Question

6. In 2007 the scores on the College Aptitude Test (C. A.T.) were distributed normally with mean 500 and standard deviation 6
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Answer #1

Answer:

6.

Given,

Mean = 500

Standard deviation = 60

P(X < x) = 0.75

P((x-u)/s < (x - 500)/60) = 0.75

since from standard normal table

z = 0.6745

(x - 500)/60 = 0.6745

x - 500 = 0.6745*60

x = 540.4694

Option D

7.

Given,

Mean = 500

Standard deviation = 60

z = (x - u)/s

= (450 - 500)/60

= -50/60

= - 0.8333

Option B

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