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.
More can be learned about the normal distribution at brainly.com/question/15181104
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Answer:
idk sorry
Step-by-step explanation:
Answer:
88.8%
Step-by-step explanation:
1.) 30-18= 12 which is the number of member that had an average of 84
2.)Then multiply 12 times 84 = 1008
3.) You multiply 18 by 92 which equals 1656
4.)Add 1008 and 1656 together and get 2664.
5.)Divide 2664 / 30 and you get the average of 88.8
Hope this helps :)
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Answer: The simplest form of ratio 6:15=2×3:5×3=2:5
Step-by-step explanation: So, 2:5 is equivalent to 6:15.
Answer:
30
Step-by-step explanation: