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?

The 99% confidence interval is from about 81.43 to 90.58 miles.
The margin of error for 95% confidence is about 2.52 miles.
The 95% confidence interval is from about 82.87 to 88.33 miles.
The margin of error for 99% confidence is about 3.59 miles.
The sample isn’t large enough to use a normal distribution.

1 answer

To analyze the statements regarding the average miles driven by delivery drivers, we can use the provided information about the average miles, standard deviation, and sample size.

  1. Sample Size and Normal Approximation: Since the sample size is 50 (which is larger than 30), we can use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed. Therefore, the statement "The sample isn’t large enough to use a normal distribution" is false.

  2. Calculating the Margin of Error for 95% Confidence:

    • For 95% confidence, we use the Z-score for 95% confidence, which is approximately 1.96.
    • The standard error (SE) is calculated as: \[ SE = \frac{\text{standard deviation}}{\sqrt{n}} = \frac{9.85}{\sqrt{50}} \approx 1.39 \]
    • The margin of error (ME) is calculated as: \[ ME = Z \times SE \approx 1.96 \times 1.39 \approx 2.72 \] So the statement "The margin of error for 95% confidence is about 2.52 miles" is false.
  3. Calculating the 95% Confidence Interval:

    • The confidence interval can be found using the mean and the margin of error: \[ \text{Lower Limit} = 85.6 - 2.72 \approx 82.88 \] \[ \text{Upper Limit} = 85.6 + 2.72 \approx 88.32 \] Therefore, the statement "The 95% confidence interval is from about 82.87 to 88.33 miles" is true.
  4. Calculating the Margin of Error for 99% Confidence:

    • For 99% confidence, the Z-score is approximately 2.576.
    • The margin of error for 99% confidence is: \[ ME = 2.576 \times 1.39 \approx 3.58 \] Hence, the statement "The margin of error for 99% confidence is about 3.59 miles" is approximately true (considering rounding).
  5. Calculating the 99% Confidence Interval:

    • The confidence interval would be: \[ \text{Lower Limit} = 85.6 - 3.58 \approx 82.02 \] \[ \text{Upper Limit} = 85.6 + 3.58 \approx 89.18 \] The provided range of "from about 81.43 to 90.58 miles" does not match our calculated confidence intervals; therefore, this statement is false.

Summary of the Statements:

  • False: The 99% confidence interval is from about 81.43 to 90.58 miles.
  • False: The margin of error for 95% confidence is about 2.52 miles.
  • True: The 95% confidence interval is from about 82.87 to 88.33 miles.
  • True: The margin of error for 99% confidence is about 3.59 miles.
  • False: The sample isn’t large enough to use a normal distribution.