The operations manager of a plant that manufactures tires wishes to compare the
ID: 3205734 • Letter: T
Question
The operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, each of which has a nominal value of 575 millimeters. A sample of five tires of each grade is selected, and the results representing the inner diameters of the tires, ordered from smallest to largest, are as follows: Compute the mean, median, and standard deviation for each grade of tire. What would be the effect on your answers in (a) and (b) if the last value for grade Y was 588 instead of 578? Explain. Which grade of tire is providing better quality? Explain.Explanation / Answer
a. Use the following formula to compute the mean.
xbarx=sigma x/n, where, x denote inner diameters of Grade X, and n is sample size of Grade X.
=(568+570+575+578+584)/5
=575
The data is sorted, and odd in number. Therefore, median is the middle most value, that is Mdx: 575.
Use following formula to cocmpute standard deviation.
sx=sqrt[1/n-1 sigma (x-xbar)^2]
=sqrt[1/5-1 {(568-575)^2+(570-575)^2+(575-575)^2+(578-575)^2+(584-575)^2}]
=6.4
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Using same formula, the mean, median and standard deviation of Grade Y is as follows:
Mean, xbary=(573+574+575+577+578)/5=575
Median, Mdy=575
Standard deviation, sy=sqrt[1/5-1 {(573-575)^2+(574-575)^2+(575-575)^2+(577-575)^2+(578-575)^2}]
=2.07
b. Due to increase in maximum value from 578 to 588, the mean will be increased, median will remain same, and the standard deviation will eventually decrease.
xbar new=(573+574+575+577+588)/5=577.40
Md new=575
s new=sqrt[1/5-1 {(573-577.40)^2+(574-577.40)^2+(575-577.40)^2+(577-577.40)^2+(578-577.40)^2}]
=6.11
c. The range for grade X is as follows:
Rx: Maximumx-Minimumx
=584-568
=16
Similarly, Ry:578-573=5
Grade Y has lower spread compared to grade X, therefore, precision of inner diameters of grade Y is high. Therefore, grade Y is providing better quality.
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