where do we expect a number of jobs to fall most of the time

Hi there

I think this question relates to z-score, if I'm not mistaken.

Where

\(z=\frac{x-μ}{σ}\)

\(0.954=\frac{x-65}{7}\)

\(0.954\left(7\right)=x-65\)

\(0.954\left(7\right)+65=x\)

\(0.954\left(7\right)+65=71.678\)

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 \(95.4 %\) of the population. So this means that \(z=2\) (see below)

The simplest way to produce a confidence interval is to use the formulas \(L=μ-zσ\) and \(U=μ + zσ\) where \(μ\) 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 \(L=65-2(7)=51\)  and \(U=65+2(7)=79\)

which means that our confidence interval is (51,79)

My answer is wrong, thanks doubleu