3.3 - Normal Distributions
Score: 0/33 0/11 answered
Question 11
Let $X$ represent the full height of a certain species of tree. Assume that $X$ has a normal probability distribution with a mean of $232.3 \mathrm{ft}$ and a standard deviation of $26.7 \mathrm{ft}$.
Find the probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$.
Round your answer to three decimal places
\[
P(245.65
Solution
Step 1 ：Let $X$ represent the full height of a certain species of tree. Assume that $X$ has a normal probability distribution with a mean of $232.3 \mathrm{ft}$ and a standard deviation of $26.7 \mathrm{ft}$.
Step 2 ：We are asked to find the probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$.
Step 3 ：We calculate the z-scores for the lower and upper bounds of the range. The z-score is calculated as \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 4 ：For the lower bound of $245.65 \mathrm{ft}$, the z-score is calculated as \(z_{lower} = \frac{245.65 - 232.3}{26.7} = 0.5\).
Step 5 ：For the upper bound of $259 \mathrm{ft}$, the z-score is calculated as \(z_{upper} = \frac{259 - 232.3}{26.7} = 1.0\).
Step 6 ：We then find the probabilities corresponding to these z-scores using the standard normal distribution. The probability for the lower bound is \(P_{lower} = 0.691\) and for the upper bound is \(P_{upper} = 0.841\).
Step 7 ：The probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$ is the difference between these two probabilities, which is \(P_{range} = P_{upper} - P_{lower} = 0.841 - 0.691 = 0.15\).
Step 8 ：Final Answer: The probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$ is \(\boxed{0.150}\).
From Solvely APP
Solution
Step 1 ：Let $X$ represent the full height of a certain species of tree. Assume that $X$ has a normal probability distribution with a mean of $232.3 \mathrm{ft}$ and a standard deviation of $26.7 \mathrm{ft}$.
Step 2 ：We are asked to find the probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$.
Step 3 ：We calculate the z-scores for the lower and upper bounds of the range. The z-score is calculated as \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 4 ：For the lower bound of $245.65 \mathrm{ft}$, the z-score is calculated as \(z_{lower} = \frac{245.65 - 232.3}{26.7} = 0.5\).
Step 5 ：For the upper bound of $259 \mathrm{ft}$, the z-score is calculated as \(z_{upper} = \frac{259 - 232.3}{26.7} = 1.0\).
Step 6 ：We then find the probabilities corresponding to these z-scores using the standard normal distribution. The probability for the lower bound is \(P_{lower} = 0.691\) and for the upper bound is \(P_{upper} = 0.841\).
Step 7 ：The probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$ is the difference between these two probabilities, which is \(P_{range} = P_{upper} - P_{lower} = 0.841 - 0.691 = 0.15\).
Step 8 ：Final Answer: The probability that the height of a randomly selected tree is between $245.65 \mathrm{ft}$ and $259 \mathrm{ft}$ is \(\boxed{0.150}\).
