The lenghts of pins produced bya machine follow anormal distribution with mean2.54cm and standard deviation 0.04cm.A pin is rejectedif its lenght is lest than 2.44cm or more than 2.60cm.
Find the percentage of pins that are accepted. I did this
upper limit =2.60-2.54/0.04 =1.5
lower limit=2.44-2.54/0.04 =-2.5 P(z)1.5-P(z)-2.5
Is this right. The second part I need help with
If it is decidedthat 2.5%of the pins are to be rejected because they are to long and 2.5% because they are too short ,determine the new range of
acceptable lenghts. Could somebdy explain how to do this. Thank,s