Question

Let X1,X2,..X10 Be a random sample from the population X, Which has a distribution of N(20,100). What is the chance that (X1+X2+..+X6) > (X7+2X8+2X9 +X10 +16) ? .....side note (the following numbers are subscripts of X)

Answer 1

So you want to find
\[P(X_1+\cdots+X_6>X_7+2X_8+2X_9+X_{10}+16) ?\]

Answer 2

(\[(X _{1} + X _{2} +....+X _{6} ) > ( X _{7} +2X _{8} + 2X _{9} + X _{10} +16 )\]

Answer 3

YES

Answer 4

One possible suggestion: Let's denote a new random variable \(Y\), defined by
\[Y=X_1+\cdots+X_6-X_7-2X_8-2X_9-X_{10}-16\]
Now the desired probability is \(P(Y>0)\). My first idea would be to find the moment generating function for \(Y\).

Answer 5

okay can we also use linear combination for this problem?

Answer 6

Yes, that's exactly what I had in mind.

Answer 7

There's no guarantee it will work, but I think it's worth a try.

Answer 8

ok thanks I'm going to to try it

Answer 9

Actually, using the MGF might be making the problem harder than it has to be. Instead, we can use the fact that a linear combination of normally distributed random variables is itself normally distributed.

In other words, given \(n\) independently and identically normally distributed random variables \(X_k\) with respective means \(\mu_i\) and variances \({\sigma_i}^2\), we have
\[Y=\sum_{i=1}^na_iX_i\]
is normally distributed with mean \(\mu=\displaystyle\sum_{i=1}^na_i\mu_i\) and variance \(\sigma^2=\displaystyle\sum_{i=1}^n {a_i}^2{\sigma_i}^2\).

Answer 10

So here we have
\[Y=X_1+\cdots+X_6-X_7-2X_8-2X_9-X_{10}-16\]
which is normally distributed with mean
\[\mu=20+\cdots+20-20-2(20)-2(20)-20-16=-16\]
and variance
\[\sigma^2=100+\cdots+100+100+2^2(100)+2^2(100)+100+0=1600\]

Answer 11

Now, knowing that \(Y\sim N(-16,1600)\), we can easily compute the probability that \(Y>0\):
\[P(Y>0)=P\left(\frac{Y-(-16)}{\sqrt{1600}}>\frac{0-(-16)}{\sqrt{1600}}\right)=P(Z>0.4)\]

Answer 12

*where \(Z\) is is normally distributed with mean \(0\) and standard deviation \(1\), of course.

Answer 13

wow thanks a lot! yea i realized MGF might of been to difficult! thanks a lot!

Answer 14

@sithsandgiggles

Answer 15

You're welcome!