# Recent content by ssd

1. ### B An elementary confusion on discrete or continuous variable

How we may handle the data is irrelevant in context of the question, as mentioned earlier. The question is not " what you would generally assume". The question is 'what is what' by definition. You 'assume' a variable to be distributed normally does not imply the variable is continuous, you just...
2. ### B An elementary confusion on discrete or continuous variable

Sounds absolutely meaningful.
3. ### B An elementary confusion on discrete or continuous variable

I was trying to go by analyzing the definition. The question is not intended for the answer that " what should we assume for convenience of inference making". Logic, yes it is the base of my understanding. Photons exist as very small discrete packets, solving and generating big questions...
4. ### B An elementary confusion on discrete or continuous variable

Agree with you. For all advanced studies or inference they are taken as continuous without problem. But, in the introductory discussion one may allow the other aspects of the idea to have a complete picture.
5. ### B An elementary confusion on discrete or continuous variable

The question is simply posed as " identity the variables as discrete or continuous. 1) Mark of a student in an examination. 2) Family income." What I think: 1) There must be a minimum gap between two possible consecutive marks that the examiner can assign. Eg. Suppose that there are N students...
6. ### A Two Bernoulli distribution- test hypothesis for biased coins

I see no relation between p1 and p2 in the proposed context of testing. pi= prob of head of coin i. Why you do not think of testing them separately. Note that you have to specify the level of significance (α) before hand. To perform a test of size exactly = α, we generally have to do a...
7. ### Sum of (n+1) terms in exponential series

No, only n+1 terms.
8. ### Sum of (n+1) terms in exponential series

Homework Statement S = 1+ x/1! +x2/2! +x3/3! +...+xn/n! To find S in simple terms. Homework Equations None The Attempt at a Solution I tried with Taylor's expansion, coshx and sinhx expansions. But cannot see consequence.
9. ### I Generating a random sample with a standard deviation

More often than not, inverse function of a CDF is not analytically solvable in terms of simple functions. We have to use a computer program for numerical solution.
10. ### I Generating a random sample with a standard deviation

Suppose, f(x)= f(x,μ,σ) is your curve with known mean (μ) and sd (σ) and f(x)≥0. Find C=∫Xf(x)dx, -∞<x<∞. Draw a 3 digited (say) random number and put a decimal point before it. Let this fraction be R. Find x by solving ∫-∞x f(x)dx/c =R. x is now a sample form f(x).
11. ### A Reasonable length of forecast horizon in a time series

Suppose we have monthly totals of observed data for last 35 years. That data is of inflow of a river in a reservoir and monthly demands from the reservoir. We are interested to check the effect of construction of a dam in the upstream. The effect is, whether the downstream reservoir will have...
12. ### B Sweets in a bag probability problem

Actually, I added the case of x=2 on the basis of the given prob condition (third item is red with prob x/2-1=0). A non existent item (here, the third sweet) can as well be drawn with prob=0 (which, in turn implies, it is impossible to draw it).
13. ### I Constructing an Index

You may use z scores with arbitrary origin and scale. Let X & Y be the two series. Calculate z from (x-μx)/σx=( z-c)/d and from (y-μy)/σy=( z-c)/d, for all observed x and y, where c & d are arbitrary. μ,σ are mean and sd etc. The z values are now comparable.
14. ### B Sweets in a bag probability problem

I think, we have to deduce that 0≤x/2-1≤1. ⇒2≤x≤4. Again three sweets are already drawn⇒3≤x≤4. Now consider Case1: x=3. Therefore, prob of third being sweet is red is either 0 or 1 (according as last sweet left in box non red or red after withdrawal of 1st two). This does not satisfy the given...
15. ### Find the minimum of c(3a+4b) when a^2+b^2 +c^2=1.

Well, I got it. a=-3/5√2 and b= -4/5√2. Actually I mentally calculated the minimum of 3a+4√(1/2-a^2) is at a=1/2 and that was the mistake. Since, our minimum could be the min of 3a-4√(1/2-a^2).