Respuesta :
Answer:
a) IQ scores of 94.2 and below represent the bottom 35%.
b) An IQ score of 110.1 represents the 3rd quartile.
c) IQ scores of 124.7 and higher are in the top 5%.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that [tex]\mu = 100, \sigma = 15[/tex]
(a) Find the IQ scores that represent the bottom 35%.
The 35th percentile and below, in which the 35th percentile is X when Z has a p-value of 0.35, so X when Z = -0.385.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.385 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = -0.385*15[/tex]
[tex]X = 94.2[/tex]
IQ scores of 94.2 and below represent the bottom 35%.
(b) Find the IQ score that represents the 3rd Quartile.
This is the 100*3/4 = 75th percentile, which is X when Z has a p-value of 0.75, so X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = 0.675*15[/tex]
[tex]X = 110.1[/tex]
An IQ score of 110.1 represents the 3rd quartile.
(c) Find the IQ score for the top 5%.
IQ scores of at least the 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = 1.645*15[/tex]
[tex]X = 124.7[/tex]
IQ scores of 124.7 and higher are in the top 5%.
