# Need help with statistics

1. Aug 13, 2010

### 619snake

1. The problem statement, all variables and given/known data
the weight of items produced by a production line is normally distributedwith a mean of 12 ounces and a standard deviation of 2 ounces.
a. what is the probability that a randomly selected item will weight between 8 and 16 ounces? (DONE)
b. what is the probability that a randomly selected item will weight over 20 ounces? (DONE)
c. Suppose that quality control requires the weight of items to be within 8 and 16 ounces. You select 7 items at random (each item is independent). What is the probability that 3 of the items will fulfill quality control requirements. (DONE BUT HAVE DOUBTS)
d. find the probability that a randomly selected item has a weight that is greater than 14 or smaller than 10. (STUCK IN THIS ONE)

2. Related formulas
if x is BIN
p(x=k) = (n!)/((n-k)!(k!)) * $$\pi$$^k * (1-$$\pi$$)^(n-k)
mean = n$$\pi$$
variance = n$$\pi$$(1-$$\pi$$)

if x is N(µ,$$\sigma$$), then z=(x-µ)/($$\sigma$$) is N(0,1)

3. The attempt at a solution
a. x~n (µ=12, $$\sigma$$=2)
p(8<x<16)
p(x<16) - p(x<8)
z=(16-12)/2 z=(8-12)/2
z=2 z= -2
p(z<2)-p(z<-2)
=.9772-.0228
=.9544

b.P(x>20)
z=(20-12)/2
z=4
p(z>4)= 1

c.x~binomial (µ=7, $$\sigma$$=.95)
p(x=3)
p(x=K)=$$(3!)/(3!)(7-3)!$$(.95)^3 (1-.95)^4
.
.
.
x=0.000187551

d. I have no idea how to deal with this one
I think I have to use the mean and standard deviation of the problem
(µ=12, $$\sigma$$=2)
P(x<10) or P(x>14)
Hope you people can help me

2. Aug 14, 2010

### danago

The events X>14 and X<10 are mutually exclusive (cannot occur at the same time) so that their probabilities are additive:

P(X>14 or X<10) = P(X>14) + P(X<10)