Question 1137508
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Part A


mu = 3 = mean
sigma = 0.5 = standard deviation


Convert the raw score x = 2.3 to its corresponding z score
z = (x-mu)/sigma
z = (2.3-3)/0.5
z = -1.40


Use <a href = "http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf">this table</a> (or similar) to find that P(Z < -1.40) = 0.0808
To find this result, turn to page 1 of that PDF, go to the row that starts with -1.4 and the column that has 0.00 at the top. The row and column intersect at 0.0808
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The probability is approximately 0.0808
which is roughly 8.08% in percent form
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Part B


mu = 800
sigma = 40


Convert the raw score x = 778 to its corresponding z score
z = (x-mu)/sigma
z = (778-800)/40
z = -0.55


Repeat for x = 834
z = (x-mu)/sigma
z = (834-800)/40
z = 0.85


So, P(778 < X < 834) is the same as P(-0.55 < Z < 0.85)


We will use this formula here
P(A < Z < B) = P(Z < B) - P(Z < A)
to help compute the area we want between the two z scores


A = -0.55
B = 0.85
P(A < Z < B) = P(Z < B) - P(Z < A)
P(-0.55 < Z < 0.85) = P(Z < 0.85) - P(Z < -0.55)
P(-0.55 < Z < 0.85) = 0.8023 - 0.2912 <font color=blue>use the same <a href = "http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf">table</a> as done in part A</font>
P(-0.55 < Z < 0.85) = 0.5111



The approximate probability is 0.5111
which in percent form would be roughly 51.11%


<a href = "http://davidmlane.com/hyperstat/z_table.html">Here</a> is a handy calculator to help check your work

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