Question 1147305: I have a homework question that I am having trouble with understanding:
Using the unit normal table, find the proportion under the standard normal curve that lies between the following values.
(a) the mean and z= 0
(b) the mean and z= 1.96
(c) z= -1.50 and z= 1.50
(d) z= -0.08 and z= -0.50
(e) z= 1.00 and z= 2.00
.........
If someone could please explain in a step by step way to solve this problem, I have already read the chapters and I am lost.
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! If you can, use a table of positive and negative z-values which lists the associated probabilities for a Normal Distribution, if you have a table with only positive z-values, you can subtract the corresponding probability from 1 and get the negative z-score probability
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For a unit normal table, z = 1 corresponds to a proportion of 0.3413 from the mean, z = 2 corresponds to a proportion of 0.4772 from the mean and z = 0.5 corresponds to a proportion of 0.19.15 from the mean
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(a) mean is 50% = 0.50 proportion(probability), a z=0 corresponds to 0.50 and 0.50 - 0.50 = 0
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(b) z = 1.96, then probability(P) = 0.9750, 0.9750 - 0.50 = 0.4750
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(c) z = -1.50 then P = 0.0668, z = 1.50 then P = 0.9332 and 0.9332 - 0.0668 = 0.8664
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(d) z = -0.08 then P = 0.4681, z = -0.50 then P = 0.3085 and 0.4681 - 0.3085 = 0.1596
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(e) z = 1.00 then P = 0.8413, z = 2 then P = 0.9772 and 0.9772 - 0.8413 = 0.1359
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Note mean of a Normal Distribution means 50% of the values are above the mean and 50% of the values are below the mean
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Note z-values correspond to standard deviations above(positive) and below(negative) from the mean
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Note z-value = (X - mean)/standard deviation, where X is the test value, for example, suppose you want to compare someone’s weight(150 pounds) to the “average” of a population, the z-score can tell you where that person’s weight is compared to the average population’s mean weight.
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Answer by Edwin McCravy(20060)  (Show Source):
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