Find the z-score boundaries that separate a normal distribution as described in each of the following. a. The middle 20% from the 80% in the tails. b. The middle 50% from the 50% in the tails. c. The middle 95% from the 5% in the tails. d. The middle 99% from the 1% in the tails.

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

a. The middle 20% from the 80% in the tails.

The middle 20% is between the 50 - (20/2) = 40th percentile and the 50 + (20/2) = 60th percentile:

40th percentile: Z has a pvalue of 0.4, so Z = -0.253.

60th percentile: Z has a pvalue of 0.6, so Z = 0.253.

The boundaries are [tex]Z = \pm 0.253[/tex].

b. The middle 50% from the 50% in the tails.

The middle 50% is between the 50 - (50/2) = 25th percentile and the 50 + (50/2) = 75th percentile:

25th percentile: Z has a pvalue of 0.25, so Z = -0.675.

75th percentile: Z has a pvalue of 0.75, so Z = 0.675.

The boundaries are [tex]Z = \pm 0.675[/tex].

c. The middle 95% from the 5% in the tails.

The middle 95% is between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile:

2.5th percentile: Z has a pvalue of 0.025, so Z = -1.96.

97.5th percentile: Z has a pvalue of 0.975, so Z = 1.96.

The boundaries are [tex]Z = \pm 1.96[/tex].

d. The middle 99% from the 1% in the tails.

The middle 99% is between the 50 - (99/2) = 0.5th percentile and the 50 + (99/2) = 99.5th percentile:

0.5th percentile: Z has a pvalue of 0.005, so Z = -2.575.

99.5th percentile: Z has a pvalue of 0.995, so Z = 2.575.

The boundaries are [tex]Z = \pm 2.575[/tex].

Lanuel Lanuel · Answer 2 · 2022-03-15T17:12:35+01:00

The z-score boundaries, Z for a middle 20% from the 80% in the tails = [tex]\pm 0.253[/tex].
The z-score boundaries, Z for a middle 50% from the 50% = [tex]\pm 0.675[/tex].
The z-score boundaries, Z for a middle 95% from the 5% = [tex]\pm 1.96[/tex].
The z-score boundaries, Z for a middle 99% from the 1% = [tex]\pm 2.575[/tex].

What is a z-core?

A z-core is also referred to as a standard score and it can be defined as a measure of the distance between a data point (raw score) and the mean, when standard deviation units are used.

In Statistics, z-scores can either be positive or negative and it is calculated by using this formula:

[tex]Z=\frac{x\;-\;u}{\delta}[/tex]

Where:

x is the observed data.

u is the mean.

[tex]\delta[/tex] is the standard deviation.

a. A middle 20% from the 80% in the tails.

Note: The middle 20% is between the 40th percentile and 60th percentile.

40th percentile (p-value = 0.4) and Z = -0.253.

60th percentile (p-value = 0.6) and Z = 0.253.

Z = [tex]\pm 0.253[/tex]

b. A middle 50% from the 50% in the tails.

Note: The middle 50% is between the 25th percentile and 75th percentile.

25th percentile (p-value = 0.25) and Z = -0.675.

75th percentile (p-value = 0.75) and Z = 0.675.

Z = [tex]\pm 0.675[/tex]

c. A middle 95% from the 5% in the tails.

Note: The middle 95% is between the 2.5th percentile and 97.5th percentile.

2.5th percentile (p-value = 0.025) and Z = -1.96.

97.5th percentile (p-value = 0.975) and Z = 1.96.

Z = [tex]\pm 1.96[/tex]

d. A middle 99% from the 1% in the tails.

Note: The middle 99% is between the 0.5th percentile and 99.5th percentile.

0.5th percentile (p-value = 0.025) and Z = -2.575.

99.5th percentile (p-value = 0.975) and Z = 2.575.

Z = [tex]\pm 2.575[/tex]

Read more on z-scores here: https://brainly.com/question/4302527