Question

The Quality Assurance Department for Cola, Inc. maintains records regarding the amount of cola in its Jumbo bottle. The actual amount of cola in each bottle is critical but varies a small amount from one bottle to the next. Cola, Inc. does not wish to yoderfill the bottles, because it will have a problem with truth in labeling. On the other hand, it cannot overfill each bottle, because it would be giving cola away, hence reducing its profits. Its records indicate that the amount of cola follows the normal probability distribution. The mean amount per bottle is 31.2 ounces and the population standard deviation is 0.4 ounces. At 8 m. today the quality technician randomly selected 16 bottles from the filling line. The mear amount of cola contained in the bottles is 31.38 ounces. Is this an unlikely result? Is it likely the process is putting too much soda in the bottles? To put it another way, is the sampling error of 0.18 ounces unusual? A normal population has a mean of 60 and a standard deviation of 12. You select a random sample of 9 Compute the probability the sample mean is: Greater than 63. Less than 56. Between 56 and 63. A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is: Less than 74. Between 74 and 76. Between 76 and 77. Greater than 77.

Answer 1

Solution! Given data mean 4260 , scd=12, nef ) probability tra sample mean > 63 l PCX>63) - PZ63) EP 2-1 63-60 P(zze.) - P1z78 45) 0.9366 : P(x+ t +-) = 0 22 66 P (x763) = 0.22.66 b) probability that sample mean 256. percee 50 = plZC56) = P(V LPG) plze 56-60g = Pl_2-1)=0-1586 IP(X<56) = 0.1586 c) probability (562x<63) =P(564 ZL 63) =pl 562 3- pl 56-60 L22 63-60) Pforts E221) 55/5/ Pl-18172 0.75) =P (Z20.75) – PLZ2-1) = 6.7733 – 0.1586 P(-22 20:75) - 0.6147 P(5622 263 = 0.6147

Siven data mean U575, Sd = 1 40 a) probability of sample mean < 74 plac74) =P ( 9 274.) - plzeta) pupel74–75) l il (24 ))=P(Z <=1.264) PIR-1264) = 0.103115 PIX <74) =0.103815 b) p 174 28276) =P(742x4 (76) =P (7422276) / +4--75 - LY76-751 64274 1.264) J40 P(Z < 1.264) - Plze-1.264) 0.8968–6.10341 P(74 < 2276) 0.79369 2)p(16<x<*+) = p (76-75 274 77-75 Jao )-1.264 CZ <2.5224 P(Z225.29)- P(221.264) 6.9942-0.8968 -0.0944 P(7629 <17)= 0.0974 d) play17) - plze 77-75 - Pl2720599) - 0.0057 140 1P18+ 4+)= 0.005717