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Question 2
suppose there is a city where the mean cost per
month of a one-bedroom apartment is $1,500, with a
standard deviation of $250.
approximately what percentage of one-bedroom
apartments in the city cost between $750 and
$1,000?

Sagot :

Using the normal distribution, it is found that 2.15% of bedroom apartments in the city cost between $750 and $1,000.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 1500, \sigma = 250[/tex]

The proportion of apartments that cost between $750 and $1,000 is given by the p-value of Z when X = 1000 subtracted by the p-value of Z when X = 750, hence:

X = 1000:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1000 - 1500}{250}[/tex]

Z = -2

Z = -2 has a p-value of 0.0228.

X = 750:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{750 - 1500}{250}[/tex]

Z = -3

Z = -3 has a p-value of 0.0013.

0.0228 - 0.0013 = 0.0215.

0.02215 = 2.15% of bedroom apartments in the city cost between $750 and $1,000.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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