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For a normal distribution of scores, what proportion of scores would fall:

a. Above a Z score of 1.00?

b. Above a Z score of 2.00?

c. Above a Z score of 3.00?

d. Below a Z score of –2.00?

e. Between the limits of –1.00 and 1.00?

User MHSaffari
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1 Answer

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Final answer:

For a normal distribution of scores, the proportion of scores falling above or below specific Z scores can be determined using a Z-table or a statistical software. The proportions are: a. About 68 percent, b. About 95 percent, c. About 99.7 percent, d. Approximately 0.0228, and e. Approximately 0.6826.

Step-by-step explanation:

a. About 68 percent of the values lie between z-scores of -1 and 1.

b. About 95 percent of the values lie between z-scores of -2 and 2.

c. About 99.7 percent of the values lie between z-scores of -3 and 3.

d. To find the proportion of scores below a Z score of -2.00, you would need to find the proportion above that z-score and subtract it from 1. Since the area under the normal curve sums up to 1, subtracting the proportion above -2.00 from 1 would give you the proportion below -2.00. By using a Z-table or a statistical software, you can find that the proportion above a Z score of -2.00 is about 0.9772. Therefore, the proportion below a Z score of -2.00 is approximately 1 - 0.9772 = 0.0228.

e. To find the proportion of scores between Z scores of -1.00 and 1.00, you would need to find the proportion above a Z score of -1.00 and subtract it from the proportion above a Z score of 1.00. By using a Z-table or a statistical software, you can find the proportion above a Z score of -1.00 is about 0.8413 and the proportion above a Z score of 1.00 is about 0.1587. Therefore, the proportion between Z scores of -1.00 and 1.00 is approximately 0.8413 - 0.1587 = 0.6826.

User Madhav Datt
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