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Chapter 2: Problem 54

Data collected over several years from college students enrolled in a business statistics class regarding their shoe size shows a roughly bell-shaped distribution, with \(\bar{x}=9.91\) and \(s=2.07\). a. Give an interval within which about \(95 \%\) of the shoe sizes fall. b. Identify the shoe size of a student which is three standard deviations above the mean in this sample. Would this be a rather unusual observation? Why?

Short Answer

Expert verified
a. [5.77, 14.05]; b. 16.12 is unusual, as it is three standard deviations above the mean.

Step by step solution

01

Understanding the Problem

We are dealing with a normal distribution (bell-shaped) with a mean (\(\bar{x}\)) of 9.91 and a standard deviation (\(s\)) of 2.07. Our task is to determine \(95\%\) confidence intervals and the value three standard deviations from the mean.
02

Calculating 95% Confidence Interval

In a normal distribution, \(95\%\) of the data falls within two standard deviations from the mean. To find this interval, we calculate: \(\bar{x} - 2s\) and \(\bar{x} + 2s\). \[\text{Lower bound} = 9.91 - 2 \times 2.07 = 5.77\] \[\text{Upper bound} = 9.91 + 2 \times 2.07 = 14.05\] Thus, the interval is [5.77, 14.05].
03

Finding Three Standard Deviations Above the Mean

To find a value that is three standard deviations above the mean, we use \(\bar{x} + 3s\). \[9.91 + 3 \times 2.07 = 16.12\] A shoe size of 16.12 indicates how large a shoe could be that is an unusual size.
04

Evaluating Unusual Observation

Values further than two standard deviations from the mean are typically considered unusual. Since 16.12 is more than two standard deviations away, it is indeed a rare event and considered unusual within this context.

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Key Concepts

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

Standard Deviation: Understanding the Spread of Data
Standard deviation is a measurement that helps determine how spread out the values in a data set are around the mean. It is essentially the average of the distances from each data point to the mean.
  • The formula to calculate standard deviation is: \[ s = \sqrt{ \frac{1}{N-1} \sum (x_i - \bar{x})^2 } \] where:
    • \(s\) represents the standard deviation
    • \(N\) is the number of observations
    • \(x_i\) represents each individual value
    • \(\bar{x}\) is the mean of the data
This concept is crucial because it tells us about the variability of the data. A smaller standard deviation means data points are relatively close to the mean, while a larger standard deviation indicates more spread.
In the exercise, the standard deviation (0.07) suggests how much most students' shoe sizes deviate from the average size of 9.91, indicating moderate variability. This aids in understanding how diverse shoe sizes are among these students.
Mean: The Central Tendency of Data
The mean, often referred to as the average, is a central value that summarizes a set of data with a single number. It is calculated by summing all the data points and dividing by the number of points.
  • Formula for calculating the mean: \[ \bar{x} = \frac{\sum x_i}{N} \] where:
    • \(\bar{x}\) stands for the mean
    • \(\sum x_i\) is the sum of all data points
    • \(N\) is the total number of data points
The mean is significant as it provides a quick glance at where the center of a data set lies. However, it may not always reflect the most typical value if the data is skewed.
In this shoe size example, the mean of 9.91 represents the typical shoe size among students taking the business statistics class. While this average gives an overall estimation, the standard deviation indicates how the shoe sizes vary around this central point.
Confidence Interval: Estimating the Range of Data
A confidence interval gives an estimated range within which we expect our true parameter, like the mean, to fall. It provides insight into the precision of our mean estimate and helps us understand the variability present in our data.
  • The general method to calculate the 95% confidence interval for a normal distribution is: \[ \bar{x} \pm 2s \] where:
    • \(\bar{x}\) is the mean
    • \(s\) is the standard deviation
    • The factor '2' corresponds to approximately 95% coverage under the normal distribution
Using this computation allows researchers to state with about 95% confidence that the true mean lies within this range.
For the provided exercise, the calculated interval of [5.77, 14.05] suggests that roughly 95% of students鈥� shoe sizes fall within this range. This tool is tremendously helpful when we want to make predictions about a larger group based on sample data, as it considers both mean and variability.

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Most popular questions from this chapter

A recent survey asked students, "On average, how many times in a week do you go to a restaurant for dinner?" Of the 570 respondents, 84 said they do not go out for dinner, 290 said once, 100 said twice, 46 said thrice, 30 said 4 times, 13 said 5 times, 5 said 6 times, and 2 said 7 times. a. Display the data in a table. Explain why the median is \(1 .\) b. Show that the mean is 1.5 . c. Suppose the 84 students who said that they did not go out for dinner had answered 7 times instead. Show that the median would still be 1 . (The mean would increase to 2.54 . The mean uses the numerical values of the observations, not just their ordering.)

Here is the five-number summary for the distribution of the net worth (in billions) of the Top 659 World's Billionaires according to Forbes magazine. $$ \begin{aligned} \text { Minimum } &=2.9, \mathrm{Q} 1=3.6, \text { Median }=4.8 \\ \mathrm{Q} 3 &=8.2, \text { Maximum }=79.2 \end{aligned} $$ a. About what proportion of billionaires have wealth (i) greater than \(\$ 3.6\) billion and (ii) greater than \(\$ 8.2\) billion? b. Between which two values are the middle \(50 \%\) of the observations found? c. Find the interquartile range. Interpret it. d. Based on the summary, do you think that this distribution was bell shaped? If so, why? If not, why not, and what shape would you expect?

According to the U.S. Census Bureau, Current Population Survey, 2015 Annual Social and Economic Supplement, the mean income for males is $$\$ 47,836.10$$ with a standard deviation of $$\$ 58,353.55$$ and the mean income for females is $$\$ 28,466$$ with a standard deviation of $$\$ 36,961.10 .$$ a. Is it appropriate to use the empirical rule for male incomes? Why? b. Compare the center and variability of the income distributions for females and males. c. Which income is relatively higher-a male's income of $$\$ 55,000$$ or a female's income of $$\$ 45,000 ?$$

Sandwiches and protein Listed in the table below are the prices of six-inch Subway sandwiches at a particular franchise and the number of grams of protein contained in each sandwich. $$ \begin{array}{lcc} \hline \text { Sandwich } & \text { Cost(\$) } & \text { Protein(g) } \\ \hline \text { BLT } & \$ 2.99 & 17 \\ \text { Ham (Black Forest, without } & & \\ \text { cheese) } & \$ 2.99 & 18 \\ \text { Oven Roasted Chicken } & \$ 3.49 & 23 \\ \text { Roast Beef } & \$ 3.69 & 26 \\ \text { Subway Club }^{\otimes} & \$ 3.89 & 26 \\ \text { Sweet Onion Chicken Teriyaki } & \$ 3.89 & 26 \\ \text { Turkey Breast } & \$ 3.49 & 18 \\ \text { Turkey Breast \& Ham } & \$ 3.49 & 19 \\ \text { Veggie Delite }^{\mathbb{B}} & \$ 2.49 & 8 \\ \text { Cold Cut Combo } & \$ 2.99 & 21 \\ \text { Tuna } & \$ 3.10 & 21 \\ \hline \end{array} $$ a. Construct a stem-and-leaf plot of the protein amounts in the various sandwiches. b. What is the advantage(s) of using the stem-and-leaf plot instead of a histogram? c. Summarize your findings from these graphs.

The table summarizes responses of 4383 subjects in a recent General Social Survey to the question, "Within the past 12 months, how many people have you known personally that were victims of homicide?" $$\begin{array}{cc}\hline \begin{array}{l} \text { Number of People You Have Known } \\ \text { Who Were Victims of Homicide }\end{array} \\\\\hline \text { Number of Victims } & \text { Frequency } \\\\\hline 0& 3944 \\\1 & 279 \\\2 & 97 \\ 3 & 40 \\\4 \text { or more } & 23 \\ \text { Total } & \mathbf{4 3 8 3} \\ \hline\end{array}$$ a. To find the mean, it is necessary to give a score to the \(" 4\) or more" category. Find it, using the score \(4.5 .\) (In practice, you might try a few different scores, such as \(4,\) \(4.5,5,6,\) to make sure the resulting mean is not highly sensitive to that choice.) b. Find the median. Note that the "4 or more" category is not problematic for it. c. If 1744 observations shift from 0 to 4 or more, how do the mean and median change? d. Why is the median the same for parts \(\mathrm{b}\) and \(\mathrm{c}\), even though the data are so different?
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