## Search found 28 matches

Wed Jul 07, 2010 2:06 am
Forum: Statistics
Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
Replies: 7
Views: 9638

### Re: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y

I couldnt understand how to evaluate P(A+B=4) as is it P(A+B=4) = P(A) + P(B) = 4 or is it some other way If X is poisson 1 and Y is poisson 1 then X+Y is Poisson 2 but how should i represent it in my solution P(A+B) = 1+1 = 2 but what about the '4' I am having a hard time how to formulate this ? s...
Tue Jul 06, 2010 2:53 am
Forum: Statistics
Topic: Need to know which formulae which has been used
Replies: 2
Views: 5193

### Re: Need to know which formulae which has been used

If you're not sure what formula this is, try working backwards from the "sample" formulas they've given you. Which one uses $\sqrt{\sigma^2}\, =\, \sigma$ in the numerator? Which one also uses $\sqrt{n}$ in the denominator?
Thanks!!
Tue Jul 06, 2010 2:50 am
Forum: Statistics
Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
Replies: 7
Views: 9638

### Re: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y

Start with this... http://en.wikipedia.org/wiki/Poisson_distribution#Properties Thanks so \mu = \lambda = 1 P(A+B=4) = ? stuck here !!! From the link I provided look at "Sums of Poisson-distributed random variables" I couldnt understand how to evaluate P(A+B=4) as is it P(A+B=4) = P(A) + P(B) = 4 o...
Mon Jul 05, 2010 4:04 pm
Forum: Statistics
Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
Replies: 7
Views: 9638

### Re: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y

Question :
Suppose X and Y are independent Poisson random variable , each with the mean 1 , obtain
i) P(X+Y=4)
ii) E[(X+Y)2]
Thanks

so $\mu$ = $\lambda$ = 1

P(A+B=4) = ? stuck here !!!
Mon Jul 05, 2010 11:41 am
Forum: Statistics
Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
Replies: 7
Views: 9638

### Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),

Question :
Suppose X and Y are independent Poisson random variable , each with the mean 1 , obtain
i) P(X+Y=4)
ii) E[(X+Y)2]
Mon Jul 05, 2010 3:44 am
Forum: Statistics
Topic: prob. of 3 red, 2 white, 1 blue (is the answer correct)
Replies: 5
Views: 6191

### Re:

Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?
Sorry silly mistakes ....
P(R).P(R).P(R) + P(W).P(W) + P(B) = 0.4289
Sun Jul 04, 2010 4:08 pm
Forum: Statistics
Topic: What is the probability that he really knows the answer
Replies: 4
Views: 5680

### Re: What is the probability that he really knows the answer

then p(knows|correct)=p(correct|knows)*p(knows)/p(correct)
=1*1/2*1/(p(correct|knows)p(knows)+p(correct|guess)p(guess)+p(correct|copy)p(copy))
=.5*1/(p(knows)+p(guess)+1/48)
=24/41
That number seems a little low.
If u know the answer can u please let me know ?
Sun Jul 04, 2010 4:03 pm
Forum: Statistics
Topic: random var. X is -1, 0, 1 w/ equal probability (check ans)
Replies: 4
Views: 5746

### Re: random var. X is -1, 0, 1 w/ equal probability (check an

Please can u tell what is wrong

My work ........
E(X) = 0

$\mu'$ 1 = 0

$\mu'$ 2= 2/3

$\mu$2 = 2/3

var(X) = $\mu'$2 - $\mu$2 = 2/3 - 2/3 = 0
$Var(X)=E(X^2)-[E(X)]^2=E(X^2)-0^2=E(X^2)=2/3$

Thanks !!!
Sun Jul 04, 2010 4:00 pm
Forum: Statistics
Topic: prob. of 3 red, 2 white, 1 blue (is the answer correct)
Replies: 5
Views: 6191

### Re: prob. of 3 red, 2 white, 1 blue (is the answer correct)

12 is not a probability