Mathematics, 19.02.2021 03:10, Jwhite8602

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.

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answered: emilylittle2442

a) The boundaries are $Z = \pm 0.253$

b) The boundaries are $Z = \pm 0.675$ .

c) The boundaries are $Z = \pm 1.96$ .

d) The boundaries are $Z = \pm 2.575$ .

Step-by-step explanation:

Z-score:

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 $Z = \pm 0.253$ .

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 $Z = \pm 0.675$ .

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 $Z = \pm 1.96$ .

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 $Z = \pm 2.575$ .

answered: Guest

hi there!

the answer will be d: 2

answered: Guest

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent

Which statement best describes the association between variable x and variable y? perfect negative

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Find the z-score boundaries that separate a normal distribution as described in each of the followin...