Question

Find the z-scores (two of them) that separate the middle 48% of the distribution from the area in the tails of the standard normal distribution.

Answer 1

Answer:

The  z-scores are   $Z_{\frac{\alpha }{2} } = 0643$ and  $Z_{\frac{\alpha }{2} } = - 0643$

Step-by-step explanation:

From the question we are told that

    The  middle of the distribution area considered is  48%

Now the rest of the area under the standard normal distribution area is  mathematically evaluated

      $\alpha = 100- 48$

      $\alpha = 52 \%$

      $\alpha = 0.52$

Now to obtain the area occupied by this proportion at the two tails we divide by 2 i.e

    $\frac{\alpha }{2} = \frac{0.52}{2} =0.26$

Now the z-score is obtained from the normal distribution table and the value is

     $Z_{\frac{\alpha }{2} } \pm 0.643$

The $\pm$ shows that we are considering the two-talil