Answered

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In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,025 and a standard deviation of $230. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 60 one-bedroom apartments and finding the mean to be at least $1,955 per month

Sagot :

Cricetus

Answer:

The probability is "0.9908". The further explanation is provided below.

Step-by-step explanation:

The given values are:

Mean,

[tex]\mu=2025[/tex]

Standard deviation,

[tex]\sigma =230[/tex]

Sample,

[tex]n=60[/tex]

According to the question,

The probability of selecting a sample of 60 one-bedroom apartments will be:

=  [tex]P(\frac{\bar{x}- \mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{1955-2025}{\sqrt{\frac{230}{\sqrt{60} } } } )[/tex]

=  [tex]P(Z\geq -2.36)[/tex]

=  [tex]1-P(Z \leq -2.36)[/tex]

After placing the values, we get

=  [tex]0.0092[/tex]

or

=  [tex]09908[/tex]

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