Question

the quality - control manager of a local fast food restaurant wishes to analyze the length of time that a car spends at the drive - through window waiting for an order. according to records obtained from the restaurant, it is determined that the mean time spent at the window is 59.4 seconds with a standard deviation of 13.6 seconds.
if a random sample of 50 cars is to be chosen, what are the mean and standard deviation of the sampling distribution for the mean time spent at the window?
\\( \mu_{\bar{x}}= \\) 59.4
\\( \sigma_{\bar{x}}= \\) 1.92 (round to 2 decimal places)
according to the 68 - 95 - 99.7 (empirical) rule,

• approximately 68% of all possible sample means are between \\( \square \\) and \\( \square \\) seconds
• approximately 95% of all possible sample means are between \\( \square \\) and \\( \square \\) seconds
• approximately 99.7% of all possible sample means are between \\( \square \\) and \\( \square \\) seconds

question help: \\( \square \\) written example \\( \square \\) message instructor

Explanation:

Step1: Recall the Empirical Rule for sampling distributions.

The sampling distribution of the sample mean \(\bar{X}\) has mean \(\mu_{\bar{x}}=\mu\) and standard deviation \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Here, \(\mu = 59.4\), \(\sigma = 13.6\), and \(n = 50\). We already calculated \(\sigma_{\bar{x}}=\frac{13.6}{\sqrt{50}}\approx1.92\) (as given).

Step2: Apply the 68% rule.

For the 68% of data, the range is \(\mu_{\bar{x}}\pm\sigma_{\bar{x}}\). So, lower bound: \(59.4 - 1.92 = 57.48\), upper bound: \(59.4+1.92 = 61.32\).

Step3: Apply the 95% rule.

For 95% of data, the range is \(\mu_{\bar{x}}\pm2\sigma_{\bar{x}}\). Lower bound: \(59.4-2\times1.92 = 59.4 - 3.84 = 55.56\), upper bound: \(59.4 + 2\times1.92=59.4 + 3.84 = 63.24\).

Step4: Apply the 99.7% rule.

For 99.7% of data, the range is \(\mu_{\bar{x}}\pm3\sigma_{\bar{x}}\). Lower bound: \(59.4-3\times1.92 = 59.4 - 5.76 = 53.64\), upper bound: \(59.4+3\times1.92 = 59.4 + 5.76 = 65.16\).

Answer:

• Approximately 68%: between \(57.48\) and \(61.32\) seconds.
• Approximately 95%: between \(55.56\) and \(63.24\) seconds.
• Approximately 99.7%: between \(53.64\) and \(65.16\) seconds.