164k views
1 vote
A forest products company claims that the amount of usable lumber in its harvested trees averages142 cubic feet and has a standard deviation of 9 cubic feet. Assume that these amounts haveapproximately a normal distribution.1. What percent of the trees contain between 133 and 169 cubic feet of lumber? Round to twodecimal places.II. If 18,000 trees are usable, how many trees yield more than 151 cubic feet of lumber?

1 Answer

2 votes


\begin{gathered} I)84\% \\ II)2857 \end{gathered}

1) Considering that the amount of lumber in this Data Set has been normally distributed, then we can start by finding this Percentage (or probability in this interval, writing out the following expressions:


\begin{gathered} P(133Now we can replace it with the Z score formula, plugging into that the Mean, the Standard Deviation, and the given values:<p></p>[tex]Z=(X-\mu)/(\sigma)

Then:


\begin{gathered} P((133-142)/(9)<(X-\mu)/(\sigma)<(169-142)/(9)) \\ P(-1Checking a Z-score table we can state that the Percentage of the trees between 133 and 169 ft³ is:<p></p>[tex]P(-1<strong>2) </strong>Now, let's check for the second part, the number of trees. But before that, let's use the same process to get a percentage that fits into that:<p></p>[tex]\begin{gathered} P(X>151)=(151-142)/(9)=1 \\ P(Z>1)=0.1587 \end{gathered}

Note that 0.1587 is the same as 15.87%. Multiplying that by the total number of trees we have:


18000*0.1587=2856.6\approx2857

Rounding it off to the nearest whole.

3) Thus, The answers are:

i.84%

ii. 2857 trees

User Ahmed
by
7.4k points