Using the normal distribution, it is found that 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for the amounts are given as follows:

The proportion is the <u>p-value of Z when X = 4250</u>, hence:


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.
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The best and most correct answer among the choices provided by your question is the fourth choice or letter D. The probability is 4/5.
<span>P ( canadian)= 3/15 = 1/5
On the other hand, P (not canadian) = 1 - 1/5 = 4/5</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
Answer:
a>-2
Step-by-step explanation: