The Hammermill Company produces paper for laser printers. The standard letter-si

Question

The Hammermill Company produces paper for laser printers. The standard letter-sized paper width is 215mm. But the actual width is random because of vibration in the rollers and cutting tools during manufacturing. The rollers and cutters can be adjusted if the paper width drifts from the correct mean. Suppose that a quality control inspector chooses 100 sheets at random and measures them with a precise instrument, obtaining a standard deviation of 1 mm. Let X denote the measured width of a sheet and µ be the population mean of X. The hypotheses to be tested are

H0 : µ = 215

H1 : µ 6= 215

Suppose that the machine was misaligned at 215.05mm and = 10%. What is the probability that the quality control inspector concludes that the machine was correctly aligned?

(a)88.493% (b)85.668% (c)84.134% (d)81.594% (e)80.234%

*Can you please show the work to solving this problem. Thank you!

Explanation / Answer

here std error of mean =std deviation/(n)1/2 =1/(100)1/2 =0.1

for 0.1 level critical z =1.645

therefore area of acceptance: P(215-1.645*0.1<X<215+1.645*0.1)=P(214.8355<X<215.1645)

hence probability of conclusding that machine is correctly aligned given mean =215.05)

=P((214.8355-215.05)/0.1<Z<(215.1645-215.05)/0.1)=P(-2.1449<Z<1.1449)=0.8739-0.0160 =0.8579 ~85.668%

option B is correct

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