Question

A manufacturer of window frames knows from long experience that 10% of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 16 window frames: a. None will need adjustment? (Round the final answer to 3 decimal places.) Probability b. At least one will need adjustment? (Round the final answer to 3 decimal places.) Probability O c. More than two will need adjustment? (Round the final answer to 3 decimal places.) Probability

Answer 1

* from the given data on to N=16 P= 10% = 0,1 (requires adjustments) So, to solve this problem we use "Binomial distributor [ P(X=K) = ncx pk gr-k | (:9=1-P) 16-0 a) Probability that none will need adjustment is 3)P(x=0) >) 16cm (O-10 C1-0-1) (i) using Binomal distribution => 1x1x 018530 = 0.18530 0185 (:. 3-decimal place) 6) probability that at least one will need adjustment is. P(X=1) = P(x=1) +P(x=2) + P(x=337 ..... PCX=16) or - |- P(X<1) -|- P(x=0) = 1-0.185 (from @) on 5) 0-815 (1:3-decimal place)

c) probability that more than two will need adjustment is P(x>2) ) PCX-3) +P(x=4)+ -. ... +P(x=16] => 1- [PCX=0) + P(x=1) + P(x-2)] By using Binomial distribution we get values of P(x=0) = 160 (0:1) coggle => 0-185 | P(x-1) = 16c, (ou Co.ggs => 0.329 P(x=2) = 1662 (0.13² (0.9 jt => 0.275 lei so A PCX32) - I-(0.185+ 0.329+ 0.275) 3) 1- 0.789 (= 0.211 (:: 3-decimal place) Please give rating if you like the answer"