Respuesta :
Answer:
[tex] z = \frac{1325-820}{217}= 2.327[/tex]
We need to take in count that 95% of the values in a normal distribution are between two deviations from the mean so then the usual values are between z=-2 and z =2.
And for this case we can conclude that if z >2 then we can conclude that this value would represent a potential outlier.
Step-by-step explanation:
We can define the variableof interest as X who represent the rent of students at Oxford, and for this case we knwo the following info for this variable
[tex]\mu=820[/tex] and [tex]\sigma=215[/tex]
We want to find the standardized score for the rent of Johns of 1325
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Replacing we got:
[tex] z = \frac{1325-820}{217}= 2.327[/tex]
For the other question about How high would the rent have to be to qualify as an outlier?.
We need to take in count that 95% of the values in a normal distribution are between two deviations from the mean so then the usual values are between z=-2 and z =2.
And for this case we can conclude that if z >2 then we can conclude that this value would represent a potential outlier.
His standardized z-score is 2.327.
The rent has to be more than 1254 to be to qualify as an outlier.
Given that: The mean rent of students at Oxnard university is $820 with a standard deviation of $217. Johns rent is $1,325.
So the z score of 1325 is = [tex]\frac{\left(1325-820\right)}{217}=2.327[/tex].
95% of the values lie between z score of -2 to 2.
So outside that interval it will be an outlier.
To find the high amount we take z>2.
[tex]z>2\\\frac{\left(x-820\right)}{217}>2\\x-820>434\\x>1254[/tex]
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