The results of a fitness trial is a random variable X which is normally distributed with mean µ and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a 98% confidence interval for µ. What is the width of this interval ?

Question

anonymous · Answer

@phi pls help

phi · Answer

if you have a random sample of 90 trials  and find the average of them, you will get mu $ \pm$ some standard deviation

anonymous · Answer

how to find the average ?

phi · Answer

the std dev would be  the population std dev /  sqrt(size of sample)
$$ \sigma= \frac{\sigma_{pop}} { \sqrt{90} } =\frac{2.4} { \sqrt{90} }  $$

phi · Answer

you don't find the average... you are told it is $ \mu$
what they want are the lower and upper bounds x so that
$$ \mu \pm x$$ represents 90% of the area under the curve.

phi · Answer

**98%

anonymous · Answer

we dont have to find mu ?

phi · Answer

no.  find the number of std deviations so that
you get 98% of the area
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