The length of time for long-distance phone calls has been found to be normally distributed, with a mean of 20.8 minutes and a standard deviation of 6.7 minutes. In a randomly selected sample of 68 calls, what is the sample average call time below which 10% of sample average call times will be shorter? (2 decimal places)
Question
The length of time for long-distance phone calls has been found to be normally distributed, with a mean of 20.8 minutes and a standard deviation of 6.7 minutes. In a randomly selected sample of 68 calls, what is the sample average call time below which 10% of sample average call times will be shorter? (2 decimal places)
Solution
To find the sample average call time below which 10% of sample average call times will be shorter, we first need to find the standard deviation of the mean. The standard deviation of the mean is the standard deviation divided by the square root of the number of observations (σ / √n), which in this case is 6.7 / √68 = 0.813 (rounded to three decimal places).
We then need to find the z-score that corresponds to the 10th percentile in a standard normal distribution. We can look up this value in a standard normal distribution table or use a calculator with a normal distribution function. The z-score that corresponds to the 10th percentile is approximately -1.28.
We then use this z-score to find the corresponding value in the distribution of call times. We use the formula:
X = μ + Zσ
where X is the value we are interested in, μ is the mean (20.8 minutes), Z is the z-score (-1.28), and σ is the standard deviation of the mean (0.813).
X = 20.8 + (-1.28)(0.813) = 19.76 minutes
So, the sample average call time below which 10% of sample average call times will be shorter is approximately 19.76 minutes.
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