Let $X$ represent the full height of a certain species of tree. Assume that $X$ has a normal probability distribution with a mean of $118.2 \mathrm{ft}$ and a standard deviation of $32.1 \mathrm{ft}$. Find the probability that the height of a randomly selected tree is between $34.7 \mathrm{ft}$ and $89.3 \mathrm{ft}$. Round your answer to at least three decimals. In the box below, type in what you entered into the calculator to get your answers for each question. If you used the online Statistics Calcuator, indicate the screens used, and any values of input variables.

Step 1 ：Let's denote the full height of a certain species of tree as \(X\). We are given that \(X\) follows a normal distribution with a mean (\(\mu\)) of 118.2 ft and a standard deviation (\(\sigma\)) of 32.1 ft.

Step 2 ：We are asked to find the probability that the height of a randomly selected tree is between 34.7 ft and 89.3 ft. Since the distribution is normal, we can use the z-score formula to standardize these values and then use the standard normal distribution to find the probability.

Step 3 ：The z-score formula is: \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：First, we calculate the z-scores for 34.7 ft and 89.3 ft. For 34.7 ft, the z-score (\(z1\)) is \(-2.601\), and for 89.3 ft, the z-score (\(z2\)) is \(-0.900\).

Step 5 ：Next, we find the area under the standard normal curve between these two z-scores. This area represents the probability we're looking for. The probability is approximately 0.179.

Step 6 ：Final Answer: The probability that the height of a randomly selected tree is between 34.7 ft and 89.3 ft is approximately \(\boxed{0.179}\).