Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 46 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 4000 ​batteries, and 33​% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?

The probability that this whole shipment will be accepted is ?

​(Round to four decimal places as​ needed.)

The company will accept ? % of the shipments and will reject ? ​% of the​ shipments, so ▼ many of the shipments will be rejected. almost all of the shipments will be accepted.

Answer 1

no. of batteries that don't meet specifications = 4000*33% = 1320

no. of batteries that meet specifications = 4000*(1 - 33%) = 2680

no. of unaccepatble batteries in sample of 46 battereis = x

acceptable cases = x = 0,1,2,3

P(x) = (1320Cx)*(2680C(46-x))/(4000C46)

P(acceptable) = P(0) + P(1) + P(2) + P(3)

= (1320C0)*(2680C(46-0))/(4000C46) + (1320C1)*(2680C(46-1))/(4000C46) + (1320C2)*(2680C(46-2))/(4000C46) + (1320C3)*(2680C(46-3))/(4000C46)

= 8.78140276*10^-9 + 2.02355513*10^-7 + 0.00000227822 + 0.00001670066

= 0.00001919001

P(shipment accepted) = 0.00001919001 = 0.0019%

very few will be accepted, many will be rejected

P(reject) = 1 - P(shipment accepted) = 100% - 0.0019% = 99.9981 %

The company will accept 0.0019% of the shipments and will reject 99.9981% of the​ shipments, so many of the shipments will be rejected.

(please UPVOTE)