discussion_9

\[E(P_{defectedn}>115)\] \(n=1900\), \(expected= n* p = 95\) , \(actual = 115\) \(\sigma = \sqrt{npq} = \sqrt{1900*.95*0.05} = 9.5\)

n_noodles <- 1900
expected_bad_noodles <- .05* n_noodles
defective_noodles <-  115

standard_dev <- sqrt(1900*.95*.05)

my_prob <-  (defective_noodles+.5 - expected_bad_noodles)/ standard_dev
my_percent <- (.4821+.4861)/2

• Chart shows that given a p value of 2.15,the probability equation would be (P_{2.1}+P_{2.2})/2 \(.5-.4841\)

.5-.4841

## [1] 0.0159

Answer

• The probability of getting at least this many defecive noodles is .0159.