Question

Amanufacturer of computer printers purchases plastic ink cartridges from a vender. When a large shipment is received, a random sample of 235 cartridges is selected, and each cartridge is inspected. If the sample proportion of defective cartridges is more than 0.02, the entire shipment is returned to the vendor. (a) What is the approximate probability that a shipment will be returned the true proportion of defective cartridges in the shipment is 0.007 (Round your answer to four decimal places.) (b) What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10% (Round your answer to four decimal places.) You may need to use the appropriate table in the appendix or technology to answer this question.

Answer 1

Solution Given Data A Random Sample (n) = 235 @ the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.06 P0.06,9-l pe 10.06 - 0.94 n235 Paking normal approximation to binomial, the sampling distribution sample proportion follows normal dibribution of ĝNN (0,02), where mean , us - P=0.06 My = 0.06 at Px (-P) 0.06*(1-0.06) - 235 variance, 0.06 x 0.94 235 0.0564 235 - 0.00024

standand deviation , a lox0.06 x 0.94 - 10.0024 = 0.015491 op = 0.015491 N N (0,06, 6.015491%) Plås 0.02) normal Distribution, u = 0.06, 0 = 0.015491, we convert this to standard nolmal using = a 0.02- 60.06) 0.015491 -0.ou = -2.5821 0.0154917 ~ - 2.58 P(23-2.58) = Area to the right of 2.58

PCÊ 50.02) = P(23-2.58) alp(Z 2-2.58) (from 2-table) = 1- 0.00491151 0.995 = 0.9951 -2,57 P(22-2.58): in a z table having area to the left of z, locate - 2.58 in the left most column move across the row to the right under column 0.00491 and get value 0.0049151 pløs 0.02) - 0.9951 99.517. 6 The approximate probability that a shipment will not be netwined it the a true proportion of detective castidges in the shipment is 0.10 p= 0.01, q=1-P = 0.9 , n=235 Taking nilmal approximation to binomial, the sampling dis bribution of sample proportion followes nõlmal district Dwn (M, a) where mean, theo =P=0.1

M = 0.1 L Px (1-P) Voviante ? 0.1 x6,9 0.09 235 235 a = 0.000382 standard deviation ap=5a = 0.000382 = 0.01954 op = 0.01954 Pun (0.1, 0.019542) P(Å£0.02) Normal Distribution U=0.), o = 0.01954 since this is contime vous distyi bulion P ( 82 0.02) = P ( P < 0.02) standort nölmal az EP-u 0.02 - (0.1) 0.01954 -0.08 0.01954 -4,0941 -4.09

0.001 -4.09 P (P < 0.02) - P(22-4.09 ) = 0.0000 3169 (From 7-table) ) belco -4.00 the one belno -4.00 the conswex=0.00003 P (2 -4.09). Since - 4.09 p( 8 , 0-02) : 0. 00003 , 0, 000697.