The CRS scores for Express entry applicants follow approximately the N(503,121) distribution.
How high must the score be, to be placed in the top 12% of all applicants.
Solution :
Given that,
mean = = 503
standard deviation = = 121
X N (503 , 121)
Using standard normal table ,
P(Z > z) = 12%
1 - P(Z < z) = 0.12
P(Z < z) = 1 - 0.12
P(Z < 1.18) = 0.88
z = 1.18
Using z-score formula,
x = z * +
x = 1.18 * 121 + 503 = 645.78
high must the score be 645.78
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