The marks on a statistics test are normally distributed with

The marks on a statistics test are normally distributed with a mean of 62 and a variance of 225. If the instructor wishes to assign B\'s or higher to the top 30% of the students in the class, what mark is required to get a B or higher?

I understand how the problem is solved, the only part that I am confused about is where z = 0.53 comes from exactly in the normal probability tables.

Solution

Mean ( u ) =62
Standard Deviation ( sd )=15
Number ( n ) = 49
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P ( Z > x ) = 0.3
Value of z to the cumulative probability of 0.3 from normal table is 0.52
P( x-u/ (s.d) > x - 62/15) = 0.3
That is, ( x - 62/15) = 0.52
--> x = 0.52 * 15+62 = 69.86

Value comes the stanadrd normal table chart correspponding to the value 0.30

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf