CE93Engineering Data Analysis
Mark Hansen, Fall 2009
CE93 Homework 2 – Solution
1. Sample median and mean values.
> median(x), mean(x), median(y), mean(y)
Data
Median
Mean
Density
2445
2444.9
Strength
60.10
60.14
These measures fall in the central regions of the corresponding normalized frequency diagrams
as shown in the following figures.
Mean=median=2445
mean=median=60.1
2. Range, IQR, Mean absolute deviations,
d
, the sample (unbiased) variance,
s
2 , the sample
(unbiased) standard deviations,
s
and the sample c.o.v.’s
d
ˆ .
>> range(x), IQR(x), mad(x), var(x), std(x), std(x)/mean(x), std(x,1)
>> range(y), IQR(y), mad(y), var(y), std(y), std(y)/mean(y), std(y,1)
Data
Range
IQR
d
s
2
s
g2012
g4632
s normalized by N
Density 77
19
12.2
255.8
16.0
0.0065 15.8
Strength 19.6
6.55
3.84
25.15
5.01
0.0834 4.95
The estimate of the standard deviation when normalized by
N
is slightly smaller than the
unbiased estimate.
The strength data has higher dispersion since it has a larger sample c.o.v.
3. Sample size
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CE93Engineering Data Analysis
Mark Hansen, Fall 2009
Answers:
>> f10=E1_2(1:10,2);
I expect (through inspection) that the mean will be about 60. I expect the median to be about 61.5.
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 Fall '10
 hansen
 Standard Deviation, Mark Hansen, CE93Engineering Data Analysis

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