1. ## probability question

Hello,
I am having trouble with my stat homework. Maybe someone can give me some guidance? I would really appreciate it!

A woman is fishing for bass on Pinelands pond. Let x=number of 'keepers' that she catches in an hour. The probability distribution for x is given in the table below

x= 0 1 2 3 4 5
p(x)= .1 .2 .3 .2 .1 .1

Questions:
a) What is the probability that she catches at most 4 keepers?
b) What is the probability that x is greater to and equal to 3?

c) Find Ux, the mean number of keepers caught
d) Find the standard deviation for X, the number of keepers caught

2. a) the probability of at most 4 is equal to the probability of three or fewer. simply add up the probabilities involved.

b) use the above and adjust for three or greater

some examples:

c) for a probability distribution x= 2 4 6 8 10, P(x)= .1 .3 .1 .2 .3 Ux + 2(.1)+4(.3)+6(.1)+8(.2)+10(.3) = 6.6

d) sigma = sqrt( sum(x-mu)^2*P(x)) so, to begin the pattern for my example in c) = sqrt((2-6.6)^2(.1) + (4-6.6)^2(.3)...)=2.83

3. c)
this is what i did, tell me if i am correct?
0(.1) + 1(.2) + 2(.3) + 3(.2) + 4(.1) + 5(.1)
mean = 2.3

4. a)
.1 + .2 + .3 + .2 + .1 = .90
is this correct also?

5. b)
.2 + .1 + .1 = .40
is this correct also?

6. d)
is the standard deviation 7.3?

abc looks fine.

8. Yeah im confused on d)
I did this:

0^2(.1) + 1^2(.2) + 2^2(.3) + 3^2(.2) + 4^2(.1) + 5^2(.1)

what did i do wrong??

sigma = sqrt( sum {( x-mu)^2*P(x) } ) as jerry mentioned

mu is the mean

10. alright i think i got it
0^2(.1) + 1^2(.2) + 2^2(.3) + 3^2(.2) + 4^2(.1) + 5^2(.1)

0+.2+1.2+1.8+1.6+2.5

=7.3 - mean squared 5.29 = 2.01

did i do it?!

11. correct. But it is not the sd.
you have calculated the variance.

12. i am terrible at statistics...ugh
back to the drawing board.

13. okay i calculated it again but came out to 2.01!!! which is the variance.
if i squared the variance (2.01)^2 = 4.0401
would that be the standard deviation??
thanks

14. Read the text book thoroughly.

sd = sqrt(var)

15. so you are saying...
take the square root of 2.01
which is 1.4177 ...is the standard deviation??