(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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??

2. Relevant 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

3. 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!!

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# Homework Help: Error analysis

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