Answer:
The minimum score necessary to be in the top 10% of the distribution is 628.
Step-by-step explanation:
Given that a normal distribution has a mean of μ = 500 and a standard deviation = σ= 100
The Significance level for this test = 10 % = 0.1
For one tailed test ∝= 0.1 the value of Z∝= ±1.28
Since it a normal distribution the test statistic used is
Z= X= u / σ/√n ( taking n= 1)
1.28 = X- 500/100/√1
128 + 500 = X
or X = 628
The minimum score necessary to be in the top 10% of the distribution is 628.
