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10h
comment Problem solving the standard deviation for a stochastic variable
Sometimes you seem to be saying $S_W=2$ and sometimes $2/\sqrt5$. The later is what you want.
11h
comment Problem solving the standard deviation for a stochastic variable
You seem to have ignored my 2nd comment: $\sqrt{\dfrac{20}{5^2}}= \dfrac{2}{\sqrt5}$
11h
answered Problem solving the standard deviation for a stochastic variable
11h
comment Calculate the mean, the median and the quartiles.
I think you may have taken the derivative of $\dfrac{\tan(z)}{\pi}+0.5$ rather than $\dfrac{\tan^{-1}(z)}{\pi}+0.5$
13h
comment Calculate the mean, the median and the quartiles.
I can see an instruction, but what in particular is your question? The median should be obvious, as should the mean if there is one.
16h
answered Pattern Recognition - How to solve this problem?
22h
comment Determining the weights of known parameters in a formula
It will depend on what you mean by "most significant". You also have other issues: e.g. multiplying all the $a_i$ by a positive constant will give the same ranking; if the $w,x,y,z$ are not comparable in some sense (similar units or similar dispersion) then comparisons of the $a_i$ may not be meaningful.
1d
answered What separates the dot product from the scalar projection?
1d
comment ways to choose 4 people including two or more male from N people with 1/3 male 2/3 female
You have an error in your final line (you have subtracted "no males" twice)
1d
comment Conditions for uniqueness of the median
An alternative would be an $m$ such that $(x-m)\left(F(x)-\frac12\right) \gt 0$ for all $x \ne m$
1d
comment A variation of a combination and a permutation, I think?
Because there is no $2$ on the left hand side.
1d
comment A variation of a combination and a permutation, I think?
So the answer to question 1 is $8 \times8 \times8 \times8 \times8 \times8 = 8^6$ and for different doors is $8 \times7 \times6 \times5 \times4 \times3 = 8!/2$. You were not that far away
1d
comment A variation of a combination and a permutation, I think?
I think you are guessing now. No, it is not $8$ choose $6$, which is just $28$.
1d
comment A variation of a combination and a permutation, I think?
The number of ways one person can go through one of the eight doors is $8$. What about two people - how many choices does the first have, and how many the second?
1d
comment A variation of a combination and a permutation, I think?
$6^8$ is also wrong. Is the number of ways one person can go through one of the eight doors $1^8$?
1d
comment A variation of a combination and a permutation, I think?
For question 2 it might be easier to calculate the number of ways nobody goes through the third door, and the number of ways exactly one person goes through the third door, and then subtract this from your $6^8$ number, or whatever you correct that to.
1d
comment A variation of a combination and a permutation, I think?
Your answer of $8!/6!$ is far too low, as the answer would not be be $8!/7!$ or $8!/8!$ if there were $7$ or $8$ people going through different doors. But factorials are the right thing to be thinking about.
1d
answered Determining density involving scaled beta distribution
1d
comment Determining density involving scaled beta distribution
I would have thought you need more information about $\theta$
1d
answered Seating arrangement probabilites