# AS Maths Statistics Random Variable Problem

#1
Hi,

I’m struggling with the following question. I’ll attach it in the next post. Any helps appreciated

Thxs
0
#2
0
#3
Bump
Help plz
0
2 years ago
#4
(Original post by dxnixl)
Bump
Help plz
What have you tried / what are you stuck with? You must know binomial formula etc?
0
#5
(Original post by mqb2766)
What have you tried / what are you stuck with? You must know binomial formula etc?
Yep I know the binomial formula and substituted r with k, but it’s impossible to simplify.
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2 years ago
#6
(Original post by dxnixl)
Yep I know the binomial formula and substituted r with k, but it’s impossible to simplify.
Which one did you try and looking at the first probability, 0.02, k must be fairly small?
Or use the guassian approximation if you've covered that?
Last edited by mqb2766; 2 years ago
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#7
(Original post by mqb2766)
Which one did you try and looking at the first probability, 0.02, k must be fairly small?
I’ve never seen a problem like this so this is all I managed
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2 years ago
#8
(Original post by dxnixl)
I’ve never seen a problem like this so this is all I managed
Is that the same question / answer?
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#9
(Original post by mqb2766)
Is that the same question / answer?
No it’s a similar question but with P(50,0.4) however it’s asking for the same thing. I just want to focus on the question in the original post

Thxs
0
2 years ago
#10
(Original post by dxnixl)
No it’s a similar question but with P(50,0.4) however it’s asking for the same thing. I just want to focus on the question in the original post

Thxs
What is the value when k=0, k=1 ...? You're being asked to evaulate when the cumulative probability is small, so it must be close to zero?
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#11
(Original post by mqb2766)
What is the value when k=0, k=1 ...? You're being asked to evaulate when the cumulative probability is small, so it must be close to zero?

I did this... but idk if that’s how the answers correct. It seems like trial and error
0
2 years ago
#12
(Original post by dxnixl)
I’ve never seen a problem like this so this is all I managed
Why is distributed binomially under the parameters ?

Question clearly states .

Anyway,

You can enter this into your calculator. The only thing you need to vary is the upper limit of the sum. You should try some small values first and see if your result is . If it is, increase the upper limit by 1 and look at the new result.
Keep doing this until you get a probability that is .

The upper limit will be precisely the value of .
0
2 years ago
#13
(Original post by dxnixl)

I did this... but idk if that’s how the answers correct. It seems like trial and error

Right answer, but your answer is really looking the probabilities, not their cumulative value. Same result, but you'd need to be clear in your working
0
#14
(Original post by mqb2766)
Right answer, but your answer is really looking the probabilities, not their cumulative value. Same result, but you'd need to be clear in your working
Could you please show a working example for all the questions above?

I don’t mind if you use a different value for p and n, it’d just give a clear idea as to how I should approach the question(s). That way, I can apply it to this and other similar questions
0
2 years ago
#15
(Original post by dxnixl)
Could you please show a working example for all the questions above?

I don’t mind if you use a different value for p and n, it’d just give a clear idea as to how I should approach the question(s). That way, I can apply it to this and other similar questions
Pretty much what you've done, but
P(x < k) = P(x=0) + P(x=1) + ... + P(x=k-1)
You've tested the values, not their sum, on the right. You need to test the cumulative sum. Same result though.
Last edited by mqb2766; 2 years ago
0
2 years ago
#16
(Original post by dxnixl)

I did this... but idk if that’s how the answers correct. It seems like trial and error
Trial and error is the best way to do questions like this. Assuming you're using your calculator (if you're not then you should be), you should be using the cumulative binomial function to find P(x<=k).
0
2 years ago
#17
If you haven't already, get yourself an a level calculator:

0
#18
(Original post by mqb2766)
Pretty much what you've done, but
P(x < k) = P(x=0) + P(x=1) + ... + P(x=k-1)
You've tested the values, not their sum, on the right. You need to test the cumulative sum. Same result though.
Ahh I think I got it! Ty! If I have an issue I’ll post here! same for Sir Cumference and RDKGames
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#19
Hey, just managed to finish my painful HW! Thankyou RDKGames AGrizzlyBearo Sir Cumference mqb2766

2
2 years ago
#20
(Original post by dxnixl)
Hey, just managed to finish my painful HW! Thankyou RDKGames AGrizzlyBearo Sir Cumference mqb2766

Note that a guestimate for Q8 would be the normal approximation (if you've covered this yet)
np = 4
sqrt(npq) ~ 2
So two standard deviations about the mean would give 95% of the mass of the distribution so
0 < X < 8
At around these critical points you'd get the tails of the distribution which is what the question was asking about. Note that the normal approximation is only a guide here ...
Last edited by mqb2766; 2 years ago
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