Respuesta :
Answer:
X = 71
Step-by-step explanation:
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 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.
According to national data, the IQ's of the 12 year-old population are normally distributed with a mean of 96 and a standard deviation of 15.
This means that [tex]\mu = 96, \sigma = 15[/tex]
Calculate the IQ below which a child would be eligible to apply for this education assistance program (X).
Bottom 5%, so the 5th percentile and below, we want to find the 5th percentile, which is X when Z has a pvalue of 0.05, so X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 96}{15}[/tex]
[tex]X - 96 = -1.645*15[/tex]
[tex]X = 71.3[/tex]
As a whole number, X = 71
