Question

If X and Y are independent random variables following N(8,2) and N(12,4*root3) respectively. Find value of 'a' such that
P (2X-Y<=a^2) = P (X + 2Y>= a)

Answer 1

normally distributed with (mean.variance) eh

Answer 2

oh but whats the method to solve?

Answer 3

i cant really say that im familiar with the 2x-y setup to determine.

would we go with a 95% spread or just means or ... i cant make sense of that part

Answer 4

well I dont know anything about.. :( just tell me whatever you think the way to solve this

Answer 5

if i were to take a guess, id think (and i could be wrong on this), that the 3sd spread is what the are relating this to ....

Answer 6

X = (8+3sqrt(2)) - (8-3sqrt(2)) = 6(2)^(1/2)
Y = (12+3(sqrt(4rt3))) - (12-3sqrt(4rt3)) = 12(3)^(1/4)

Answer 7

then its just mathing out the innards and comparing results, but thats just a guess is all

Answer 8

i see some content in the internet that suggest to subtract means and add variances .... but im unclear as to why

Answer 9

o.O but I didnt get it, whatever you have done. Basically it is related with the area under normal distribution . and that 2X+Y and X+2Y are the lines. The area under normal curve before or after these lines as per the sign must be found out and equated. and for that we have to find Standard normal variate Z. I dont know how to find it out in case of 2 variables X and Y. but for single variable it is given byt Z= (X - m) / SD where m=mean and SD= standard deviation

Answer 10

are there, or was there, any options to choose from?

Answer 11

ive got a site here that says: the mean of the sum of the distribution of random variables is the sum of the means, u(x+y) = u(x)+u(y)

the variance is the difference var(x+y) = var(x)-var(y)

in this manner we can setup a distribution: Z = X +- Y, N(u(x+-y),var(x-+y))

Answer 12

2X = X+X, so maybe

Z1 = 2X - Y = ((8+8)-12,(2-2)+4rt(3)) = (4,4rt(3))

Z2 = X + 2Y = (8+(12+12),2-(4rt(3)-4rt(3))) = (32,2)

Answer 13

yep that should be correct

Answer 14

I found one similar problem in my book. in that for standard deviation they took sqrt od multi squares and adding those for 2 ie. sd of Z1=root (4*4 + 1* 16*3)=root64 =8

Answer 15

sd for Z2 = root(1*4 + 4*16*3) =root 196=14

Answer 16

i wish i could say this makes sense now, but im still on the fence ....

Answer 17

still I cant find solution for a which is required.

Answer 18

whats the title of your textbook?

Answer 19

its applied maths 4 by G. V. Kumbhojkar

Answer 20

let me do some reading, if noone else can determine an answer ill see what i can learn on it

Answer 21

okay thats fine. actually I think we have reached upto mean and variance correctly. only we have to find standard normal variate by snv= (U-m)/ sd. In the textbook problem which i was looking at they have taken the result as U I mean in this case a^2 and a and found the area in between. but here we cant find snv as a is unknown and then cant find area as snv is unknown. Pretty confusing. hope you'll get the answer.

Answer 22

https://onlinecourses.science.psu.edu/stat414/node/172

this looks very promising to me

Answer 23

X = N(8,2)
Y = N(12,4*root3)

2X = (2(8),2^2(2))
-Y = (-1(12),(-1)^2(4 rt(3)))
-------------------------
= ( 4 , 8 + 4 rt(3) ) for the distribution setup

\[Z_1\le\frac{a^2-4}{\sqrt{8+4\sqrt3}}\]

Answer 24

X + 2Y>= a

X = N(8,2)
Y = N(12,4*root3)

X = (1(8),1^2(2))
2Y = (2(12),(2)^2(4 rt(3)))
-------------------------
= ( 32 , 2+16 rt(3) )

\[Z_2\ge\frac{a-32}{\sqrt{2+16\sqrt3}}\]

Answer 25

since these are spose to equate, then Z1 = Z2, when a=?

Answer 26

\[a^2\sqrt{2+16\sqrt{3}}-4\sqrt{2+16\sqrt{3}} = a\sqrt{8+4\sqrt{3}}-32\sqrt{8+4\sqrt{3}}\]

lets clean this up alittle

\[a^2k-4k = am-32m\]
\[a^2k-am+32m-4k = 0\]
and using the quadratic formula to solve for a
\[a=\frac{m\pm\sqrt{m^2-4(k)(32m-4k)}}{2k}\]

Answer 27

make sure the sds are correct, and then that how i would solve for "a"

Answer 28

@amistre64 everything is correct upto mean but for variance we have to take variance= ∑Ci^2 *σi ^2 which you havent taken. then sd= sqrt(variance). and finally we are not suppose to equate Z1 and Z2 but their probabilities which are depending on the value of a.

Answer 29

for variance you have taken only Ci^2 and not σi^2.

Answer 30

o^2 is variance, and variance is given in the notation:\[N(\mu,\sigma^2)\]
simga^2 is variance

Answer 31

library is closing early today, so im outta time :/

Answer 32

oh thats fine not a problem. Thanks a lot for your support we'll solve this later.

Answer 33

Tapping on my phone. But after thinking last night we are creating 2 new normal disteibutions W and T we then take and standardize them both to Z in order to compare their probabilities. The symmetry of a normal distribution is such that the left and right tails of -z and z are equal.