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Stat 503
Homework #2
(80pts)
1. (10pts) The mean and SD of a set of 47 body temperature measurements were as
follows:
y = 36.497 °C
s = 0.172 °C
If the measurements were converted to °F, and
the conversion from Celsius to Fahrenheit is
y’ = 1.8 y + 32.
(a)
(5pts) What would be the new mean and SD?
The conversion from Celsius to Fahrenheit is F = y’ = 1.8 y + 32.
The new mean would be
y’ = 1.8
y + 32 = 1.8(36.497) + 32 = 65.6946 + 32 = 97.6946
97.7.
The new SD would be s’ = 1.8 s = 1.8(0.172) = 0.3096
0.31
(b)
(5pts) What would be the new coefficient of variation?
The new coefficient of variation would be s’/
y’ = 0.3096/97.6946 = 0.003169
0.0032.
2.
(10pts) A researcher measured the average daily gains (in kg/day) of 20 beef cattle; typical
values were
1.39, 1.57,1.44,…
The mean of the data was 1.461 and the standard deviation was .178.
(a)
(5 pts) Express the mean and std in lb/day. ( Hint: 1 kg=2.2 lb)
' (2.2)
yy
, hence, the new mean = (2.2)1.461=3.214 lb/day
The SD=(2.2)0.178=0.392
(b) (5 pts) What is the mean and std after we standardize the data?
Mean=0 and std=1
3.
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 Fall '08
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