# Im stuck in this statistics problem

1. Aug 12, 2010

### 619snake

1. The problem statement, all variables and given/known data

the weight of items produced by a production line is normally distributed with a mean of 12 ounces and a standard deviation of 2 ounces.
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 items will fulfill quality control requirements.

What I have to the moment is:
X~n(12,2)
P(x<16) - P(x<8)
z=(16-12)/2
z=2

z= (8-12)/2
z=-2

p(z<2)-P(z<-2)
=.9772-.0228
=.9544

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

Those where other requirements of the problem. I'm stuck at the supposing, cant decide if it is binomial or not, and how to start the procedure

2. Aug 25, 2010

### presbyope

So you calculated the probability of success for each trial to be 0.9544? That takes care of the supposing part. Now figure out what the 3 out of 7 means.

.. personally I can't tell. Is it asking for _exactly_ 3 out of 7?