Question

In an effort to cut costs and improve profits, many U.S. companies have been turning to outsourcing. In fact, according to Purchasing magazine, 54% of companies surveyed outsourced some part of their manufacturing process in the past two to three years. Suppose 575 of these companies are contacted a. What is the probability that 336 or more companies outsourced some part of their manufacturing process in the past two to three years? b. What is the probability that 286 or more companies outsourced some part of their manufacturing process in the past two to three years? c. what is the probability that 50% or less of these companies outsourced some part of their manufacturing process in the past two to three years? (Round the values of z to 2 decimal places. Round the intermediate values to 4 decimal places. Round your answers to 4 decimal places.) b. C.

Answer 1

XBinomial (n - 575,p 0.54) Binomial can be approximated to normal with: μ-np = 575 * 0.54 310.5 σ = v np(1- p)-V575 * (0.54) (1-0.54)-11.95 1151 ะ 11.951 P(X 〉 336) Normal Distribution, 11-31 0.5, ơ-11.951151, Since this is continuous distribution, P(X 2 336) - P(X > 336) we convert this to standard normal using z- 336- (310.5) 11.951151 2.1336862.13 P(Z > 2.13)-Area to the right of 2.13 0.0166 2.13 P(X > 336) P(Z2.13) 1 - P(Z<2.13) 1 - 0.9834 0.0166 PZ2.13): in a z-table having area to the left of z, locate 2.1 in the left most column Move across the row to the right under column 0.03 and get value 0.9834 Using technology, answer is 0.016434255535573 P(X > 336) 0.0166-1.66% B PX286

Normal Distribution, μ = 310.5, σ = 11.951151, Since this is continuous distribution, P(X 2 286) - P(X > 286) we convert this to standard normal using 286 - (310.5) 11.951151 ~一2.050012 ~-2.05 P(Z >-2.05) = Area to the right of-2.05 0.9798 -2.05 P(X > 286) = P(Z >-2.05) = 1-P(Z <-2.05)-1-0.0202 0.9798 P(Z <-2.05): in a z-table having area to the left of z, locate-2 in the left most column Move across the row to the right under column 0.05 and get value 0.0202 Using technology, answer is 0.979818361571631 P(X> 286)= 0.9798 = 97.98% n 575, p = 0.54 mean of sampling distribution, Hp p 0.54 Mp 0.54

p × (1-p) 0.54 × 0.46 575 variance, σ-- 0.000432 standard deviation of sampling distribution, σρ- 0.020785 0.54 x 0.46 575 νσ2- p 0.020785 Pp0.5) Normal Distribution, μ-0.54, σ-0.020785, Since this is continuous distribution, P(p < 0.5)-P(p< 0.5) we convert this to standard normal using 0.5 - (0.54) ー·020785 -1.924501 -1.92 P(Z<-1.92) - Area to the left of -1.92 0.0274 1.92 P(p< 0.5)-P(Z<-1.92) 0.0274 (from z-table) P(Z <-1.92): in a z-table having area to the left of z, locate-1.9 in the left most column Move across the row to the right under column 0.02 and get value 0.0274 Using technology, answer is: 0.027145914183427 P(p<0.5)= 0.0274 = 2.74% noaThank youe