The probability that the average selling price of the 35 homes will be within $2000 of the true mean is 0.5753
First, we need to find the z-score and this is expressed according to the formula;
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
x is the sample space
[tex]\mu[/tex] is the mean value
[tex]\sigma[/tex] is the standard deviation
Given the following parameters
Mean = $275,000
Standard deviation = $10,500
If the true mean is $273,000, then;
[tex]z=\dfrac{273,000-275000}{10,500} \\z = \frac{-2000}{10500}\\z= -0.1905[/tex]
If the true mean is $277,000
[tex]z=\dfrac{277,000-275000}{10,500} \\z = \frac{2000}{10500}\\z= 0.1905[/tex]
This means that the z-score is between -0.1095 and 0.1095
Pr( -0.1095 ≤ z ≤ 0.1095)
According to the z table, the probability that the average selling price of the 35 homes will be within $2000 of the true mean is 0.5753
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