# Discrete probability disturbution

So in this question I,m given the discrete random variable (25,0.1)
I,m also told the probability that P(X<a)<0.99
and p(X<=a)>0.99
I could easily do this when I guessed values in my calculator and I got a=6 but I,m very confused about the how you can find it in another way as I don’t think you can do that in a exam.
This is for binomial distrubution
(edited 8 months ago)

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Original post by realed
So in this question I,m given the random variable (25,0.1)
I,m also told the probability that P(X<a)<0.99
and p(X<=a)>0.99
I could easily do this when I guessed values in my calculator and I got a=6 but I,m very confused about the how you can find it in another way as I don’t think you can do that in a exam.

I believe that some calculators (perhaps the Casio CG-50?) have an inverse binomial distribution function. Otherwise you will need to use the kind of trial and error approach that you've mentioned. In an exam you would normally list the probabilities for a few values of a (say 4, 5, 6 and 7), and form a conclusion (with justification) based on the probabilities found.
(edited 8 months ago)
hey is this for a level edexcel maths?
Original post by realed
So in this question I,m given the random variable (25,0.1)
I,m also told the probability that P(X<a)<0.99
and p(X<=a)>0.99
I could easily do this when I guessed values in my calculator and I got a=6 but I,m very confused about the how you can find it in another way as I don’t think you can do that in a exam.

Have I missed something - you haven't said what distribution it is, only the parameters (25, 0.1) ?
Original post by davros
Have I missed something - you haven't said what distribution it is, only the parameters (25, 0.1) ?

I think it's binomial as it's discrete random variable
Original post by Bookworm524
I think it's binomial as it's discrete random variable

Yeah they put "discrete" in the title, then old_engineer mentioned inverse Normal, but obvs it's going to make a difference what the actual distribution is in terms of how you try to approach the answer
Original post by old_engineer
I believe that some calculators (perhaps the Casio CG-50?) have an inverse normal distribution function. Otherwise you will need to use the kind of trial and error approach that you've mentioned. In an exam you would normally list the probabilities for a few values of a (say 4, 5, 6 and 7), and form a conclusion (with justification) based on the probabilities found.

Thanks, also how do you use the inverse normal thing that you,ve mentioned
(edited 8 months ago)
Original post by study1234okay
hey is this for a level edexcel maths?

This is for OCR Mei further maths(stats) but i think it,s the same thing as the binomial distribution in edexcel stats
(edited 8 months ago)
Original post by davros
Have I missed something - you haven't said what distribution it is, only the parameters (25, 0.1) ?

I forget to say it,s binomial in the description
Original post by realed
I forget to say it,s binomial in the description

So in the past you might have had cumulative binomial tables to help you; these days I suspect a calculator that can do cumulative distributions would be accepted in the exam. Is this an actual "current spec" question, or is it from an old paper / textbook when the spec might have been different?
A back of the envelope approximation would be to use the normal approximation (its on the boundary of what would be ok) so mean (np) 2.5 and std dev (sqrt(np(1-p)) 1.5 so mean + 2 or 3 std devs ~ 6 roughly corresponds to 0.99 cumulative probability. Relatively easy to check the value on your calc then.
Original post by davros
So in the past you might have had cumulative binomial tables to help you; these days I suspect a calculator that can do cumulative distributions would be accepted in the exam. Is this an actual "current spec" question, or is it from an old paper / textbook when the spec might have been different?

This is from the new spec
Original post by mqb2766
A back of the envelope approximation would be to use the normal approximation (its on the boundary of what would be ok) so mean (np) 2.5 and std dev (sqrt(np(1-p)) 1.5 so mean + 2 or 3 std devs ~ 6 roughly corresponds to 0.99 cumulative probability. Relatively easy to check the value on your calc then.

Mean+2=4.5 so how does that correspond to 6
Original post by realed
Mean+2=4.5 so how does that correspond to 6

Its rough, but
2.5 + 3 = 5.5 (2 std devs)
2.5 + 4.5 = 7 (3 std devs)
The probability of 0.99 lies between 0.975 (2 std devs) and 0.9985 (3 std devs). Thinking about it, the numbers of the binomial have almost been chosen to make this simple (variance = 25*0.1*0.9 = 2.25 = 1.5^2).
(edited 8 months ago)
Original post by mqb2766
IIts rough, but
2.5 + 3 = 5.5 (2 std devs)
2.5 + 4.5 = 7 (3 std devs)
The probability of 0.99 lies between 0.975 (2 std devs) and 0.9985 (3 std devs). Thinking about it, the numbers of the binomial have almost been chosen to make this simple (variance = 25*0.1*0.9 = 2.25 = 1.5^2).

Ok but why do you decide to just do 2 or 3 standard deviations. You could do 100 or 20000
when you know the answer it makes makes sense but, how do you know this from the question to just use 2 or 3 standard deviations.
(edited 8 months ago)
Original post by realed
Ok but why do you decide to just do 2 or 3 standard deviations. You could do 100 or 20000
when you know the

Those values correspond to the stated probability values so
P(X<mean + 2 std dev) = 0.975
P(X<mean + 3 std dev) = 0.9985
which bracket the value 0.99 youre after. You should know that for a normal distribution,
* 68% lies in the interval: mean +/- 1 std dev (so each tail has ~16%)
* 95% lies in the interval: mean +/- 2 std dev (each tail is ~2.5%)
* 99.7% lies in the interval: mean +/- 3 std dev (each tail is 0.15%)
Original post by mqb2766
Those values correspond to the stated probability values so
P(X<mean + 2 std dev) = 0.975
P(X<mean + 3 std dev) = 0.9985
which bracket the value 0.99 youre after. You should know that for a normal distribution,
* 68% lies in the interval: mean +/- 1 std dev (so each tail has ~16%)
* 95% lies in the interval: mean +/- 2 std dev (each tail is ~2.5%)
* 99.7% lies in the interval: mean +/- 3 std dev (each tail is 0.15%)

Firstly when I put but x is less than 5.5 in my calculator i get an argument error, also when I put x is less than or equal to 7 I get 0.9977 but I assume that with this you look for the highest standard deviation below the answer and the lowest standard deviation above the answer and you find the smallest that a can be that satisfies this equation.
(edited 8 months ago)
Original post by realed
Firstly when I put but x is less than 5.5 in my calculator i get an argument error, also when I put x is less than or equal to 7 I get 0.9977 but I assume that with this you look for the highest standard deviation below the answer and the lowest standard deviation above the answer and you find the smallest that a can be that satisfies this equation.

Ill drop out. You cant put a fraction in your binomial for x (so 5 or 6, not 5.5) and as in the first post, the normal approximation of the binomial is rough here because its on the boundary of when it can be used, so the probabilities will not be exact. For me, it said your answer for the binomial was around 6 (maybe 5 or 7) and would be easy to verify on your calculator using the actual binomial distribution.
Original post by mqb2766
I’ll drop out. You cant put a fraction in your binomial for x (so 5 or 6, not 5.5) and as in the first post, the normal approximation of the binomial is rough here because its on the boundary of when it can be used, so the probabilities will not be exact. For me, it said your answer for the binomial was around 6 (maybe 5 or 7) and would be easy to verify on your calculator using the actual binomial distribution.

Ok but if you have a fraction like 5.5 can you find the values for 5 and 6 and then work out the mean .Also., thanks for helping me I really appreciate it.
(edited 8 months ago)
Surely you just use the cumulative binomial function on the calculator and "trial and error"? I'd certainly find this quicker than messing around with normal approximations and you're not left with any uncertainty about whether you're approximation is valid.