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Chapter 3: Problem 114

The following table gives the frequency distribution of the times (in minutes) that 50 commuter students at a large university spent looking for parking spaces on the first day of classes in the Fall semester of 2012 . \begin{tabular}{lc} \hline \multicolumn{1}{c} { Time } & Number of Students \\ \hline 0 to less than 4 & 1 \\ 4 to less than 8 & 7 \\ 8 to less than 12 & 15 \\ 12 to less than 16 & 18 \\ 16 to less than 20 & 6 \\ 20 to less than 24 & 3 \\ \hline \end{tabular} Find the mean, variance, and standard deviation. Are the values of these summary measures population parameters or sample statistics?

Short Answer

Expert verified
The mean time spent by the students looking for parking spaces is 12.4 minutes. The variance of this time is 15.84 square minutes and the standard deviation is 3.98 minutes. These are all population parameters, not sample statistics.

Step by step solution

01

Calculation of Mean

First, find the midpoints of each interval by adding the two boundaries and dividing by 2. Then, multiply each midpoint by the respective frequency, and compute the total of these products. The mean is this total divided by the sum of the frequencies. For the given table, the midpoints are \(2.0\), \(6.0\), \(10.0\), \(14.0\), \(18.0\), and \(22.0\) respectively. The respective frequency times midpoint (fi*xi) are \(2.0*1 = 2.0\), \(6.0*7 = 42.0\), \(10.0*15 = 150.0\), \(14.0*18 = 252.0\), \(18.0*6 = 108.0\), and \(22.0*3 = 66.0\). Summing up, we get \(2.0 + 42.0 + 150.0 + 252.0 + 108.0 + 66.0 = 620.0\). As the total number of students is 50, the mean (碌) is \(620.0/50 = 12.4\) minutes.
02

Calculation of Variance

To calculate the variance, first compute the square of each interval鈥檚 midpoint. Multiply this square by the corresponding frequency. That will give you a new set of products. Calculate the total of these products and divide the result by total frequency, subtract the square of mean. For this dataset, the squared midpoints multiplied by frequencies result in \(4.0*1 = 4.0\), \(36.0*7 = 252.0\), \(100.0*15 = 1500.0\), \(196.0*18 = 3528.0\), \(324.0*6 = 1944.0\), \(484.0*3 = 1452.0\). The sum thereof results in \(4.0 + 252.0 + 1500.0 + 3528.0 + 1944.0 + 1452.0 = 8680.0\). So, the variance (蟽^2) is \(8680.0/50 - (12.4^2) = 15.84\) square minutes.
03

Calculation of Standard Deviation

The standard deviation is simply the square root of the variance. So standard deviation (蟽) will be \(sqrt{15.84} = 3.98\) minutes.
04

Identify Population or Sample

The data given represents the time spent by all 50 commuter students on the first day of a particular semester. So, this data is a population, not a sample. Therefore, the mean, variance, and standard deviation calculated are population parameters and not sample statistics.

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Key Concepts

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

Frequency Distribution
When we study descriptive statistics, we often start with frequency distributions. A frequency distribution is an overview of all distinct values in a dataset and how frequently each value occurs. This information is typically organized in a table, making it easy to see patterns and trends.
In the original exercise, the frequency distribution shows us how many students took varying amounts of time looking for parking spaces. For example:
  • 1 student found parking in less than 4 minutes.
  • 7 students took between 4 and 8 minutes.
  • 15 students took between 8 and 12 minutes.
Such tables help us see the distribution of time among the students, laying the groundwork for calculating other statistics like the mean, variance, and standard deviation.
Mean Calculation
To find the average or mean of data using a frequency distribution, we make use of midpoints. Midpoints simplify data because they represent the "center" of each interval. Here's how you compute the mean from a frequency distribution:
For each interval:
  • Find the midpoint: it's the average of the lower and upper boundaries.
  • Multiply the midpoint by its respective frequency.
Afterwards, sum up all these results and divide by the total number of data points.
In our exercise, after doing the math, we obtained a mean (\( \mu \) of 12.4 minutes, which indicates, on average, how long students take to find parking. This average provides a general idea of the "center" of the dataset.
Variance Calculation
The variance measures the spread of data points around the mean. To calculate variance in a frequency distribution:
  • Square each midpoint to "ignore" its sign and highlight deviations.
  • Multiply each squared midpoint by its corresponding frequency.
  • Sum these products and divide by the total frequency.
  • Finally, subtract the square of the mean from this result to find the variance.
The variance (\( \sigma^2 \) represents the degree of variation in our case. From the calculation, we found it to be 15.84 square minutes. This number tells us about the consistency (or inconsistency) in students' parking search times. A higher variance would indicate more variability in search times, while a lower number signifies more concentrated data around the mean.
Standard Deviation
The standard deviation is another way to describe the spread of a set of data. It provides insight into the average distance between each data point and the mean. Essentially, it's the square root of the variance.
After calculating the variance, take the square root to find the standard deviation. In this scenario, we arrived at a standard deviation of 3.98 minutes.
So, what does this number mean? It implies that, on average, a student's parking search time deviates from the mean by about 3.98 minutes. This value helps us understand the range or "spread" of times more intuitively compared to variance, as it's in the same unit as the original data (minutes in this case).

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Most popular questions from this chapter

The following table gives the total points scored by each of the top 16 National Basketball Association (NBA) scorers during the \(2010-11\) regular season (source: www.nba.com). \begin{tabular}{lclc} \hline Name & Points Scored & Name & Points Scored \\ \hline Kevin Durant & 2161 & Kevin Martin & 1876 \\ LeBron James & 2111 & Blake Griffin & 1845 \\ Kobe Bryant & 2078 & Russell Westbrook & 1793 \\ Derrick Rose & 2026 & Dwight Howard & 1784 \\ Amare Stoudemire & 1971 & LaMarcus Aldridge & 1769 \\ Carmelo Anthony & 1970 & Dirk Nowitzki & 1681 \\ Dwyane Wade & 1941 & Brook Lopez & 1673 \\ Monta Ellis & 1929 & Danny Granger & 1622 \\ \hline \end{tabular} a. Calculate the mean and median. Do these data have a mode? Why or why not? b. Compute the range, variance, and standard deviation for these data.

The following table gives the frequency distribution of the number of errors committed by a college baseball team in all of the 45 games that it played during the \(2011-12\) season. \begin{tabular}{cc} \hline Number of Errors & Number of Games \\ \hline 0 & 11 \\ 1 & 14 \\ 2 & 9 \\ 3 & 7 \\ 4 & 3 \\ 5 & 1 \\ \hline \end{tabular} Find the mean, variance, and standard deviation. (Hint: The classes in this example are single valued. These values of classes will be used as values of \(m\) in the formulas for the mean, variance, and standard deviation.)

In the Olympic Games, when events require a subjective judgment of an athlete's performance, the highest and lowest of the judges' scores may be dropped. Consider a gymnast whose performance is judged by seven judges and the highest and the lowest of the seven scores are dropped. a. Gymnast A's scores in this event are \(9.4,9.7,9.5,9.5,9.4,9.6\), and \(9.5\). Find this gymnast's mean score after dropping the highest and the lowest scores. b. The answer to part a is an example of (approximately) what percentage of trimmed mean? c. Write another set of scores for a gymnast \(B\) so that gymnast \(A\) has a higher mean score than gymnast B based on the trimmed mean, but gymnast B would win if all seven scores were counted. Do not use any scores lower than \(9.0\).

Briefly explain the difference between a population parameter and a sample statistic. Give one example of each.

A sample of 3000 observations has a bell-shaped distribution with a mean of 82 and a standard deviation of \(16 .\) Using the empirical rule, find what percentage of the observations fall in the intervals \(\bar{x} \pm 1 s, \bar{x} \pm 2 s\), and \(\bar{x} \pm 3 s .\)
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