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2. We measure 42 one-year old hatchery-raised trout and
find a mean length of 173mm with standard deviation of 27
mm. The distribution is fairly symmetric with no outliers.
A. We want to find a central interval which contains half
the lengths. What distribution (Normal or t) should we use?
If t, what degrees of freedom?
B. Using that distribution and the web app, what two fish
lengths have the middle half of the data values between
them?
C. What length cuts off the longest 10% of fish?
D. A fish of length 200 mm is at what percentile of the
distribution?
Solution
a)
we will use Z distribution ( Normal)
b)
P( z > z ) = 0.50
z = 0
0 = x - 173 / (27 / srqt(2) )
x = 173
c)
P( z > z ) = 0.10
z = 1.28
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1.28 = x - 173 / 27
x = 207.56
d)
z = 200 - 173 / 27
z = 1
percentile 0.1587 = 15.87%

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2. We measure 42 one-year old hatchery-raised trout and find a mean length of 173mm with standard deviation of 27 mm. The distribution is fairly symmetric with no outliers. A. We want to find a “central” interval which contains half the lengths. What distribution (Normal or t) should we use? If t, what degrees of freedom? B. Using that distribution and the web app, what two fish lengths have the middle half of the data values between them? C. What length cuts off the longest 10% of fish? D. A fish of length 200 mm is at what percentile of the distribution? Solution a) we will use Z distribution ( Normal) b) P( z > z ) = 0.50 z=0 0 = x - 173 / (27 / srqt(2) ) x = 173 c) P( z > z ) = 0.10 z = 1.28 1.28 = x - 173 / 27 x = 207.56 d) z = 200 - 173 / 27 z=1 percentile 0.1587 = 15.87% Name: Description: ...
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