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Chapter 23: Problem 3

The resistances of 50 resistors are measured and the results recorded as follows: $$ \begin{array}{cc} \hline \text { Resistance }(\Omega) & \text { Frequency } \\ \hline 5.0 & 17 \\ 5.5 & 12 \\ 6.0 & 10 \\ 6.5 & 6 \\ 7.0 & 5 \\ \hline \end{array} $$ Calculate the standard deviation of the measurements.

Short Answer

Expert verified
Answer: The standard deviation of the resistance measurements is approximately 0.66Ω.

Step by step solution

01

Calculate the mean of the resistances

To calculate the mean resistance, multiply each resistance value by its frequency and sum those products, then divide the sum by the total number of resistances. Mean = \(\frac{(5.0\times17)+(5.5\times12)+(6.0\times10)+(6.5\times6)+(7.0\times5)}{50}\) Mean = \(\frac{85+66+60+39+35}{50}\) Mean = \(\frac{285}{50}\) Mean = \(5.7\) The mean resistance value is 5.7Ω.
02

Calculate the variance

To calculate the variance, find the squared difference between each resistance value and the mean (5.7), then multiply the squared difference by the respective frequency, sum all the results and divide by the total number of resistances. Variance = \(\frac{(5.0-5.7)^2\times17+(5.5-5.7)^2\times12+(6.0-5.7)^2\times10+(6.5-5.7)^2\times6+(7.0-5.7)^2\times5}{50}\) Variance = \(\frac{(-0.7)^2\times17+(-0.2)^2\times12+(0.3)^2\times10+(0.8)^2\times6+(1.3)^2\times5}{50}\) Variance = \(\frac{0.49\times17+0.04\times12+0.09\times 10+0.64\times6+1.69\times5}{50}\) Variance = \(\frac{8.33+0.48+0.90+3.84+8.45}{50}\) Variance = \(\frac{22}{50}\) Variance = \(0.44\)
03

Calculate the standard deviation

Finally, the standard deviation is the square root of the variance. Standard Deviation = \(\sqrt{0.44}\) Standard Deviation ≈ \(0.66\) The standard deviation of the resistance measurements is approximately 0.66Ω.

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Most popular questions from this chapter

Components are manufactured by three machines, \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). Machine A makes \(30 \%\) of the components, machine B makes \(25 \%\) of the components and machine \(\mathrm{C}\) makes the rest. Of those components made by machine A, \(6 \%\) are substandard; when made by machine B, \(3 \%\) are substandard; and when made by machine C, \(5 \%\) are substandard. A component is picked at random. Calculate the probability that it is (a) substandard (b) made by machine B given it is substandard (c) made by either machine \(\mathrm{A}\) or machine \(\mathrm{B}\) (d) substandard and made by machine B (e) substandard, given it is made by machine \(\mathrm{A}\) (f) made by machine A or is substandard.

The probability that a car will not develop a major fault within the first 3 years of its life is \(0.997\). Calculate the probability that of 20 cars selected at random (a) 19 will not develop any major faults in the first 3 years (b) 19 or more will not develop any major faults in the first 3 years.

Out of 6000 components, 39 fail within 12 months of manufacture. (a) Calculate the probability that a component picked at random fails within 12 months of manufacture. (b) A batch contains 2000 components. How many of these would you expect to fail within 12 months?

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