Respuesta :
Answer:
So the z-scores that separate the unusual IQ scores from those that are usual are Z = -2 and Z = 2.
The IQ scores that separate the unusual IQ scores from those that are usual are 84 and 148.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 116, \sigma = 16[/tex]
What are the z scores that separate the unusual IQ scores from those that are usual?
If Z<-2 or Z > 2, the IQ score is unusual.
So the z-scores that separate the unusual IQ scores from those that are usual are Z = -2 and Z = 2.
What are the IQ scores that separate the unusual IQ scores from those that are usual?
Those IQ scores are X when Z = -2 and X when Z = 2. So
Z = -2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2 = \frac{X - 116}{16}[/tex]
[tex]X - 116 = -2*16[/tex]
[tex]X = 84[/tex]
Z = 2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2 = \frac{X - 116}{16}[/tex]
[tex]X - 116 = 2*16[/tex]
[tex]X = 148[/tex]
The IQ scores that separate the unusual IQ scores from those that are usual are 84 and 148.
Answer:
[tex]\mu -2*\sigma = 116-2*16= 84[/tex]
[tex]\mu +2*\sigma = 116+2*16= 148[/tex]
Step-by-step explanation:
We know that the distirbution for the IQ have the following parameters:
[tex]\mu=116, \sigma=16[/tex]
The z score is given by this formula
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And replacing the limits we got:
[tex] X = \mu \pm z \sigma[/tex]
For this case we know that a value to be unusual if its z score is less than minus2 or greater than 2, so then we can find the limits with this formulas:
[tex]\mu -2*\sigma = 116-2*16= 84[/tex]
[tex]\mu +2*\sigma = 116+2*16= 148[/tex]
