A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees is 19 inches with a standard deviation of 2.3 inches. Assume the distribution of diameters is roughly mound-shaped. (a) What fraction of the trees will have diameters between 14.4 and 25.9 inches? 0.75 Incorrect: Your answer is incorrect. (b) What fraction of the trees will have diameters greater than 21.3 inches?

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boffeemadrid boffeemadrid · Answer 1 · 2019-10-25T07:46:53+02:00

Answer:

97.5 %

15.7 %

Step-by-step explanation:

[tex]\mu[/tex] = Mean = 19

[tex]\sigma[/tex] = Standard deviation = 2.3

a)

[tex]\mu+x\sigma=25.9\\\Rightarrow 19+x\times 2.3=25.9\\\Rightarrow x=\frac{25.9-19}{2.3}\\\Rightarrow x=3[/tex]

[tex]\mu-x\sigma=14.4\\\Rightarrow 19-x\times 2.3=14.4\\\Rightarrow x=\frac{19-14.4}{-2.3}\\\Rightarrow x=2[/tex]

From the empirical rule we get that the percentage of trees between 14.4 and 25.9 inches is 13.6+34.1+34.1+13.6+2.1 = 97.5 %

b)

[tex]\mu+x\sigma=21.3\\\Rightarrow 19+x\times 2.3=21.3\\\Rightarrow x=\frac{21.3-19}{2.3}\\\Rightarrow x=1[/tex]

The fraction of tree under 21.3 inches is 34.1.

Hence the fraction of trees above the diameter of 21.3 inches is 13.6+2.1 = 15.7 %