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%