A test has normal distribution with mean μ =100 and standard deviation σ =14
Let X be the score obtained by person in test. Here we have to find test score x such that person would be in top 1%
It means x is such that probability above x is 0.01 and below is 0.9
So we will find z score such that area below z is 0.9
In z score table there is no accurate probability value 0.9 so we take probability value close to 0.9 which is 0.8997 and corresponding z score is 1.28
Now we will find x value from z=1.28
x = (z * standard deviation) + mean
x = (1.28 * 14) + 100
x = 117.92 ~ 118
The score value for a person to be in top 1% is 118.
