SOLUTION: A small computing center has found that the number of jobs submitted per day to its computers has a distribution that is approximately mound-shaped and symmetric, with a mean of 7

Algebra ->  Probability-and-statistics -> SOLUTION: A small computing center has found that the number of jobs submitted per day to its computers has a distribution that is approximately mound-shaped and symmetric, with a mean of 7      Log On


   

Question 163008: A small computing center has found that the number of jobs submitted per day to its computers has a distribution that is approximately mound-shaped and symmetric, with a mean of 72 jobs and a standard deviation of 6.
Where do we expect approximately 95% of the distribution to fall?
(Compute the interval that will contain 95% of the data).
Z= (0.95-72)/6= -71.05/6= -11.86 ? (I wasn’t sure about this one or it might be 6 because the distribution is known)
Is this correct or how do I work it out- I would greatly appreciate the help- thanks
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's denote the confidence interval as (L,U) where "L" is the lower limit of the confidence interval and "U" is the upper limit of the confidence interval

It's very helpful to remember that within 2 standard deviations of the mean lies 95% of the population. So this means that z=2 (see below)


The simplest way to produce a confidence interval is to use the formulas L=mu-z%2Asigma and U=mu%2Bz%2Asigma where mu is the mean and sigma is the standard deviation, and "z" is the number of standard deviations away from the mean (ie the z score).


Note: there are other ways to calculate the confidence interval


So L=72-2%2A6=72-12=60 and U=72%2B2%2A6=72%2B12=84 which means that our confidence interval is (60,84)