On an exam, the average score is 76 with a standard deviation of 6 points What is the probability that an individual chosen at random will have a score below 67 on this exam

P [ X < 67 ] =  0,66,81      or    66,81 %

Step-by-step explanation:

We assume Normal Distribution  N ( μ ; σ )    N ( 76 ; 6 )

z score for 67 is :

z(s) =  (  X - μ  ) /σ

z(s) =  (  67 - 76 ) / 6

z(s) =  - 9 / 6

z(s) = - 1,5

with 1,5 we fnd n z-table area undr the curve  α = 0,6681

Then  P [ X < 67 ] =  0,66,81      or    66,81 %

$15 each

step-by-step explanation:

$45/3 children= $15 each

other answer is correct. i had used the radius of the radius somehow. a website: