Question 14 14. The population mean outstanding bill for delinquent customer acc
ID: 2936867 • Letter: Q
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Question 14 14. The population mean outstanding bill for delinquent customer accounts for a national department store chain is $187.50 with a standard deviation of S54.50. In a simple random sample of 50 delinquent accounts, what is the probability that the sample mean outstanding bill is over $200? a) 0.0667 b) 0.0526 c) 0.4090 d) 0.5910 e) None of the answers is correct to the fourth decimal place. Question 15 15. A facility for bottling soft drinks uses two fill and seal machines (1 and 2). It is as- sumed that the two machines have the same overall average fill amount (so, we will take this as truth). As part of quality control, a random sample of size 30 is taken from each machine and the sample means are compared. If the standard deviations of the machines are the same and are known to be 0.3 oz, what is the probability that sample mean for machine 1 is more than 0.25 oz above the sample mean for machine 2? a) 0.0392 b) 0.0113 c) 0.0006 d) 0.9994 e) None of the answers is correct to the fourth decimal place.Explanation / Answer
14) std error of mean =std deviation/(n)1/2 =54.50/(50)1/2 =7.7075
hence probability=P(X>200)=P(Z>(200-187.50)/7.7075)=P(Z>1.6218)=0.0526
hence option b is correct
15)
here std error of mean difference =(0.32/30+0.32/30)1/2 =0.0775
therefore probability =P(X>0.25)=P(Z>(0.25-0)/0.0775)=P(Z>3.2275)=0.0006
hence option C is correct
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