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Chapter 5: Problem 39

A forester measured 27 of the trees in a large woods that is up for sale. He found a mean diameter of 10.4 inches and a standard deviation of 4.7 inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies. a. Draw the Normal model for tree diameters. b. What size would you expect the central \(95 \%\) of all trees to be? c. About what percent of the trees should be less than an inch in diameter? d. About what percent of the trees should be between 5.8 and 10.4 inches in diameter? e. About what percent of the trees should be over 15 inches in diameter?

Short Answer

Expert verified
a. The normal model would have mean at \(10.4\) and standard deviation of \(4.7\).\nb. The central \(95\%\) of all trees would be between approximately \(1\) and \(20\) inches.\nc. Almost \(0\%\) of the trees should have a diameter of less than \(1\) inch.\nd. About \(34\%\) of the trees should be between \(5.7\) and \(10.4\) inches.\ne. About \(16\%\) of trees should be over \(15\) inches.

Step by step solution

01

Sketching the Normal Model

The Normal Model will be a bell-shaped curve with the vertical axis signifying relative frequency and the horizontal axis representing our tree diameters. At the center of the curve, label the value \(10.4\) (the mean). Label one standard deviation above the mean (mean+std.dev) as \(15.1\) and one below (mean-std.dev) as \(5.7\).
02

Determining central 95% of all tree sizes

When we talk about 'central \(95\%'\), we're referring to the data within two standard deviations either side of the mean. That will range from \(10.4-2*4.7\) to \(10.4+2*4.7\), which is approximately between \(1\) and \(20\) inches in diameter.
03

Calculating percentage of trees less than an inch in diameter

For trees of less than \(1\) inch in diameter, this falls below \(2\) standard deviations below the mean. In a normal distribution, the percentage of data beyond \(2\) standard deviations is roughly \(5 \%'\). Because we're considering the left tail of the distribution, it's about \(2.5 \%'\). So, we can say it is very unlikely and almost \(0 \%'\) of the trees will have less than \(1\) inch in diameter.
04

Calculate percentage of trees between 5.7 and 10.4 inches

This range covers all the trees from one standard deviation below the mean to the mean itself. In a Normal distribution, \(68 \%'\) of the data will fall within one standard deviation from the mean, and half of that (i.e. \(34 \%'\)) falls on either side of the mean. Consequently, about \(34 \%'\) of the trees should be between \(5.7\) and \(10.4\) inches in diameter.
05

Calculate percentage of trees above 15 inches

Trees with a diameter over \(15\) inches fall to the right of one standard deviation above the mean. Hence about \(16 \%'\) of the trees (half of \(32 \%'\) which lies outside the \(68 \%'\) from the mean in normal distribution) should have a diameter greater than \(15\) inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
The standard deviation is a key concept in understanding the spread of data points around the mean, often referred to as the 'average'. In the context of a normal distribution, it helps in determining how much variation exists from the mean value.
  • A smaller standard deviation means the data points tend to be closer to the mean.
  • A larger standard deviation indicates a wider spread of data.

In our exercise, the standard deviation is 4.7 inches. This means that, on average, the diameter of the trees varies by 4.7 inches from the mean (10.4 inches). This information is crucial when predicting the range within which most tree diameters will fall. By using the standard deviation, we can easily calculate the range for central percentages of data, such as the 68%, 95%, and 99.7%, which represent 1, 2, and 3 standard deviations from the mean, respectively.
Mean
The mean, also known as the average, is calculated by summing up all data points and dividing by the number of points. It provides a central value that represents the entire dataset.
  • In the exercise, the mean tree diameter is 10.4 inches.

This value is at the center of the normal distribution curve and serves as a reference point for measuring the spread of the data through standard deviation. Understanding the mean is fundamental, as it helps you interpret the overall level of your data and how other measures like standard deviation, can describe the variability around this central point.
Z-Score
A z-score measures how many standard deviations a data point is from the mean. It is a standardized way to compare individual data points across different datasets or relative to the mean in a single dataset.
  • A positive z-score indicates the data point is above the mean.
  • A negative z-score indicates it is below the mean.

For example, if a tree has a diameter of 15.1 inches, its z-score can be calculated as follows: \( z = \frac{(15.1 - 10.4)}{4.7} \) This calculation will show how many standard deviations 15.1 inches is above the average diameter. Z-scores are vital in assessing the probability of data points appearing within a certain range of the distribution.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics, describing how the mean of a sample group will have a normal distribution, even if the original dataset is not normally distributed, provided the sample size is sufficiently large.
  • The CLT allows predictions and conclusions about population parameters.
  • It is useful for calculating probabilities and constructing confidence intervals.

In the context of our exercise, the CLT justifies assuming a normal distribution for tree diameters from just a sample of 27 trees. This means that even if the full population of trees in the woods doesn鈥檛 perfectly align with a normal distribution, our sample mean of 10.4 inches will approximate this normality, so long as we analyze a sufficiently large and representative sample.

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Most popular questions from this chapter

A company's customer service hotline handles many calls relating to orders, refunds, and other issues. The company's records indicate that the median length of calls to the hotline is 4.4 minutes with an IQR of 2.3 minutes. a. If the company were to describe the duration of these calls in seconds instead of minutes, what would the median and IQR be? b. In an effort to speed up the customer service process, the company decides to streamline the series of pushbutton menus customers must navigate, cutting the time by 24 seconds. What will the median and IQR of the length of hotline calls become?

Most people think that the "normal" adult body temperature is \(98.6 \circ \mathrm{F}\) That figure, based on a 19th-century study, has recently been challenged. In a 1992 article in the Journal of the American Medical Association, researchers reported that a more accurate figure may be \(98.2 \circ \mathrm{F}\). Furthermore, the standard deviation appeared to be around \(0.7 \circ \mathrm{F}\). Assume that a Normal model is appropriate. a. In what interval would you expect most people's body temperatures to be? Explain. b. What fraction of people would be expected to have body temperatures above \(98.6 \circ \mathrm{F}\) c. Below what body temperature are the coolest \(20 \%\) of all people?

A tire manufacturer believes that the treadlife of its snow tires can be described by a Normal model with a mean of 32,000 miles and standard deviation of 2500 miles. a. If you buy one of these tires, would it be reasonable for you to hope it will last 40,000 miles? Explain. b. Approximately what fraction of these tires can be expected to last less than 30,000 miles? c. Approximately what fraction of these tires can be expected to last between 30,000 and 35,000 miles? d. Estimate the IQR of the treadlives. e. In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If the dealer is willing to give refunds to no more than 1 of every 25 customers, for what mileage can he guarantee these tires to last?

Suppose your statistics professor reports test grades as \(z\) -scores, and you got a score of 2.20 on an exam. a. Write a sentence explaining what that means. b. Your friend got a z-score of -1 . If the grades satisfy the Nearly Normal Condition, about what percent of the class scored lower than your friend?

Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of \(6.2 \mathrm{mpg}\) for highway driving. Assume that a Normal model can be applied. a. Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b. In what interval would you expect the central \(68 \%\) of autos to be found? c. About what percent of autos should get more than 31 mpg? d. About what percent of cars should get between 31 and \(37.2 \mathrm{mpg} ?\) e. Describe the gas mileage of the worst \(2.5 \%\) of all cars.
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