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Chapter 3: Problem 19

The following frequency distribution shows the average number of pupils per teacher in the 50 states of the United States. Find the variance and standard deviation for the data. $$ \begin{array}{rr} \text { Class limits } & \text { Frequency } \\ \hline 9-11 & 2 \\ 12-14 & 20 \\ 15-17 & 18 \\ 18-20 & 7 \\ 21-23 & 2 \\ 24-26 & \frac{1}{50} \end{array} $$

Short Answer

Expert verified
Variance is 4.523; standard deviation is 2.126.

Step by step solution

01

Calculate the Class Midpoints

To find the class midpoint for each interval, average the lower and upper limits of each class. For example, for the interval 9-11, the midpoint is \((9 + 11) / 2 = 10\). Calculate the midpoints for all classes:- 9-11: Midpoint = 10- 12-14: Midpoint = 13- 15-17: Midpoint = 16- 18-20: Midpoint = 19- 21-23: Midpoint = 22- 24-26: Midpoint = 25
02

Calculate the Weighted Mean

First, multiply each class midpoint by its respective frequency to find the "weighted value" for each class. Then, divide the sum of these weighted values by the total frequency (which is 50). For instance, the weighted value for the class 9-11 is \(10 \times 2 = 20\). Calculate the sum:\[\text{Weighted Sum} = 10 \times 2 + 13 \times 20 + 16 \times 18 + 19 \times 7 + 22 \times 2 + 25 \times \frac{1}{50} = 10 \times 2 + 260 + 288 + 133 + 44 + 0.5 \]The total is 725.5, divided by 50 gives the weighted mean:\[\bar{x} = \frac{725.5}{50} = 14.51\]
03

Calculate the Variance

For each class, subtract the weighted mean from the class midpoint, square the result, multiply by the class frequency, and then sum all these values. Finally, divide by the total frequency to get the variance. For instance, for the class 9-11:\[((10 - 14.51)^2 \times 2 = 40.96)\]Calculate similarly for all classes and sum:\[\text{Variance} \approx \frac{((10-14.51)^2 \times 2) + ((13-14.51)^2 \times 20) + ((16-14.51)^2 \times 18) \+ ((19-14.51)^2 \times 7) + ((22-14.51)^2 \times 2) + ((25-14.51)^2 \times \frac{1}{50})}{50}\]\[= \frac{226.1602}{50} \approx 4.523\]
04

Calculate the Standard Deviation

The standard deviation is the square root of the variance. Calculate it for the variance found:\[\sigma = \sqrt{4.523} \approx 2.126\]
05

Conclusion

The variance is approximately 4.523, and the standard deviation is approximately 2.126.

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Key Concepts

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

Frequency Distribution
When we talk about a frequency distribution, we are essentially summarizing data by showing how frequently each value or range of values occurs in a given dataset. In our exercise, each class limit represents a different range in a group. For example, the class limit 9-11 indicates that the number of pupils per teacher falls between 9 and 11 for 2 states. Here's how the concept breaks down:
  • The frequency for each class tells us how many states have a pupil per teacher count within that range.
  • This method of distribution makes it easier to visualize and analyze large data sets by dividing the data into manageable intervals.
Each class's frequency helps give a quick snapshot of how data is spread across different intervals, leading up to further calculations like weighted mean, variance, and standard deviation.
Standard Deviation
The standard deviation is a powerful statistical tool that helps us understand how spread out the values in a dataset are around the mean. It is defined as the square root of the variance. In simple terms, standard deviation measures the extent to which each data point deviates from the mean. Here's a closer look:
  • A small standard deviation means that most of the numbers are close to the mean 鈥� the data points are tightly clustered.
  • A larger standard deviation suggests that the data points are spread out over a wider range of values.
In our exercise, we calculated the standard deviation to be approximately 2.126. This means the number of pupils per teacher in each state varies just a bit from the weighted mean of 14.51.
Weighted Mean
The weighted mean is like an average, but it gives different importance, or weight, to each data point based on its frequency. It's a perfect way to get a more accurate average when dealing with data that has different levels of significance, like the number of occurrences in our exercise.
  • In our exercise, each class midpoint is multiplied by its frequency to get a weighted value.
  • The sum of these weighted values is then divided by the total frequency, giving us the weighted mean.
Typically more representative than a simple mean, the weighted mean in our problem was calculated to be 14.51, indicating an average of around 14.5 students per teacher when accounting for the difference in frequency across states.
Class Midpoints
Class midpoints play a crucial role when calculating distributions like variance or weighted mean. They are the values in the middle of each class interval and are used because they represent each class as a single value.
  • For the class 9-11, the midpoint is calculated as \(\frac{9 + 11}{2} = 10\).
  • This process is repeated for each class, providing a neat way to handle grouped data.
Knowing the midpoints helps in simplifying further calculations, such as finding the weighted mean, which requires midpoints to calculate the weighted value, and variance, where the midpoint is compared to the mean.

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Most popular questions from this chapter

An employment evaluation exam has a variance of \(250 .\) Two particular exams with raw scores of 142 and 165 have \(z\) scores of -0.5 and \(0.955,\) respectively. Find the mean of the distribution.

Another instructor gives four 1 -hour exams and one final exam, which counts as two 1 -hour exams. Find a student's grade if she received \(62,83,97,\) and 90 on the 1 -hour exams and 82 on the final exam.

The hourly compensation costs (in U.S. dollars) for production workers in selected countries are represented below. Find the mean and modal class for the data. $$\begin{array}{lc}\text { Class } & \text { Frequency } \\\\\hline 2.48-7.48 & 7 \\\7.49-12.49 & 3 \\\12.50-17.50 & 1 \\\17.51-22.51 & 7 \\\22.52-27.52 & 5 \\\27.53-32.53 & 5\end{array}$$

Team batting averages for major league baseball in 2015 are represented below. Find the variance and standard deviation for each league. Compare the results. $$ \begin{array}{crcc} \text { NL } & & \text { AL } & \\ \hline 0.242-0.246 & 3 & 0.244-0.249 & 3 \\ 0.247-0.251 & 6 & 0.250-0.255 & 6 \\ 0.252-0.256 & 1 & 0.256-0.261 & 2 \\ 0.257-0.261 & 11 & 0.262-0.267 & 1 \\ 0.262-0.266 & 11 & 0.268-0.273 & 3 \\ 0.267-0.271 & 1 & 0.274-0.279 & 0 \end{array} $$

The mean and standard deviation of the number of hours the employees work in the music store per week are, respectively, 18.6 and 3.2 hours. If the owner increases the number of hours each employee works per week by 4 hours, what will be the new mean and standard deviation of the number of hours worked by the employees?
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