Using the normal distribution, it is found that 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The mean and the standard deviation for the amounts are given as follows:
[tex]\mu = 3456, \sigma = 478[/tex]
The proportion is the p-value of Z when X = 4250, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4250 - 3456}{478}[/tex]
Z = 1.66
Z = 1.66 has a p-value of 0.9515.
Hence 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
More can be learned about the normal distribution at https://brainly.com/question/15181104
#SPJ1