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A delivery company is logging the number of miles each employee drives each day. The company found an average of about 85.6 miles. Assume that the standard deviation is 9.85 miles and that a random sample of 50 drivers was used to determine the average. Which statements are true about this situation?
1.)The 99% confidence interval is from about 81.43 to 90.58 miles.
2.)The 95% confidence interval is from about 82.87 to 88.33 miles.
3.)The margin of error for 95% confidence is about 2.52 miles.
4.)The sample isn’t large enough to use a normal distribution.
5.)The margin of error for 99% confidence is about 3.59 miles.
There are multiple answers

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The correct answer is: (the margin of error for 99% confidence is about 3.59 miles.
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The delivery company's statistics on the average driving miles show that the true statements are:

  • 95% confidence interval ... 82.87 to 88.33 miles.
  • Margin of error 99% confidence ... 3.59 miles.

What is the 99% confidence interval?

Z score for 99% confidence interval is 2.576 so this can be found as:

= 85.6 ± (2.576 x 9.85) / √50

= (82.01 to 89.19)

What is the 95% confidence interval?

Z score for 95% confidence interval is 1.960 so this can be found as:

= 85.6 ± (1.96 x 9.85) / √50

= (82.87 to 88.33)

What is the 95% confidence margin of error?

With a Z score of 1.960:

= (1.96 x 9.85) / √50

= 2.70 miles

What is the 99% confidence margin of error?

= (2.576 x 9.85) / √50

= 3.59 miles

The correct options are therefore 2 and 5.

Find out more on the margin of error at https://brainly.com/question/10218601.

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