Answer:
The minimum score needed to be accepted is 525.3.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 500, \sigma = 100](https://tex.z-dn.net/?f=%5Cmu%20%3D%20500%2C%20%5Csigma%20%3D%20100)
If the state college only accepts students who score in the top 60% on the SAT, what is the minimum score needed to be accepted?
The 60th percentile, which is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.253. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.253 = \frac{X - 500}{100}](https://tex.z-dn.net/?f=0.253%20%3D%20%5Cfrac%7BX%20-%20500%7D%7B100%7D)
![X - 500 = 0.253*100](https://tex.z-dn.net/?f=X%20-%20500%20%3D%200.253%2A100)
![X = 525.3](https://tex.z-dn.net/?f=X%20%3D%20525.3)
The minimum score needed to be accepted is 525.3.