Scores on a test are normally distributed with a mean of 65 and a standard deviation of 10. Find the score to the nearest whole number which separates the bottom 81% from the top 19%.
A. 88
B. 68
C. 56
D 74
X ~ N ( µ = 65 , σ = 10 )
P ( X < x ) = 81% = 0.81
To find the value of x
Looking for the probability 0.81 in standard normal table to
calculate Z score = 0.8779
Z = ( X - µ ) / σ
0.8779 = ( X - 65 ) / 10
X = 73.779 ≈ 74
P ( X < 74 ) = 0.81
Scores on a test are normally distributed with a mean of 65 and a standard deviation...
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