Question

About _____% of the area under the curve of the standard normal distribution is outside the interval z=[−0.53,0.53] (or beyond 0.53 standard deviations of the mean).

Answer 1

About 59.62%

Step-by-step explanation:

Looking at the z-score table:

We can see that the z-score for |0.53| is 0.2019

Since we want the area outside the interval [-0.53, 0.53]:

`Area\text{ }outside=1-area\text{ }inside`

The area inside is twice the z-score for |0.53|: 2 · 0.2019 = 0.4038

Thus:

`Area\text{ }outside=1-0.4038=0.5962`

To percentage, we multiply by 100:

`0.5962=59.62\%`