Answer: 99.73%
Step-by-step explanation:
Given : Mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma=2[/tex]
Let X be the random variable that represents the data values.
Formula for Z-score : [tex]z=\dfrac{X-\mu}{\sigma}[/tex]
For x=94, we have
[tex]z=\dfrac{94-100}{2}=-3[/tex]
For x=106, we have
[tex]z=\dfrac{106-100}{2}=3[/tex]
The probability that the samples are between 94 and 106:-
[tex]P(-3<x<3)=1-2P(z<-3)\\\\=1-2(0.0013499)=0.9973002\approx0.9973=99.73\%[/tex]
Hence, the percent of the samples are expected to be between 94 and 106 = 99.73%
