SOLUTION: in a test to 1000 students, the average score was 42 and the standard deviation 24.find;
(a) the number of students exceeding a score of 50.
(b) the number of students lying be
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-> SOLUTION: in a test to 1000 students, the average score was 42 and the standard deviation 24.find;
(a) the number of students exceeding a score of 50.
(b) the number of students lying be
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Question 1129621: in a test to 1000 students, the average score was 42 and the standard deviation 24.find;
(a) the number of students exceeding a score of 50.
(b) the number of students lying between 30 and 54.
(c) the value of score exceeded by the top 100 students Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
=(50-42)/24 or here z>(1/3) That is 0.3696 or 370 students (using z=+0.33)
Between 30 and 54 is between z of -0.5 and +0.5, which is a probability of 0.3829 or 383 students (that is 12 points on either side of the mean and 12/24 is 1/2, dividing by the sd)
Top 100 students is at the 90th percentile z=+1.28
using the formula above z*sd=30.72
That is x-mean, so x=mean+30.72 or a score 72.72 or 73 to the nearest integer.
