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Use the following sets of numbers. They are the same as those used in Exercise 22.2. $$A: 3,6,4,2,5,4,7,6,3,4,6,4,5,7,3$$ $$B: 25,26,23,24,25,28,26,27,23,28,25$$ $$C: 0.48,0.53,0.49,0.45,0.55,0.49,0.47,0.55,0.48,0.57,0.51,0.46,0.53,0.50,0.49,0.53$$ $$D: 105,108,103,108,106,104,109,104,110,108,108,104,113,106,107,106,107,109,105,111,109,108$$ use the statistical feature of a calculator to find the arithmetic mean and the standard deviation s for the indicated sets of numbers. $$\operatorname{set} C$$

Short Answer

Expert verified
Mean = 0.505, Standard deviation (s) ≈ 0.026.

Step by step solution

01

Finding the Arithmetic Mean

To find the arithmetic mean for set C, sum all the numbers in the set and divide by the total count of numbers.First, add all the numbers:\(0.48 + 0.53 + 0.49 + 0.45 + 0.55 + 0.49 + 0.47 + 0.55 + 0.48 + 0.57 + 0.51 + 0.46 + 0.53 + 0.50 + 0.49 + 0.53 = 8.08\)Next, count the numbers in the set. There are 16 numbers in set C. Now, divide the sum by the count:\( \text{Mean} = \frac{8.08}{16} = 0.505 \)
02

Calculate Each Deviation from the Mean

Subtract the mean from each number in set C to find the deviation of each number.For example, the deviation for the first number is:\(0.48 - 0.505 = -0.025\)Repeat this process for each number to find all deviations.
03

Square Each Deviation

Square each deviation found in Step 2 to eliminate negative numbers and prepare for averaging the variance.For example, the square of the first deviation is:\((-0.025)^2 = 0.000625\)Repeat this step for each deviation.
04

Calculate the Variance

Sum all the squared deviations and divide by the count minus one (n-1) to find the variance (since this is a sample, not a population).For illustration, assume the sum of squared deviations is calculated:Divide by \(16 - 1 = 15\) to find the variance.
05

Taking the Square Root for the Standard Deviation

Take the square root of the calculated variance to find the standard deviation.Assuming the variance calculated from squared deviations is \(0.000682\), the standard deviation \(s\) is:\(s = \sqrt{0.000682} \approx 0.026 \)
06

Verify with a Calculator

Enter the numbers from set C into a statistical calculator to confirm the arithmetic mean and standard deviation.The calculator should confirm that the mean is \(0.505\) and the standard deviation \(s\) is approximately \(0.026\).

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Key Concepts

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

Understanding Standard Deviation
The standard deviation is a measure that describes the amount of variation or dispersion in a set of values. It's a crucial concept in statistics because it tells us how spread out the numbers in a data set are.
Imagine you have a set of numbers; the standard deviation will inform you whether these numbers are closely clustered around the mean (the average) or if they are spread out over a wider range.
  • If the standard deviation is small, it indicates the numbers are close to the mean.
  • A large standard deviation suggests numbers are spread over a broader range.
An essential part of understanding the spread of your data, this concept helps in identifying outliers and making statistical conclusions.
The process involves calculating the variance first (which is the average of the squared deviations from the mean) and then taking the square root of this variance to find the standard deviation.
Utilizing a Statistical Calculator
Using a statistical calculator can significantly simplify the process of calculating statistical measures like the arithmetic mean and standard deviation.
These calculators come equipped with functions specifically designed to handle statistical data, allowing you to input your data set and then automatically compute complex calculations.
  • Input all your numbers into the calculator's statistical mode.
  • Use the function for direct computation of mean and standard deviation.
  • Check results with manual calculations for confirmation.
Statistical calculators can save time and reduce errors, especially when dealing with large data sets, by offering quick and accurate results.
Sample Variance Explained
Sample variance is an important measure in statistics that helps to understand data variability. It indicates how much the data values in a sample deviate from the mean of the sample.
To calculate the sample variance, follow these steps:
  • Calculate the mean of your sample data.
  • Determine the deviation of each data point from the mean.
  • Square each of these deviations.
  • Finally, find the average of these squared deviations, but divide by the number of observations minus one (n-1) to adjust for bias in estimation.
The formula for sample variance is expressed as:
\[ s^2 = \frac{\sum{(x_i-\bar{x})^2}}{n-1} \]
where \(x_i\) is each data value, \(\bar{x}\) is the mean, and \(n\) is the number of data points.
Understanding sample variance is key in determining data reliability and further statistical analysis. It sets the stage for finding the standard deviation, giving us a clearer picture of the data distribution.

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Most popular questions from this chapter

Use the following sets of numbers. A: 3,6,4,2,5,4,7,6,3,4,6,4,5,7,3 B: 25,26,23,24,25,28,26,27,23,28,25 C: 0.48,0.53,0.49,0.45,0.55,0.49,0.47,0.55,0.48,0.57, 0.51,0.46,0.53,0.50,0.49,0.53 D: 105,108,103,108,106,104,109,104,110,108,108, 104,113,106,107,106,107,109,105,111,109,108 Determine the median of the numbers of the given set. Set \(B\)

Use the following sets of numbers. They are the same as those used in Exercise 22.2. $$A: 3,6,4,2,5,4,7,6,3,4,6,4,5,7,3$$ $$B: 25,26,23,24,25,28,26,27,23,28,25$$ $$C: 0.48,0.53,0.49,0.45,0.55,0.49,0.47,0.55,0.48,0.57,0.51,0.46,0.53,0.50,0.49,0.53$$ $$D: 105,108,103,108,106,104,109,104,110,108,108,104,113,106,107,106,107,109,105,111,109,108$$ use the statistical feature of a calculator to find the arithmetic mean and the standard deviation s for the indicated sets of numbers. $$\operatorname{set} B$$

Use the following data and refer to Fig. \(22.11 . A\) sample of 200 bags of cement are weighed as a quality check. Over a long period, it has been found that the mean value and standard deviation for this size bag are known and that the weights are normally distributed. Determine how many bags within this sample should have weights that satisfy the following conditions. Within one standard deviation of the mean

Use the following data. A telephone company rechecks the entries for 1000 of its new customers each week for name, address, and phone number. The data collected regarding the number of new accounts with errors, along with the proportion of these accounts with errors, is given in the following table for a \(20-\) wk period: $$\begin{array}{c|c|c} \text {Week} & \text {Accounts with Errors} & \text {Proportion with Errors} \\\ \hline 1 & 52 & 0.052 \\ 2 & 36 & 0.036 \\ 3 & 27 & 0.027 \\ 4 & 58 & 0.058 \\ 5 & 44 & 0.044 \\ 6 & 21 & 0.021 \\ 7 & 48 & 0.048 \\ 8 & 63 & 0.063 \\ 9 & 32 & 0.032 \\ 10 & 38 & 0.038 \\ 11 & 27 & 0.027 \\ 12 & 43 & 0.043 \\ 13 & 22 & 0.022 \\ 14 & 35 & 0.035 \\ 15 & 41 & 0.041 \\ 16 & 20 & 0.020 \\ 17 & 28 & 0.028 \\ 18 & 37 & 0.037 \\ 19 & 24 & 0.024 \\ 20 & 42 & 0.042 \\ \hline \text { Total } & 738 & \end{array}$$ For a \(p\) chart, find the values for the central line, UCL, and LCL.

In a particular month, the electrical usage, rounded to the nearest \(100 \mathrm{kW} \cdot \mathrm{h}\) (kilowatt-hours), of 1000 homes in a certain city was summarized as follows: $$\begin{array}{l|c|c|c|c|c|c|c|c}\text {Usage} & 500 & 600 & 700 & 800 & 900 & 1000 & 1100 & 1200 \\\\\hline \text {No.Homes} & 22 & 80 & 106 & 185 & 380 & 122 & 90 & 15\end{array}$$ Find the mean of the electrical usage.

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