the score on an achievement of test given to 1,00,000 students are normally distributed with mean 500 and standard deviation of 100 what should be the score of student be the place him among the top 10% of all students.
Thank you
You can play around with Z table stuff at
davidmlane.com/hyperstat/z_table.html
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To find the score that places a student among the top 10% of all students, you need to calculate the z-score associated with the top 10% tail of the standard normal distribution.
First, let's calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = score of the student
μ = mean of the distribution (in this case, 500)
σ = standard deviation of the distribution (in this case, 100)
The top 10% corresponds to a z-score that leaves 90% to the left. Looking up this value in a standard normal distribution table or using a calculator, we find it to be approximately 1.28.
Now, we can solve for x:
1.28 = (x - 500) / 100
Rearranging the equation:
1.28 * 100 = x - 500
128 = x - 500
x = 128 + 500
x ≈ 628
Therefore, a student would need a score of around 628 to place in the top 10% of all students.