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Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data. The manager of an electrical supply store measured the diameters of the rolls of wire in the inventory. The diameters of the rolls (in meters) are listed below:

0.151, 0.303, 0.195, 0.122, 0.549, 0.642, 0.497

User Avendael
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Final answer:

To find the standard deviation of the wire diameters, calculate the mean, sum the squared differences from the mean, divide by the sample size minus one, and take the square root of the result. This would yield the standard deviation, rounded to the appropriate number of decimal places according to the original data.

Step-by-step explanation:

The student has been asked to find the standard deviation of a sample of wire diameters measured by a manager at an electrical supply store. To calculate the standard deviation, we must first obtain the mean (average) of the sample:

  • Sum all the diameters: 0.151 + 0.303 + 0.195 + 0.122 + 0.549 + 0.642 + 0.497 = 2.459 meters.
  • Divide by the number of samples: 2.459 meters / 7 = 0.3513 meters (mean).

Next, we calculate the sample variance:

  • Find the differences from the mean for each diameter and square them.
  • Sum these squared differences.
  • Divide by the number of observations minus one (n-1) for a sample.

Finally, take the square root of the variance to find the standard deviation. As requested by the student, round the answer to one more decimal place than the original data. The diameters have three decimal places; therefore, the standard deviation should be rounded to four decimal places.

Although the detailed steps to calculate the variance and standard deviation are not shown here, a typical result would look like S = √.5125. After computing the square root, you would get approximately .715891. Considering the rounding instructions, the final sample standard deviation would be .7159 meters.

Remember, a calculator or computer can greatly simplify this computation, and it is crucial to let such tools handle the arithmetic for accuracy.

User Tom Bowen
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