Question 1038752
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You'll need to calculate two z scores.


For the raw score of x = 63, the z score is...


z = (x - mu)/sigma
z = (x - 76)/10
z = (63 - 76)/10
z = -13/10
z = -1.30


For the raw score of x = 90, the z score is...


z = (x - mu)/sigma
z = (x - 76)/10
z = (90 - 76)/10
z = 14/10
z = 1.4


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So we need to find the area under the standard normal bell curve (Z distribution curve) between z = -1.3 and z = 1.4


First find the area to the left of z = -1.3


Use this <a href = "http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf">table</a> locate the row that starts with "-1.3" (page 1) and the column that has 0.00 at the top. The row and column intersect at the value 0.0968


So this means P(Z < -1.3) = 0.0968
Basically, the area to the left of z = -1.3 is 0.0968. Let's make M = 0.0968. We'll use this value of M later.


Now use the table again to find that
P(Z < 1.4) = 0.9192
this value of 0.9192 is found by looking at the row that starts with 1.4 (page 2) and has 0.00 at the top of the column.
Let's make N = 0.9192


Now subtract the values of N and M (big minus small)
N - M = 0.9192 - 0.0968 = 0.8224



So the area under the curve between z = -1.3 and z = 1.4 is <font color=red size=5>0.8224</font>


This means the proportion of those between 63 dollars and 90 dollars is <font color=red size=5>0.8224</font> (which is equivalent to roughly 82.24%). This proportion is approximate because the table values are approximate.



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