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The mean time taken to learn the basics of a software program by all students is 200 minutes with a standard deviation of 20 minutes. a. Using Chebyshev's theorem, find the minimum percentage of students who learn the basics of this software program in i. 160 to 240 minutes ii. 140 to 260 minutes *b. Using Chebyshev's theorem, find the interval that contains the times taken by at least \(84 \%\) of all students to learn this software program.

Short Answer

Expert verified
Minimum percentage of students who learn the software program basics: i. in 160 to 240 minutes is 75%, ii. in 140 to 260 minutes is 89%; The time interval that contains at least 84% of students to learn this software program is about 172.8 to 227.2 minutes.

Step by step solution

01

Chebyshev's theorem application

Let's recall Chebyshev's theorem: For any number k greater than 1, at least \(1-1/k^2 \% \) of data from a data set will fall within k standard deviations from the mean. In the case of i, he interval 160 to 240 minutes is 2 standard deviations away from the mean (i.e., k=2). So, according to Chebyshev's theorem, at least \(1-1/2^2 = 1-1/4 = 0.75 \, or \, 75\% \) of the students will take between 160 and 240 minutes to learn the software program.
02

Continuation of Chebyshev's theorem application

In the case of ii, the interval 140 to 260 minutes is 3 standard deviations away from the mean (i.e., k=3). So, according to Chebyshev's theorem, at least \(1-1/3^2 = 1-1/9 \approx 0.89 \, or \, 89\% \) of the students will take between 140 and 260 minutes to learn the software program.
03

Find Time Interval for an 84% of Students

In order to find the time range that accommodates at least 84% of the students, we need to find an appropriate k value that will result in the desired percentage when applied to Chebyshev's theorem. Solving the equation \(1-1/k^2 = 0.84\) for k, we get approximately \(k \approx 1.36\). Then, we can find the time interval that is 1.36 standard deviations away from the mean on both sides using the formula \(interval = mean \pm k * standard \, deviation\), which gives us \(200 \pm 1.36 * 20 = 200 \pm 27.2\), or approximately between 172.8 minutes and 227.2 minutes.

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Key Concepts

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

Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. Essentially, standard deviation tells us how spread out the values in a data set are from the mean.
The computation involves finding the square root of the variance, which is the average of the squared differences from the mean.

In our problem, the standard deviation is given as 20 minutes. This means that most of the students' learning times are clustered around the mean (200 minutes) with an average deviation of 20 minutes. This is important, as it gives us an idea of how consistent the learning times are among students. A small standard deviation implies a close cluster around the mean, whereas a large standard deviation indicates a wider spread.
Mean
The mean, often referred to simply as the average, is a measure of central tendency. In statistics, it provides a single value that represents the center of a data set, allowing us to understand the overall "typical" value.

To calculate the mean, you sum up all the values in your data set and divide by the number of values. For this exercise, the mean time taken to learn the basics of the software is given as 200 minutes.

This value acts as a central anchor from which we can measure variations, such as in standard deviation calculations, to understand how much individual data points vary from the average learning time.
Interval Calculation
Interval calculation is a process of finding a range within a data set where a certain percentage of data points will fall within. It is particularly important when dealing with random variables and distributions in statistics.

With Chebyshev's theorem, we can calculate intervals within which at least a certain percentage of values are expected to fall. This theorem holds regardless of the data distribution shape.
  • The interval for 2 standard deviations from the mean (160 to 240 minutes) ensures that at least 75% of students fall within this range.
  • For 3 standard deviations (140 to 260 minutes), at least 89% of students are included.
  • For at least 84% of students, an interval 1.36 standard deviations from the mean (172.8 to 227.2 minutes) is calculated.
These intervals help us understand the spread of students' learning times and how they cluster around the mean.
Percentage of Students
When analyzing data, understanding the percentage of data points within specific intervals can provide valuable insights. Chebyshev's theorem helps us determine the minimum percentage of observations that lie within a certain distance from the mean based on the standard deviation.

For instance:
  • With 2 standard deviations (160 to 240 minutes), 75% of students fall within this learning time.
  • With 3 standard deviations (140 to 260 minutes), the coverage extends to 89% of students.
  • To ensure at least 84% of students are covered, an interval of 172.8 to 227.2 minutes is calculated, indicating the range where most students fall.
This brings a better understanding of how data points are dispersed in terms of students' learning times and how these intervals relate to the total percentage of students.

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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 give the number of patients who visited a walk-in clinic on each of 20 randomly selected days. \(\begin{array}{llllllllll}23 & 37 & 26 & 19 & 33 & 22 & 30 & 42 & 24 & 26 \\\ 28 & 32 & 37 & 29 & 38 & 24 & 35 & 20 & 34 & 38\end{array}\) a. Calculate the range, variance, and standard deviation for these data. b. Calculate the coefficient of variation.

Although the standard workweek is 40 hours a week, many people work a lot more than 40 hours a week. The following data give the numbers of hours worked last week by 50 people. \(\begin{array}{llllllllll}40.5 & 41.3 & 41.4 & 41.5 & 42.0 & 42.2 & 42.4 & 42.4 & 42.6 & 43.3 \\ 43.7 & 43.9 & 45.0 & 45.0 & 45.2 & 45.8 & 45.9 & 46.2 & 47.2 & 47.5 \\ 47.8 & 48.2 & 48.3 & 48.8 & 49.0 & 49.2 & 49.9 & 50.1 & 50.6 & 50.6 \\ 50.8 & 51.5 & 51.5 & 52.3 & 52.3 & 52.6 & 52.7 & 52.7 & 53.4 & 53.9 \\ 54.4 & 54.8 & 55.0 & 55.4 & 55.4 & 55.4 & 56.2 & 56.3 & 57.8 & 58.7\end{array}\) a. The sample mean and sample standard deviation for this data set are \(49.012\) and \(5.080\), respectively. Using Chebyshev's theorem, calculate the intervals that contain at least \(75 \%\), \(88.89 \%\), and \(93.75 \%\) of the data. b. Determine the actual percentages of the given data values that fall in each of the intervals that you calculated in part a. Also calculate the percentage of the data values that fall within one standard deviation of the mean. c. Do you think the lower endpoints provided by Chebyshev's theorem in part a are useful for this problem? Explain your answer. d. Suppose that the individual with the first number \((54.4)\) in the fifth row of the data is a workaholic who actually worked \(84.4\) hours last week and not \(54.4\) hours. With this change, the summary statistics are now \(\bar{x}=49.61\) and \(s=7.10 .\) Recalculate the intervals for part a and the actual percentages for part b. Did your percentages change a lot or a little? e. How many standard deviations above the mean would you have to go to capture all 50 data values? Using Chebyshev's theorem, what is the lower bound for the percentage of the data that should fall in the interval?

The range, as a measure of spread, has the disadvantage of being influenced by outliers. Illustrate this with an example.

In a survey of 640 parents of young children, 360 said that they will not want their children to play football because it is a very dangerous sport, 210 said that they will let their children play football, and 70 had no opinion. Considering the opinions of these parents, what is the mode?

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