Three sigma rule

The height of trees in a given stand is known to be a quantity with a normal probability distribution with a mean value of 15 m and a variance of 5 m². Determine the interval in which there will be tree heights in such a stand with a probability of 90%

Final Answer:

x₁ =  11.334
x₂ =  18.666

Step-by-step explanation:

$$\begin{gathered} m=15\text{m} \\ \sigma =\sqrt{5}=\sqrt{5}\text{m}\doteq 2.2361\text{m} \\ p=90\%=\frac{90}{100}=\frac{9}{10}=0.9 \\ s=3.666 \\ {x}_1=m-s=15-3.666=11.334 \end{gathered}$$

$x_2=m+s=15+3.666=18.666$

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