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 ?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

@phi pls help
  • phi
if you have a random sample of 90 trials and find the average of them, you will get mu \( \pm\) some standard deviation
how to find the average ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

  • 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
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
**98%
we dont have to find mu ?
  • phi
no. find the number of std deviations so that you get 98% of the area |dw:1438259659150:dw|
  • phi
the tails use 2% , 1% on each end
okay
  • phi
after you get the value for x (# of std deviations) you calculate the interval as \( 2 \cdot x\cdot \sigma\) where \( \sigma= \frac{2.4}{\sqrt{90}} \)
but we do not have sample mean.. how to calculate ?
  • 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
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)

Not the answer you are looking for?

Search for more explanations.

Ask your own question