Question

The diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard devia- tion of 1.5 inches. a. What proportion of the trees will have diameters between 3 and 5 inches? b. What proportion of the trees will have diameters less than 3 inches? c. Your Christmas tree stand will expand to a diameter of 6 inches. What proportion of the trees will not fit in your Christmas tree stand?

Short answer

Answer: Approximately 49.72% of trees have diameters between 3 and 5 inches, 25.14% have diameters less than 3 inches, and 9.18% have diameters greater than 6 inches.

Step-by-step solution

Identify the given information

We know that the diameters of the trees are normally distributed with a mean (μ) of 4 inches and a standard deviation (σ) of 1.5 inches.

Calculate the z-scores

In each part of the problem, we need to calculate the z-scores corresponding to the given diameters using the formula z = (x - μ) / σ: a) x = 3 and x = 5: Calculate the z-scores for these diameters. b) x = 3: Calculate the z-score for this diameter. c) x = 6: Calculate the z-score for this diameter.

Find the proportions from the standard normal distribution table

Use the calculated z-scores to find the corresponding proportions from the standard normal distribution table. a) Find the proportion of trees with diameters between z-scores calculated for x = 3 and x = 5. b) Find the proportion of trees with diameters less than the z-score calculated for x = 3. c) Find the proportion of trees with diameters greater than the z-score calculated for x = 6.

Calculate the z-scores

a) z1 = (3 - 4) / 1.5 = -0.67 z2 = (5 - 4) / 1.5 = 0.67 b) z3 = -0.67 (since it's the same as the lower bound in part a) c) z4 = (6 - 4) / 1.5 = 1.33

Find the proportions from the standard normal distribution table

a) Proportion of trees with diameters between 3 and 5 inches: P(-0.67 < z < 0.67) = P(z < 0.67) - P(z < -0.67) ≈ 0.7486 - 0.2514 = 0.4972 b) Proportion of trees with diameters less than 3 inches: P(z < -0.67) ≈ 0.2514 c) Proportion of trees with diameters greater than 6 inches (not fitting in the stand): P(z > 1.33) = 1 - P(z < 1.33) ≈ 1 - 0.9082 = 0.0918 So, the proportions of trees are: a) Approximately 49.72% of trees have diameters between 3 and 5 inches. b) Approximately 25.14% of trees have diameters less than 3 inches. c) Approximately 9.18% of trees have diameters greater than 6 inches and will not fit in the Christmas tree stand.

Key concepts

Standard Deviation

When analyzing the variation of data, standard deviation is a key statistical measure that helps us understand how spread out the numbers are around the mean. In the context of Douglas firs grown at a Christmas tree farm, let's think of the diameter of trees as our data set.The standard deviation indicates the average distance of each tree's diameter from the mean diameter. A smaller standard deviation means that most trees have a diameter close to the mean, resulting in uniformly-sized trees. Conversely, a larger standard deviation points to greater variety in tree sizes. With a mean diameter of 4 inches and a standard deviation of 1.5 inches, we can infer that while many trees will be around 4 inches, we can expect quite a few that are as small as 2.5 inches or as large as 5.5 inches.An important aspect of understanding and using the standard deviation is that it allows us to calculate probabilities for normally distributed data, particularly when used in conjunction with z-scores, which we'll explore in the next section.

Z-Score Calculation

Moving deeper into the realm of statistics, the z-score calculation is a method used to determine how many standard deviations an element is from the mean. This calculation is especially useful when dealing with normally distributed data, such as the diameters of our Christmas trees.To calculate a z-score, we use the formula \[ z = \frac{(x - \mu)}{\sigma} \]where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For our Douglas firs, if we want to find the z-score for a tree with a 3-inch diameter, we substitute \( x = 3 \), \( \mu = 4 \), and \( \sigma = 1.5 \) into the formula to get \[ z_1 = \frac{(3 - 4)}{1.5} = -0.67 \].The negative z-score tells us that a 3-inch tree is below the mean diameter. The z-score is a dimensionless quantity that we can use to compare across different normal distributions or to find probabilities using the standard normal distribution table, which we will discuss in the final section. The z-score is key to converting individual data points into standardized scores that can be easily compared and interpreted.

Standard Normal Distribution Table

To make practical use of our calculated z-scores, we turn to the standard normal distribution table. This table, also known as the z-table, provides the probability that a standard normal random variable will be less than a given z-score.For example, if we look at the z-score of -0.67, the table will show the proportion of data below this score. It's based on the standard normal distribution, which has a mean of 0 and a standard deviation of 1. Any normally distributed dataset, like our tree diameters, can be compared to this standardized form.Using the table, we find these probabilities for the given z-scores: \[ P(-0.67 < z < 0.67) = P(z < 0.67) - P(z < -0.67) \]By referring to the table, we can deduce that the proportion of trees between 3 and 5 inches in diameter is roughly 49.72%. The table is essentially the toolkit for interpreting z-scores, turning them into understandable probabilities. Without it, z-scores would merely indicate relative positioning, but with the table, we unlock the potential to make data-driven predictions and decisions about our Christmas tree farm.