Answer: (A) 1.3%
Step-by-step explanation:
Given: The monthly rent for one-bedroom apartments in a city is approximately normal with mean $936 and standard deviation $61.
i.e. [tex]\mu=936\ \ \ , \sigma=61[/tex]
Let X denotes the monthly rent for one-bedroom apartments.
Required probability = [tex]P(X\leq800)=P(\dfrac{X-\mu}{\sigma}\leq\dfrac{800-936}{61})[/tex]
[tex]=P(Z\leq\dfrac{-136}{61})\ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=P(Z\leq-2.23)\\\\=1-P(Z\leq 2.23)\ \ \ [P(Z<-z)=1-P(Z<z)]\\\\= 1- 0.9871\ \ \ \ [\text{By p-value table}]\\\\= 0.0129\approx1.29\% \approx 1.3\%[/tex]
Hence, the correct option is (A) 1.3%
1.3% of one-bedroom apartments in the city have a monthly rent of at most $800
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} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
Given that mean (μ) = $936 and standard deviation(σ) = $61.
Let x represent the monthly rent, hence:
The percentage of apartments with a monthly rent at most $800 is:
x < $800
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{800-936}{61}\\\\z=-2.23[/tex]
From the normal distribution table, P(x < 800) = P(z < -2.23) = 0.0129 = 1.4%
Therefore 1.3% of one-bedroom apartments in the city have a monthly rent of at most $800
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