Question

Suppose that a box of electrical fuses contains 3% defectives, and a sample of 100 fuses was obtained from the box. a. Find the probability of obtaining less than 2 defective fuses. (Round to four decimal places.) b. Find the expected number of defective fuses in the sample. (Round to the nearest whole number.)

Answer 1

= 1-0.03 100-0 P= 3.). defective = 0.03, 89 = 1- p. n=100 = 0.97 use Binomial expression - P(x=x) = "C* p** q n-2 ca) PI less than a defective fuses) P(x*_) = P(x=0) + P(x= 1) = 1000*0.039* 0.971007 100*0.03'*0.4.40-1 = 0.971007 100 * 0.03 * 0.9799 = 0.0475525 +0.1470696 - 0.1946 Aug (b) Expected number of defective feses =E(X) E(x) = .np = 100*0.03 = 3 ans Scanned with CamScanner