Explore BrainMass

# Finding probabilities based on a sample

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month.

a. what is the value of the population mean? what is the best estimate of this value?
b. explain why we need to use the t distribution. what assumption do yo uneed to make?
c. for a 95 percent confidence interval, what is the value of t?
d. develop the 95 percent confidence interval for the population mean
e. would it be reasonable to conclude that the population mean is 21 eggs? what about 25 eggs?
---------
Details: A normal distribution has a mean of 50 and a standard deviation of 4.
a. compute the probability of a value between 44.0 and 55.0
b. compute the probability of a value greater than 55.0
c. compute the probability of a value between 52.0 and 55.0

https://brainmass.com/statistics/probability/finding-probabilities-based-on-a-sample-16577

#### Solution Preview

Please see the attached file for details.

--------------------------------------------------------------------------

ect: normal distribution & standard deviation
Details: A normal distribution X has a mean of 50 and a standard deviation of 4.
a. compute the probability of a value between 44.0 and 55.0
Solution. Using the following website, you can compute these probabilities.
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html

P(44.0<=X<=55.0)=0.827541

b. compute the probability of a value greater than 55.0
Solution. Using the following website, you can compute these probabilities.
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html

...

#### Solution Summary

The solution contains detailed explanation of finding probabilities based on a sample from a normal distribution.

\$2.49