1. The problem statement, all variables and given/known data 1. Let Y_1, Y_2, ... Y_n be a random sample from a normal distribution with mean = 2 and variance = 4. How large must n be in order that P(1.9 <= Y <= 2.1) >= 0.99 (there is a bar _ above the Y in the problem.) 2. When a machine is improperly adjusted, it has a probability of .15 of producing a defective item. Each day the machine is run until three defective items are produced. If this occurs, it is stopped and checked for adjustment. What is the probability that an improperly adjusted machine will produce five or more items before being stopped? What is the average number of items an improperly adjusted machine will produce before being stopped. 2. Relevant equations 1. (Y - mean)/[(s.d.)/sqrt(n)] 2. Negative binomial formula: p_x(k) = (k-1, r-1) * p^r * (1-p)^(k-r), k = r, r+1, ... (k-1, r-1) is a binomial. 3. The attempt at a solution 1. I got: P( (1.9-2) / (4 / sqrt(n) ) <= (Y - 2) / ( 4 / sqrt (n) ) <= (2.1 - 2) / ( 4 / sqrt (n) ) ) >= .99 p(-sqrt(n) * .1/4 <= Z <= sqrt(n) * .1/4) >= .99 At this point I checked the table for cumulative areas under the standard distribution and got 2.4 = sqrt(n) * (.1/4) n = 10? 2. I don't know how to set this one up... p = .15 that an improperly adjusted machine will produce a defective item. and we want P( X >= 5) or P( X >= 5) = 1 - P( 2 defects out of the first 3 made and the 3rd defective one on the 4th) that is, P(X >= 5) = 1 - P( X < 5) then r = 3 and k = 3,4,5,6 P(X >= 5) 1 - [ (sum from k=3 to 6) (k-1, 2) * (.15)^3 * (.85)^(k-3)] I know k has to start at 3, but whats the limit of k here, 4 trials? Any input is greatly appreciated.