Respuesta :
Answer: D.0.0853
Step-by-step explanation:
We assume that the daily revenue totals for the next 30 days will be follow a normal distribution.
Given : Population mean : [tex]\mu=\$7200[/tex]
Standard deviation : [tex]\sigma=\ $1200[/tex]
Sample size : n=30
Let X be the random variable that represents the daily revenue .
Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x=$7500
[tex]z=\dfrac{7500-7200}{\dfrac{1200}{\sqrt{30}}}\approx1.37[/tex]
By using the standard normal distribution table for z , we have
[tex]P(x>7500)=P(z>1.37)=1-P(z\leq1.37)[/tex]
[tex]=1- 0.9146565= 0.0853435\approx0.0853[/tex]
Hence, the probability that the mean daily revenue for the next 30 days will exceed $7500= 0.0853=
The probability that the mean daily revenue for the next 30 days will exceed $7500 is 0.0853
What is z score?
The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / (standard deviation / √ sample)
Given mean $7200 and standard deviation $1200. Hence:
For > 7500:
z = (7500 - 7200) / (1200/√30) = 1.37
P(z > 1.37) = 1 - P(z < 1.37) = 1 - 0.9147 = 0.0853
The probability that the mean daily revenue for the next 30 days will exceed $7500 is 0.0853
Find out more on z score aT: https://brainly.com/question/25638875