in certain manufacturing processes parts must be machined tonwithin specified tolerences. the parts must have a mean length of of six inches plus or minus 0.26 inches .A SAMPLING PROCEDURE HAS BEEN ESTABLISHED to determine wether the ten parts are within the specified tolerences. if for a sample of ten parts selected at random the mean plus 1.5 standard deviation are within the specified limits of 5.74 inches to 6.26 inches than the parts are meeting the specified tolerences this means that a z score of 1.5 and a z score of -1.5 must be within the limits of 5.74 inches and 6.26 inches the length in inches represent the 10 parts as follows
5.94, 6.20,5.86,6.08,6.14,6.12,6.01,5.92,5.90,6.23
procedure
a. calculate the sample mean and standard deviation s
b. convert z=1.5 and z= -1.5 into inches
c. compare these values with the specified limits of 5.74 inches to 6.26 inches and determine if the sample meets the stated conditions
We do not do your work for you, but here is some help.
Mean(μ) = ΣX/N
Standard Deviation(SD) = Square root of (Σd^2/N), where d = the deviation of each score from the mean.
Z = (X-μ)/SD
Insert the values from above for Z = 1.5 and -1.5.
Then you can make the comparison.