Discrete probability disturbution

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.

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) ?

I think it's binomial as it's discrete random variable

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.

hey is this for a level edexcel maths?

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

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.

Mean+2=4.5 so how does that correspond to 6

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

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.

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.