.66x10^-7
nevermind
This is the method we learnt
hence why I didn't recognise it
ok I understand this
but problem
I forgot how to do variables.
I have a
main umbra
This is the method we learnt
That's not correct.
P(X > 1) = 1 - P(X ≤ 1), not just P(X ≤ 1).
raven cape
erm
its what Physics and Maths Tutor is Telling people so I'm not sure
oh
yeah no they missed the one off
they did it in the working
main umbra
I forgot how to do variables.
b.
P(Y ≥ 22) = P(4X + 1 ≥ 22)
After that, do some algebraic manipulation to isolate X.
c.
Y + X = 4X + 1 + X = 5X + 1
The rest is similar to the approach in (b).
raven cape
b.
P(Y ≥ 22) = P(4X + 1 ≥ 22)
After that, do some algebraic manipulation to isol...
ah P(X=> 21/4) then
main umbra
Yes. Now you just need to calculate it.
raven cape
ok thanks
main umbra
You're welcome!
raven cape
ok now I have that I'm not sure what I would put in X~B(N, p)
well exactly what I need to work out
they are all bigger than 21/4
main umbra
No. Only 6 is.
main umbra
No. The distribution here isn't binomial.
It's just the distribution desribed by the given table.
raven cape
so 3/21
or 1/7
as 6 is the only value over 5.25
well 18/21
wait
since its smaller values less than 5.25 it would be 1 - 18/21 (3, 4 and 5) = 3/21
my brain is frying a bit
main umbra
Not sure what you mean.
P(Y ≥ 22) = P(4X + 1 ≥ 22) = P(X ≥ 21/4) = P(X = 6) = k/21 (with whatever k you found).
main umbra
Yes.
main umbra
I recommend taking a break if you're tired. Have some more practice later.
raven cape
ok finally D
Chance of Landing on 5 = 7/21
N = 40, X > 10 P = 7/21
p(X > 10) = 1 - p(X <= 10) right
wait
main umbra
Yes.
raven cape
is it not p(X > 10) = p(X >= 11)
main umbra
Same thing for this distribution, doesn't matter.