## anonymous one year ago 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 ?

1. anonymous

@phi pls help

2. phi

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

3. anonymous

how to find the average ?

4. phi

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

5. 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.

6. phi

**98%

7. anonymous

we dont have to find mu ?

8. phi

no. find the number of std deviations so that you get 98% of the area |dw:1438259659150:dw|

9. phi

the tails use 2% , 1% on each end

10. anonymous

okay

11. phi

after you get the value for x (# of std deviations) you calculate the interval as $$2 \cdot x\cdot \sigma$$ where $$\sigma= \frac{2.4}{\sqrt{90}}$$

12. anonymous

but we do not have sample mean.. how to calculate ?

13. anonymous

@phi

14. 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

15. 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)