In an intelligence test administered to 1000 students the average score was 42 and standard deviation 24. Find
The number of students who scored 50 and above (3 Marks)
The number of students lying between 30 and 54 (3 Marks)
The value of score exceeded by the top 100 students
Answers
Given : In an intelligence test administered to 1000 students the average score was 42 and standard deviation 24
To find : The number of students who scored 50 and above (3 Marks)
The number of students lying between 30 and 54 (3 Marks)
The value of score exceeded by the top 100 students
Solution:
Mean = 42
SD = 24
Z score = ( value - Mean)/SD
Value = 50
Z score = ( 50 - 42)/24 = 8/24 = 1/3 = 0.333
for Z score 0.33 , 63 % students scored below 50
Hence above 50 = 37% = 370 students
370 students scored 50 and above
The number of students lying between 30 and 54
Z score for 30 = (30 - 42)/24 = -0.5 => 30.85 % below this
Z score for 54 = (54 - 42)/24 = 0.5 => 69.15 % below this
Between = 69.15 - 30.85 = 38.3 % = 383 Students
383 Students lying between 30 and 54
The value of score exceeded by the top 100 students
=> 900 Students achieved score less than them
=> 90 % below that => z score = 1.282
1.282 = ( Value - 42)/24
=> Value = 72.8
=> Score of top 100 students is above 73
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