The voltage (in volts) of a given circuit is a random variable $ X $ that is normally distributed with the parameters $ μ = 120 $ and $ σ ^ 2 = 4 $
If three independent measurements are taken, what is the probability that the three measurements are between $ 116 $ and $ 118 $ volts?
My idea is to first get a probability of success $ p $, which I will calculate by standardizing $ X $, and then finding the probability that $ X $ is between $ 116 $ and $ 118 $.
Since I need to count the number of measurements, each one with probability of success $ p $ and each measurement attempt is done independently, I would do it with another variable $ Y $ ~ $ B (3, p) $
The answer to the question would be $ P (Y = 3) $, but in this case $ n = y = 3 $ then $ P (Y = 3) = p ^ 3 $
$Z=\dfrac{X-μ}{σ}=\dfrac{X-120}{2}\Rightarrow p = (116<X<118) = P(\dfrac{116-120}{2}<Z<\dfrac{118-120}{2}) = $
$P(-2<Z<-1) = \Phi(-1) - \Phi(-2) = 0,13786 - 0,01831 = 0,11955 \Rightarrow p^3 = 0,001708633$
Is the correct way I'm thinking the solution to the exercise?
Thank you very much.
Your reasoning is correct, not your computation. You are indeed looking for $p^3$ where $$ p=P(116<X<118)=P(-2<Z<-1)=P(Z<-1)-P(Z<-2)=0.158655-0.0227501=0.135905. $$ (you can get these number from standard normal tables). The result is $$ p^3=0.135905^3=0.0025102 $$ or about $0.25\%$.