Anormal distribution has a mean of µ = 100 with σ = 20. if one score is randomly selected from this distribution, what is the probability that the score will be less than x = 84?

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 =

Z =

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

nope

step-by-step explanation:

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .