A forester measured the trees in one part of the forest. She found the mean diam

Question

A forester measured the trees in one part of the forest. She found the mean diameter to be 10.1 inches and a standard deviation of 3.2 inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies. A) What sizes would you expect the middle 95% of all trees to be between? (Round to 1 decimal place) to inches in diameter B) What sizes would you expect the middle 80% of all trees to be between? (Round to 1 decimal place) to inches in diameter C) What proportion of trees are less than 6.1 inchcs in diameter? (Give your answer to 4 decimal places) D) The top 15% of trees arc larger than what diameter? (Round to 1 decimal plaoc) inchcs in diameter

Explanation / Answer

a)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.95      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.025      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.959963985      
By symmetry,          
z2 =    1.959963985      
          
As          
          
u = mean =    10.1      
s = standard deviation =    3.2      
          
Then          
          
x1 = u + z1*s =    3.828115249   [ANSWER]  
x2 = u + z2*s =    16.37188475   [ANSWER]

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b)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.8      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.1      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.281551566      
By symmetry,          
z2 =    1.281551566      
          
As          
          
u = mean =    10.1      
s = standard deviation =    3.2      
          
Then          
          
x1 = u + z1*s =    5.99903499   [ANSWER]  
x2 = u + z2*s =    14.20096501   [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    6.1      
u = mean =    10.1      
          
s = standard deviation =    3.2      
          
Thus,          
          
z = (x - u) / s =    -1.25      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.25   ) =    0.105649774 [ANSWER]

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d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1-0.15 =   0.85      
          
Then, using table or technology,          
          
z =    1.036433389      
          
As x = u + z * s,          
          
where          
          
u = mean =    10.1      
z = the critical z score =    1.036433389      
s = standard deviation =    3.2      
          
Then          
          
x = critical value =    13.41658685   [ANSWER]  

  
  

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