What is the confidence level (written as a percentage) for the interval x overbar minus 2.14 sigma divided by startroot n endroot less-than-or-equal-to mu less-than-or-equal-to x overbar plus 2.14 sigma divided by startroot n endroot?

The interval notation is: m +/- z · (σ / √n)
where: 
m = mean
z = z-score
σ = standard deviation
n = sample 

In your case, you have: x +/- 2.14 · (σ / √n)

Comparing the two formulae, you get z = +/-2.14

Therefore, you are looking for P(-2.14 < z < +2.14) = P(z < 2.14) - P(z < -2.14)

Now, look at a z-table to find that 
P(z < 2.14) = 0.9838    and that 
P(z < -2.14) = 0.0162

Therefore:
P(-2.14 < z < +2.14) = 0.9838 - 0.0162 = 0.9676

Hence, the confidence level written as a percentage is 96.76%

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