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
be the point on the ground to with the 18 foot guy is anchored and 
the point on the ground to with the 21 foot guy is anchored.