Respuesta :
Answer:
a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.
b) 156 would be expected to have a test score that would meet the colleges requirement
c) The lowest score that would meet the colleges requirement would be decreased to 26.388.
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:
[tex]\mu = 21.5, \sigma = 4.7[/tex]
a. Find the lowest test score that a student could get and still meet the colleges requirement.
This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.175 = \frac{X - 21.5}{4.7}[/tex]
[tex]X - 21.5 = 1.175*4.7[/tex]
[tex]X = 27.0225[/tex]
The lowest test score that a student could get and still meet the colleges requirement is 27.0225.
b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?
Top 12%, so 12% of them.
0.12*1300 = 156
156 would be expected to have a test score that would meet the colleges requirement
c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?
It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.04 = \frac{X - 21.5}{4.7}[/tex]
[tex]X - 21.5 = 1.04*4.7[/tex]
[tex]X = 26.388[/tex]
The lowest score that would meet the colleges requirement would be decreased to 26.388.
