Question
In: Statistics and Probability
INDIAN STUDENTS DATA FILE (MINITAB) Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3...
INDIAN STUDENTS DATA FILE (MINITAB)
| Variable | N | N* | Mean | SE Mean | StDev | Minimum | Q1 | Median | Q3 | Maximum |
| GRE Score | 500 | 0 | 316.47 | 0.505 | 11.3 | 290 | 308 | 317 | 325 | 340 |
| TOEFL Score | 500 | 0 | 107.19 | 0.272 | 6.08 | 92 | 103 | 107 | 112 | 120 |
| CGPA | 500 | 0 | 8.5764 | 0.027 | 0.6048 | 6.8 | 8.1225 | 8.56 | 9.04 | 9.92 |
- What is the probability of students between CGPA 8 and 9?
- What is the probability of students above GRE 325 score?
- What is the GRE score middle 50% students?
- What is the TOEFL score of top 10% students?
- What is the CGPA of top 10% students?
Solutions
Expert Solution
a) probability of students between CGPA 8 and 9 = 68%
( because the mean is 8.5764 with ths standard deviation 0.6048. And according to empirical rule 68% of the data lie between 0ne standard deviation of the mean )
b)probability of students above GRE 325 score = 25%
( Because the third quartile is 325 that means that 75% of the data is below 325 and 25% is above 325)
c)GRE score middle 50% students = Median = 317
d)TOEFL score of top 10% students
Mean =107.19
Standard deviation, 
= 6.08
For top 10% we need to find the 90th percentile
P( Z < z ) = 0.9
P( Z < 1.282) = 0.9
z = 1.282
= 1.282
= 1.282
X = 107.18 + 6.08 * 1.282
X= 114.97
Hence , TOEFL score of top 10% students is 114.97
e)
CGPA of top 10% students
Mean =8.5764
Standard deviation,
= 0.6048
For top 10% we need to find the 90th percentile
P( Z < z ) = 0.9
P( Z < 1.282) = 0.9
z = 1.282
= 1.282
= 1.282
X = 8.5764 + 0.6048 * 1.282
X= 9.35
Hence CGPA of top 10% students is 9.35
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