Answer: 40.68 %
Step-by-step explanation:
We need to find the interval between the values 0.2 and 2.2
0.2 ≤ z ≤ 2.2
Since in the standardized normal curve the mean 5.8 will be 0
Area for z values ≥ 0.2 ⇒ 0.0793 and area for z ≤ 2.2
area between 0 and 0.2 is 0.0793 ; and area between 0 to 2.2 is 0.4861
this last value includes the previous value at the left of z = 0.2
As we are looking for participants in the interval 0.2 ≤ z ≤ 2.2 we need to subtract areas 0.4861 - 0.0793 = 0.4068 express in % is 40.68 %
