If the mean exam score of a class was 75%, with a standard deviation of 15%, what percent of students would be expected score at or higher than 92%? Assume that the distribution of the scores is normal and the variable is random.
Solution :
Given ,
mean =
= 0.75
standard deviation =
= 0.15
P(x > 0.92) = 1 - P(x< 0.92)
= 1 - P[(x -
)
/
< (0.92-0.75) / 0.15]
= 1 - P(z <1.13 )
Using z table
= 1 - 0.8708
= 0.1292
answer=12.92%
If the mean exam score of a class was 75%, with a standard deviation of 15%,...
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