The average value and the standard deviation has to be determined and the number of values that fall within one standard deviation of the average value has to be given. Concept Introduction: Standard Deviation: A series of measurements is equal to the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements. S D = ∑ | x − μ | 2 N where , x is values in the data set, μ i s average of the data set (mean value), N is no .of data points .
The average value and the standard deviation has to be determined and the number of values that fall within one standard deviation of the average value has to be given. Concept Introduction: Standard Deviation: A series of measurements is equal to the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements. S D = ∑ | x − μ | 2 N where , x is values in the data set, μ i s average of the data set (mean value), N is no .of data points .
The average value and the standard deviation has to be determined and the number of values that fall within one standard deviation of the average value has to be given.
Concept Introduction:
Standard Deviation: A series of measurements is equal to the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements.
SD=∑|x−μ|2Nwhere,xis values in the data set,μisaverage of the data set (mean value), N is no.of data points.
Expert Solution & Answer
Answer to Problem 67RIL
The average value is 5.24 %
The standard deviation obtained is 0.0506%
Seven of the ten values fall within the region of 5.19 ≤ x ≤ 5.29.
Explanation of Solution
The average value and the standard deviation is calculated as,
SD=∑|x−μ|2Nwhere,xis values in the data set,μisaverage of the data set (mean value), N is no.of data points.Given : x = 5.22%,5.28%,5.22%,5.30%,5.19%, 5.23%,5.33%,5.26%,5.15%,5.22%.Find mean value (μ):μ=5.22%+5.28%+5.22%+5.30%+5.19%+5.23%+5.33%+5.26%+5.15%+5.22%10 = 5.24 %DeterminationMeasured valuesDifferencebetweenMeasurementandAverage(x−μ)Square of Difference|x−μ|215.22%−0.024×10−425.28%0.041.6×10−335.22%−0.024×10−445.30%0.063.6×10−355.19%−0.052.5×10−365.23%−0.011×10−475.33%0.098.1×10−385.26%0.024×10−495.15%−0.098.1×10−3105.22%−0.024×10−4∑|x−μ|2=4×10−4+1.6×10−3+4×10−4+3.6×10−3+2.5×10−3+1×10−4+8.1×10−3+4×10−4+8.1×10−3+4×10−4 = 0.0256 %SD=0.025610=0.0506%Seven of the ten values fall within the region of 5.19 ≤ x ≤ 5.29.
As shown above, the mean value of the given data and also find the standard deviation value by substituting the obtained values in the equation.
Conclusion
The average value was 5.24 %
The standard deviation obtained was 0.0506%
Seven of the ten values fall within the region of 5.19 ≤ x ≤ 5.29.
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A group of students took turns using a laboratory balanceto weigh the water contained in a beaker. The results theyreported were 111.42 g, 111.67 g, 111.21 g, 135.64 g,111.02 g, 111.29 g, and 111.42 g.(a) Should any of the data be excluded before the average is calculated?(b) From the remaining measurements, calculate the average value of the mass of the water in the beaker.(c) Calculate the standard deviation s and, from it, the 95% confidence limit.
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Which of the following is not a way of expressing data reliability:
Statistical Tests
Sampling
Calibration of Equipment
Chemical literature