Given: Mean of Normal distribution μ = 100, standard deviation σ =20
We have to find the probability that the score will be less than X=84
P(X < 84)
The z-score for x=84 is
Z = [tex] \frac{x-mean}{standard deviation} [/tex]
Z = [tex] \frac{84-100}{20} [/tex]
Z = -0.8
We have to find probability that Z < -0.8
Using Z score table to find the probability below z=-0.8 we get
P(Z < -0.8) = 0.2119
Hence the probability that the score will be less than x = 84 is 0.2119
