Question

Find the -scores that bound the middle 96 of the area under the standard normal curve by using the TI-84 calculator.

Answer 1

Answer:

$\alpha=1-0.96 =0.04$

And $\alpha/2= 0.02$ and if we use the following function on the Ti84 plus we got:

invNorm(0.02,0,1)

invNorm(1-0.02,0,1)

And the values with the middle 96% of the values are:

$z_{\alpha/2}= \pm 2.054$

Step-by-step explanation:

For this case we want to find the limits with the middle 96% of the area below the normal curve, then the significance level would be:

$\alpha=1-0.96 =0.04$

And $\alpha/2= 0.02$ and if we use the following function on the Ti84 plus we got:

invNorm(0.02,0,1)

invNorm(1-0.02,0,1)

And the values with the middle 96% of the values are:

$z_{\alpha/2}= \pm 2.054$