Answer:
The minimum score of those who received C's is 67.39.
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 = 73, \sigma = 11[/tex]
If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?
This is X when Z has a pvalue of 1-0.695 = 0.305. So it is X when Z = -0.51.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.51 = \frac{X - 73}{11}[/tex]
[tex]X - 73 = -0.51*11[/tex]
[tex]X = 67.39[/tex]
The minimum score of those who received C's is 67.39.
