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6-7 PROBABILITY PLOTS 219
MIND-EXPANDING EXERCISES
6-87. Consider the airfoil data in Exercise 6-12. 6-94. Suppose that we have a sample x 1 , x 2 , p , x n and
Subtract 30 from each value and then multiply the re- we have calculated x n and s n 2 for the sample. Now an
2
sulting quantities by 10. Now compute s for the new (n 1)st observation becomes available. Let x n 1 and
2
2
data. How is this quantity related to s for the original s n 1 be the sample mean and sample variance for the
data? Explain why. sample using all n 1 observations.
n
6-88. Consider the quantity g i 1 1x i a2 2 . For what (a) Show how x n 1 can be computed using x n and x n 1 .
value of a is this quantity minimized? n1x n 1 x n 2 2
(b) Show that 2 2
6-89 Using the results of Exercise 6-87, which of the ns n 1 1n 12s n n 1
n 2 n 2
two quantities g i 1 1x x2 and g i 1 1x 2 (c) Use the results of parts (a) and (b) to calculate the
i
i
will be smaller, provided that x ? new sample average and standard deviation for the
6-90. Coding the Data. Let y a bx , i i data of Exercise 6-22, when the new observation is
i
1, 2, . . . , n, where a and b are nonzero constants. Find the x 38 64 .
relationship between and , and between syx x and s . 6-95. The Trimmed Mean. Suppose that the data are
y
6-91. A sample of temperature measurements in a fur- arranged in increasing order, T% of the observations are
nace yielded a sample average ( F ) of 835.00 and a sam- removed from each end and the sample mean of the re-
ple standard deviation of 10.5. Using the results from maining numbers is calculated. The resulting quantity is
Exercise 6-90, what are the sample average and sample called a trimmed mean. The trimmed mean generally
x
standard deviations expressed in C ? lies between the sample mean and the sample median
6-92. Consider the sample x 1 , x 2 , . . . , x n with sample x . Why?
mean x and sample standard deviation s. Let (a) Calculate the 10% trimmed mean for the yield data
z i 1x i x 2 s, i 1, 2, . . . , n. What are the values of in Exercise 6-17.
the sample mean and sample standard deviation of the ? z i (b) Calculate the 20% trimmed mean for the yield data
6-93. An experiment to investigate the survival time in Exercise 6-17 and compare it with the quantity
found in part (a).
in hours of an electronic component consists of placing
(c) Compare the values calculated in parts (a) and (b)
the parts in a test cell and running them for 100 hours
with the sample mean and median for the yield
under elevated temperature conditions. (This is called an
data. Is there much difference in these quantities?
“accelerated” life test.) Eight components were tested
Why?
with the following resulting failure times:
6-96. The Trimmed Mean. Suppose that the sample
75, 63, 100 , 36, 51, 45, 80, 90 size n is such that the quantity nT 100 is not an integer.
The observation 100 indicates that the unit still func- Develop a procedure for obtaining a trimmed mean in
this case.
tioned at 100 hours. Is there any meaningful measure of
location that can be calculated for these data? What is its
numerical value?
IMPORTANT TERMS AND CONCEPTS
In the E-book, click on any Normal probability plot Sample mean CD MATERIAL
term or concept below to Population mean Sample standard Exponential probability
go to that subject. Population standard deviation plot
Box plot deviation Sample variance Goodness of fit
Frequency distribution Population variance Stem-and-leaf diagram Weibull probability plot
and histogram Random sample Time series plots
Median, quartiles and
percentiles