- #1
stunner5000pt
- 1,461
- 2
Homework Statement
A student makes 10 measurements of length x and gets the results all in mm
46,48,44,38,45,47,58,44,45,43
Using cahuvenenet's criterion should he accept or reject the measurement of 58??
Homework Equations
[itex] \overline{x} [/itex] = average
[itex] sigma_{x} [/itex] = standard deviation
[itex] x_{sus} [/itex] = the measurement we want to reject or accept
[tex] t_{sus} = \frac{x_{sus}-\overline{x}}{\sigma_{x}} [/tex]
the number of standard deviationsfrom which x sus differes from x bar
n(worse than [itex]x_{sus} [/itex]) = N P(outside [itex] t_{sus} \sigma_{x} [/itex])
is n < 0.5 then 58 is rejected
if n > 0.5 then 58 is accepted
The Attempt at a Solution
well the average
x bar = 45.8
standrad deviation = 5.1
[tex] t_{sus} = 2.4 [/tex] standard deviations
then
P(putside 2.4[itex] \sigma [/itex]) = 1 - P(within 2.4 [itex] \sigma [/itex])
= 1 - 0.984
the 0.984 is taken from a table which shows the percent probability
P(within [itex] t\sigma[/itex])= \int_{X-t\sigma}^{X+t\sigma} f_{X,\sigma} (x) dx [/tex], as a function of t
but why is .984?? Why is it that the probability should be chonse to be 0.01 and not 0.00 ... or 0.02??
thanks for your help in advance!