Respuesta :
Answer:
a) 159.44
b) 147.33
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 problem, we have that:
Verbal portion of GRE:
[tex]\mu = 151, \sigma = 7[/tex]
Quantitative portion of GRE:
[tex]\mu = 153, \sigma = 7.67[/tex]
a) Find the score of a student who scored in the 80th percentile on the Quantitative Reasoning section of the exam. (please round to two decimal places)
Value of X when Z has a pvalue of 0.8. So X when Z = 0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 153}{7.67}[/tex]
[tex]X - 153 = 7.67*0.84[/tex]
[tex]X = 159.44[/tex]
b)Find the score of a student who scored worse than 70% of the test takers in the Verbal Reasoning section of the exam please round to two decimal places)
Value of Z when Z has a pvalue of 1-0.7 = 0.3. So X when Z = -0.525. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.525 = \frac{X - 151}{7}[/tex]
[tex]X - 151 = -0.525*7[/tex]
[tex]X = 147.33[/tex]
