Answer: 0.2119
Step-by-step explanation:
We assume that the random variable X is normally distributed.
Given : Population mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma=20[/tex]
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Then, z-score corresponds to 116
[tex]z=\dfrac{116-100}{20}=0.8[/tex]
By using the standard normal distribution table for z , we have
[tex]P(x>116)=P(z>0.8)=1-P(z\leq0.8)[/tex]
[tex]=1-0.7881446\approx0.2119[/tex]
Hence, the required probability = 0.2119