Accepting/Rejecting Measurement of 58mm Using Cahvenenet's Criterion

[stunner5000pt]

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

$\overline{x}$ = average
$sigma_{x}$ = standard deviation
$x_{sus}$ = the measurement we want to reject or accept
$$t_{sus} = \frac{x_{sus}-\overline{x}}{\sigma_{x}}$$
the number of standard deviationsfrom which x sus differes from x bar
n(worse than $x_{sus}$) = N P(outside $t_{sus} \sigma_{x}$)

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
$$t_{sus} = 2.4$$ standard deviations

then
P(putside 2.4$\sigma$) = 1 - P(within 2.4 $\sigma$)
= 1 - 0.984
the 0.984 is taken from a table which shows the percent probability
P(within $t\sigma$)= \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!

 

[OlderDan]

I'm not sure exactly what you are asking, but the .984 is the probablility that a measurement made on normally distributed data will be within 2.4 standard deviations of the mean. I've never heard of cahuvenenet's criterion, but then I'm hardly an expert in things statistical. .984 is the area under the normal distribution function from -2.4 to +2.4 standard deviations. There are several calculators online for finding and graphing these areas. Here is one of them:

http://www.math.csusb.edu/faculty/stanton/probstat/normal_distribution.html