anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@phi pls help
phi
  • phi
if you have a random sample of 90 trials and find the average of them, you will get mu \( \pm\) some standard deviation
anonymous
  • anonymous
how to find the average ?

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phi
  • phi
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
  • phi
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
  • phi
**98%
anonymous
  • anonymous
we dont have to find mu ?
phi
  • phi
no. find the number of std deviations so that you get 98% of the area |dw:1438259659150:dw|
phi
  • phi
the tails use 2% , 1% on each end
anonymous
  • anonymous
okay
phi
  • phi
after you get the value for x (# of std deviations) you calculate the interval as \( 2 \cdot x\cdot \sigma\) where \( \sigma= \frac{2.4}{\sqrt{90}} \)
anonymous
  • anonymous
but we do not have sample mean.. how to calculate ?
anonymous
  • anonymous
@phi
phi
  • phi
they are asking for What is the width of this interval ? the "middle" of the interval is \( \mu\) the upper bound of the interval is \( \mu+x\) the lower bound is \( \mu - x \) the width of this interval is upper bound - lower bound which is \( \mu+x - ( \mu-x) \) = 2x so you need to find x
phi
  • phi
x is the # of std dev so that you get 98% of the area times sigma (which you calculate from the info they gave us)

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