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2) The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500? l1=7200 P(XZ 1500) 7500 30,98

2) The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500? l1=7200 P(XZ 1500) 7500 30,98

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**Transcription for Educational Website:** 2) The owner of a computer repair shop has determined that their daily revenue has a mean of $7200 and a standard deviation of $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500? - \(\mu = 7200\) - \(n = 30\) - Standard Deviation \(\sigma = 1200\) - \(P(\overline{X} > 7500)\) - \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{1200}{\sqrt{30}} \approx 309.8\) 3) A simple random sample of size \(n = 1180\) is obtained from a population whose size is \(N = 1,200,000\) and where a population proportion with a specified characteristic is \(p = 0.64\). Describe the sampling distribution.
Transcribed Image Text:**Transcription for Educational Website:** 2) The owner of a computer repair shop has determined that their daily revenue has a mean of $7200 and a standard deviation of $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500? - \(\mu = 7200\) - \(n = 30\) - Standard Deviation \(\sigma = 1200\) - \(P(\overline{X} > 7500)\) - \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{1200}{\sqrt{30}} \approx 309.8\) 3) A simple random sample of size \(n = 1180\) is obtained from a population whose size is \(N = 1,200,000\) and where a population proportion with a specified characteristic is \(p = 0.64\). Describe the sampling distribution.
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