Question

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.

Answer #1

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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