# probability question

#### letsstophere

##### New Member
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

#### jerryb

##### New Member
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

#### letsstophere

##### New Member
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

#### letsstophere

##### New Member
a)
.1 + .2 + .3 + .2 + .1 = .90
is this correct also?

#### letsstophere

##### New Member
b)
.2 + .1 + .1 = .40
is this correct also?

#### letsstophere

##### New Member
d)
is the standard deviation 7.3?
please someone check my answers. Thank you very much!

#### letsstophere

##### New Member
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??

#### vinux

##### Dark Knight
follow the formula
sigma = sqrt( sum {( x-mu)^2*P(x) } ) as jerry mentioned

mu is the mean

#### letsstophere

##### New Member
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?!

#### vinux

##### Dark Knight
correct. But it is not the sd.
you have calculated the variance.

#### letsstophere

##### New Member
i am terrible at statistics...ugh
back to the drawing board.

#### letsstophere

##### New Member
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

#### vinux

##### Dark Knight
Read the text book thoroughly.

sd = sqrt(var)

#### letsstophere

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

#### letsstophere

##### New Member
wanted to let you guys know i got a 100 on this take home quiz. thanks a lot!!