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:
Rounding it off to the nearest whole.
3) Thus, The answers are:
i.84%
ii. 2857 trees