SOLUTION: find the percent of the total area under the standard curve between the given z-scores. z = 2.18 and z = 3.45
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-> SOLUTION: find the percent of the total area under the standard curve between the given z-scores. z = 2.18 and z = 3.45
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Question 165391: find the percent of the total area under the standard curve between the given z-scores. z = 2.18 and z = 3.45 Found 2 solutions by gonzo, Fombitz:Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! between z = 2.18 and z = 3.45 area under the curve is .014348
there's an online calculator that helps you to solve problems like this.
it is at the following internet address:
http://davidmlane.com/hyperstat/z_table.html
check it out.
the normal curve has a mean of 0 and a standard deviation of 1.
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here's a definition of z score that might be helpful.
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Z scores are a special application of the transformation rules. The z score for an item, indicates how far and in what direction, that item deviates from its distribution's mean, expressed in units of its distribution's standard deviation. The mathematics of the z score transformation are such that if every item in a distribution is converted to its z score, the transformed scores will necessarily have a mean of zero and a standard deviation of one.
Z scores are sometimes called "standard scores". The z score transformation is especially useful when seeking to compare the relative standings of items from distributions with different means and/or different standard deviations.
Z scores are especially informative when the distribution to which they refer, is normal. In every normal distribution, the distance between the mean and a given Z score cuts off a fixed proportion of the total area under the curve. Statisticians have provided us with tables indicating the value of these proportions for each possible Z score.
You can put this solution on YOUR website! We can find the area under the curve from to and also from to , subtract them and that will give us the area between and . We then multiply by 100 to get the percentage. I used EXCEL's NORMSDIST function with the given z scores to calculate areas.
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1.43% of the curve's area is between z=2.18 and z=3.45.
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