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* im in a hurry * A recent internet survey showed per day lodging expenses at disney world in orlando range from less than $100 for a campsite to more than $2000 for a top luxury room . The cost - per day average of the top 10 most popular hotels at Disney World is $348

a - If the standard deviation is $80 , find the z score for a per - day expense of $400

b - if the data are approximately normally distributed with mean $348 and standard deviation $80 , what percent of the hotels have per day expenses greater than $268?

Sagot :

Raphealnwobiidn

The z score for a per - day expense of $400 is 0.65 while about 84.13% of the hotels is greater than $268

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\\mu=mean,x=raw\ score,\sigma=standard \ deviation\\\\Given\ that:\\\\\mu=348,\sigma=80[/tex]

a) For an expense of $400:

[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{400-348}{80}\\\\z=0.65[/tex]

The z score for a per - day expense of $400 is 0.65

b) For an expense greater than $268:

[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{268-348}{80}\\\\z=-1[/tex]

P(x > 268) = P(z > -1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413 = 84.13%

About 84.13% of the hotels is greater than $268

Find out more at: https://brainly.com/question/15016913

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