n
= 10;
x
~
N
(300,10
2
);
0.10;
0.95 ; one-
sided From Appendix VIII:
K
= 2.355
UTL
x
KS
300
2.355(10)
323.55
8.46.
Sample of 50 measurements on a normally distributed quality characteristic has a mean of 35 and a
standard deviation of 2. Using a confidence probability of 0.95, find a value such that 90% of the future
measurements on this quality characteristic will lie above it.
n
= 50;
x
~
N
(35,2
2
);
0.10;
0.95 ; one-
sided From Appendix VIII:
K
= 1.646
x
KS
35
1.646(2)
31.708
8.47.
A sample of 20 measurements on a normally distributed quality characteristic had
x
350 and
s
10 .
Find an upper natural tolerance limit that has probability 0.90 of containing 95% of the distribution of
this quality characteristic.
n
= 20;
x
~
N
(350,10
2
);
0.05;
0.90 ; one-
sided From Appendix VIII:
K
= 2.208
UTL
x
KS
350
2.208(10)
372.08
8.48.
How large a sample is required to obtain a natural tolerance interval that has probability 0.90 of containing
95%
of the distribution? After the data are collected, how would you construct the interval?
0.05;
0.90
2
2
7.779
1
,4
0.10,4
1
2
2
1
2
0.05
7.779
n
1
,4
77
2
4
2
0.05
4
After the data are collected, a natural tolerance interval would be the smallest to largest observations.

8.49.
A random sample of
n
= 40 pipe sections resulted in a mean wall thickness of 0.1264 in. and a standard
deviation
of 0.0003 in. We assume that wall thickness is normally distributed.
(a) Between what limits can we say with 95% confidence that 95% of the wall thicknesses should fall?
= 0.05;
= 0.95; and two-
sided From Appendix VII:
K
=
2.445
TI on
x
:
x
KS
0.1264
2.445(0.0003)
[0.1257,0.1271]
(b)
Construct a 95% confidence interval on the true mean thickness. Explain the difference between
this
interval and the one constructed in part (a).
0.05;
t
/2,
n
1
t
0.025,39
2.023
CI on
x
:
x
t
/2,
n
1
S
n
0.1264
2.023
0.0003
40
[0.1263,0.1265]
Part (a) is a tolerance interval on individual thickness observations; part (b) is a confidence interval on
mean thickness. In part (a), the interval relates to individual observations (random variables), while in
part (b) the
interval refers to a parameter of a distribution (an unknown constant).
8.50.
Find the sample size required to construct an upper nonparametric tolerance limit that contains at least 95%
of
the population with probability at least 0.80. How would this limit actually be computed from sample
data?
0.05;
0.80
log(1
)
log(1
0.80)
n
31.4
32 log(1
)
log(1
0.05)
The largest observation would be the nonparametric upper tolerance limit.

#### You've reached the end of your free preview.

Want to read all 57 pages?

- Fall '16
- Managment, Industrial Engineering, Normal Distribution, Quality Engineering, USL