Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

At a certain restaurant, the distribution of wait times between ordering a meal and receiving the meal has mean 11.4 minutes and standard deviation 2.6 minutes. The restaurant manager wants to find the probability that the mean wait time will be greater than 12.0 minutes for a random sample of 84 customers. Assuming the wait times among customers are independent, what describes the sampling distribution of the sample mean wait time for random samples of size 84?

Sagot :

Using the Central Limit Theorem, the sampling distribution of the sample mean wait time for random samples of size 84 has mean of 11.4 minutes and standard deviation of 0.28 minutes.

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem, the population has:

  • Mean of 11.4 minutes, thus [tex]\mu = 11.4[/tex].
  • Standard deviation of 2.6 minutes, thus [tex]\sigma = 2.6[/tex]

Samples of 84 are taken, thus, by the Central Limit Theorem:

[tex]n = 84, s = \frac{2.6}{\sqrt{84}} = 0.28[/tex]

The sampling distribution of the sample mean wait time for random samples of size 84 has mean of 11.4 minutes and standard deviation of 0.28 minutes.

A similar problem is given at https://brainly.com/question/16364846