Answer:
a) P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %
b) P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %
c) P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %
d) z(s) = 0,84
Step-by-step explanation:
Normal Distribution    N (  μ₀  ;   σ )   is  N ( 26,5 ; 4,8 )
a) P [ X < 24 mm ] = ( X - μ₀ ) / σ
P [ X < 24 mm ] = (24 - 26,5)/ 4,8 = - 0,5208 ≈ - 0,52
P [ X < 24 mm ] = - 0,52  
And from z-table we find area for z score
P [ X < 24 mm ] = 0,3015          or     P [ X < 24 mm ] =  30,15 %
b)P [ X > 32 mm ]  =  1  - P [ X < 32 mm ] 
P [ X < 32 mm ]  = ( 32 - 26,5 ) / 4,8
P [ X < 32 mm ]  = 5,5/4,8  =  1,1458 ≈ 1,15
P [ X < 32 mm ]  = 1,15
And from z-table  we get
P [ X < 32 mm ]  = 0,8749
Then:
P [ X > 32 mm ]  =  1  -  0,8749
P [ X > 32 mm ]  = 0,1251          or        P [ X > 32 mm ]  = 12,51 %
c) P [  25 < X < 30 ] = P [ X < 30 ] - P [ X < 25 ]
P [ X < 30 ]  = 30 - 26,5 / 4,8   =  0,73
From  z-table    P [ X < 30 ]  =  0,7673
P [ X < 25 ] = 25 - 26,5 / 4,8  = - 0,3125  ≈  - 0,31
From z-table  P [ X < 25 ] =  0,2709
Then
 P [  25 < X < 30 ] =  0,7673 -  0,2709
 P [  25 < X < 30 ] = 0,4964      or     P [  25 < X < 30 ] = 49,64 %
d) If  20 %
z- score for 20% is from z-table
z(s) = 0,84