Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-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 be between $\$ 7000$ and $\$ 7500$ ?
A) $0.7333$
B) $0.2667$
C) $0.9147$
D) $0.8186$

Question

Quizplus · Accepted Answer

The central limit theorem states that the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, if the sample size is large enough. In this case, the sample size is 30, which is large enough to invoke the central limit theorem. Therefore, the distribution of the sample mean will be approximately normal with mean $\$ 7200$ and standard deviation $\$ 1200/\sqrt{30} = \$219.09$.To find the probability that the mean daily revenue for the next 30 days will be between $\$ 7000$ and $\$ 7500$, we can use the standard normal distribution (z-distribution). The z-score for $\$ 7000$ is $(7000 - 7200)/219.09 = -0.91$, and the z-score for $\$ 7500$ is $(7500 - 7200)/219.09 = 1.37$. Using a standard normal distribution table, we find that the probability of a z-score between -0.91 and 1.37 is 0.7333.Therefore, the probability that the mean daily revenue for the next 30 days will be between $\$ 7000$ and $\$ 7500$ is 0.7333.

Describe the Distribution of the Sample Mean-Samples from a Population $7200\$ 7200$7200

Describe the Distribution of the Sample Mean-Samples from a Population $\$ 7200$