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Chapter 3: Problem 50

The lengths (in seconds) of the eight most recent songs played on \(98.9\) FM WCLZ and WCLZ.com (Portland, ME) at \(1: 28\) p.m. on Wednesday July 20,2011 were as follows: \(\begin{array}{llllllll}251 & 252 & 213 & 182 & 244 & 259 & 262 & 216\end{array}\) Calculate the range, variance, and standard deviation.

Short Answer

Expert verified
The range of the song durations is 80 seconds, the variance is 822.9199 square seconds, and the standard deviation is 28.6909 seconds.

Step by step solution

01

Calculate the Range

The range is the difference between the highest and lowest values in a set of data. Here, the highest value is 262 seconds and the lowest is 182 seconds. Therefore, the range is \( 262 - 182 = 80 \) seconds.
02

Calculate the Variance

Variance is the sum of squares of differences between all numbers and means. First, calculate the mean of the data set: \( (251 + 252 + 213 + 182 + 244 + 259 + 262 + 216) / 8 = 234.875 \). The difference from the mean for each value are: \( 16.125, 17.125, -21.875, -52.875, 9.125, 24.125, 27.125, -18.875 \). Their squares respectively are: \( 259.015625, 293.015625, 479.015625, 2798.015625, 83.265625, 581.015625, 736.015625, 356.015625 \). The variance is found by calculating the average of these squared differences: \( (259.015625 + 293.015625 + 479.015625 + 2798.015625 + 83.265625 + 581.015625 + 736.015625 + 356.015625) / 8 = 822.9199 \)
03

Calculate the Standard Deviation

The standard deviation is simply the square root of the variance: \( \sqrt{822.9199} = 28.6909 \)

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Key Concepts

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

Understanding the Range
The range is one of the simplest ways to measure spread in a dataset.
It tells us how much the numbers in the data set differ from each other.
In simple terms, it is the difference between the highest value and the lowest value.

To find the range:
  • Identify the largest number in the data set, which is called the maximum.
  • Identify the smallest number in the data set, known as the minimum.
  • Subtract the minimum from the maximum.
In our example of song lengths, the maximum value is 262 seconds, and the minimum is 182 seconds.
Thus, the range is calculated as follows: \[ 262 - 182 = 80 \]The range provides a quick snapshot of how spread out the data is, but it can be affected by outliers.
Exploring Variance
Variance gives a more detailed understanding of data spread than the range.
It considers how far each number in the set is from the mean.
This tells us how data points differ from the average value (mean) in squared units.

To calculate variance, follow these steps:
  • First, find the mean (average) of the data set.
  • Subtract the mean from each number to get the deviation of each number.
  • Square each deviation to eliminate negatives and find the squared differences.
  • Finally, find the average of these squared differences to get the variance.
In our song length dataset:- The mean is 234.875 seconds.- The squared differences from the mean are calculated and averaged, resulting in the variance:\[ \frac{259.02 + 293.02 + 479.02 + 2798.02 + 83.27 + 581.02 + 736.02 + 356.02}{8} = 822.92 \]Variance is essential for understanding data variability, but it is not in the same units as the original data. This is where standard deviation comes into play.
Clarifying Standard Deviation
Standard deviation takes the calculation of variance a step further.
It allows us to understand the spread of data in the same units as the data itself.
By taking the square root of the variance, we convert the squared units back to the original units.

In practical terms:
  • It measures how much data points tend to spread from the average value.
  • The closer the standard deviation is to zero, the less variability in the data set.
  • A higher standard deviation indicates more spread out points from the mean.
In our given dataset, the variance was calculated as 822.92.
Therefore, the standard deviation is:\[ \sqrt{822.92} = 28.69 \]This means, on average, song lengths deviate by about 28.69 seconds from the mean length of 234.875 seconds.
Standard deviation offers a helpful insight irrespective of data type, making it easier to comprehend compared to variance.

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Most popular questions from this chapter

Prepare a box-and-whisker plot for the following data: \(\begin{array}{llllllll}36 & 43 & 28 & 52 & 41 & 59 & 47 & 61 \\ 24 & 55 & 63 & 73 & 32 & 25 & 35 & 49 \\ 31 & 22 & 61 & 42 & 58 & 65 & 98 & 34\end{array}\) Does this data set contain any outliers?

The following data represent the 2011 guaranteed salaries (in thousands of dollars) of the head coaches of the final eight teams in the 2011 NCAA Men's Basketball Championship. The data represent the 2011 salaries of basketball coaches of the following universities, entered in that order: Arizona, Butler, Connecticut, Florida, Kansas, Kentucky, North Carolina, and Virginia Commonwealth. (Source: www.usatoday.com). \(\begin{array}{llllllll}1950 & 434 & 2300 & 3575 & 3376 & 3800 & 1655 & 418\end{array}\) Compute the range, variance, and standard deviation for these data.

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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 \(\bar{x} \pm 3 s .\)

The following data give the numbers of new cars sold at a dealership during a 20 -day period. 8 1 \(\begin{array}{lrlrllllll}8 & 5 & 12 & 3 & 9 & 10 & 6 & 12 & 8 & 8 \\ 4 & 16 & 10 & 11 & 7 & 7 & 3 & 5 & 9 & 11\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. Where does the value of 4 lie in relation to these quartiles? b. Find the (approximate) value of the 25 th percentile. Give a brief interpretation of this percentile. c. Find the percentile rank of 10 . Give a brief interpretation of this percentile rank.
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